Vasile Cîrtoaje
C
j
MATHEM
MATICAL
INEQUA
Q ALITIES
Volume 2
SYMM
METRIC
RATIONAL AND NONRATIONAL
RATIONAL AND ALITIES
INEQUA
UNIVERSITY OF PLLOIESTI, ROMANIA
2015
Contents
1 Symmetric Rational Inequalities
1
1.1 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
2 Symmetric Nonrational Inequalities
257
2.1 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257
2.2 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273
3 Symmetric Power-Exponential Inequalities
419
3.1 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 419
3.2 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 425
4 Bibliography
473
i
ii
Vasile Cîrtoaje
Chapter 1
Symmetric Rational Inequalities
1.1
Applications
1.1. If a, b are nonnegative real numbers, then
1
1
1
+
≥
.
(1 + a)2 (1 + b)2
1 + ab
1.2. If a, b, c are nonnegative real numbers, then
a2 − bc
b2 − ca
c2 − a b
+
+
≥ 0.
3a + b + c 3b + c + a 3c + a + b
1.3. If a, b, c are positive real numbers, then
4a2 − b2 − c 2 4b2 − c 2 − a2 4c 2 − a2 − b2
+
+
≤ 3.
a(b + c)
b(c + a)
c(a + b)
1.4. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
(a)
(b)
(c)
a2
2a2
a2
1
1
1
3
+ 2
+ 2
≥
;
+ bc b + ca c + a b
a b + bc + ca
1
1
1
2
+ 2
+ 2
≥
.
+ bc 2b + ca 2c + a b
a b + bc + ca
1
1
1
2
+ 2
+ 2
>
.
+ 2bc b + 2ca c + 2a b
a b + bc + ca
1
2
Vasile Cîrtoaje
1.5. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
a(b + c) b(c + a) c(a + b)
+ 2
+ 2
≥ 2.
a2 + bc
b + ca
c + ab
1.6. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
a2
b2
c2
a
b
c
+
+
≥
+
+
.
2
2
2
2
2
2
b +c
c +a
a +b
b+c c+a a+b
1.7. Let a, b, c be positive real numbers. Prove that
1
1
1
a
b
c
+
+
≥ 2
+ 2
+ 2
.
b+c c+a a+b
a + bc b + ca c + a b
1.8. Let a, b, c be positive real numbers. Prove that
1
1
1
2a
2b
2c
+
+
≥ 2
+ 2
+ 2
.
b+c c+a a+b
3a + bc 3b + ca 3c + a b
1.9. Let a, b, c be positive real numbers. Prove that
(a)
(b)
b
c
13 2(a b + bc + ca)
a
+
+
≥
−
;
b+c c+a a+b
6
3(a2 + b2 + c 2 )

‹
p
a
b
c
3
a b + bc + ca
+
+
− ≥ ( 3 − 1) 1 − 2
.
b+c c+a a+b 2
a + b2 + c 2
1.10. Let a, b, c be positive real numbers. Prove that

‹2
1
1
1
a+b+c
.
+
+
≤
a2 + 2bc b2 + 2ca c 2 + 2a b
a b + bc + ca
1.11. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
a2 (b + c) b2 (c + a) c 2 (a + b)
+ 2
+ 2
≥ a + b + c.
b2 + c 2
c + a2
a + b2
Symmetric Rational Inequalities
1.12. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
3(a2 + b2 + c 2 )
a2 + b2 b2 + c 2 c 2 + a2
+
+
≤
.
a+b
b+c
c+a
a + b + c)
1.13. Let a, b, c be positive real numbers. Prove that
a2
1
1
9
1
+ 2
+ 2
≥
.
2
2
2
+ ab + b
b + bc + c
c + ca + a
(a + b + c)2
1.14. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
a2
b2
c2
1
+
+
≤ .
(2a + b)(2a + c) (2b + c)(2b + a) (2c + a)(2c + b) 3
1.15. Let a, b, c be positive real numbers. Prove that
P
(a)
P
(b)
a
1
≤
;
(2a + b)(2a + c)
a+b+c
a3
1
≤
.
2
2
2
2
(2a + b )(2a + c )
a+b+c
1.16. If a, b, c are positive real numbers, then
X
1
2
1
≥
+
.
2
(a + 2b)(a + 2c) (a + b + c)
3(a b + bc + ca)
1.17. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
1
1
1
4
+
+
≥
;
2
2
2
(a − b)
(b − c)
(c − a)
a b + bc + ca
(a)
(b)
(c)
a2
1
1
1
3
+ 2
+ 2
≥
;
2
2
2
− ab + b
b − bc + c
c − ca + a
a b + bc + ca
a2
1
1
1
5
+ 2
+ 2
≥
.
2
2
2
+b
b +c
c +a
2(a b + bc + ca)
3
4
Vasile Cîrtoaje
1.18. Let a, b, c be positive real numbers, no two of which are zero. Prove that
(a2 + b2 )(a2 + c 2 ) (b2 + c 2 )(b2 + a2 ) (c 2 + a2 )(c 2 + b2 )
+
+
≥ a2 + b2 + c 2 .
(a + b)(a + c)
(b + c)(b + a)
(c + a)(c + b)
1.19. Let a, b, c be positive real numbers such that a + b + c = 3. Prove that
a2
1
1
1
+ 2
+ 2
≤ 1.
+b+c b +c+a c +a+b
1.20. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that
a2 − bc b2 − ca c 2 − a b
+ 2
+ 2
≥ 0.
a2 + 3
b +3
c +3
1.21. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that
1 − bc 1 − ca 1 − a b
+
+
≥ 0.
5 + 2a 5 + 2b 5 + 2c
1.22. Let a, b, c be positive real numbers such that a + b + c = 3. Prove that
a2
1
1
1
3
+ 2
+ 2
≤ .
2
2
2
+ b +2 b +c +2 c +a +2 4
1.23. Let a, b, c be positive real numbers such that a + b + c = 3. Prove that
1
1
1
1
+
+
≤ .
4a2 + b2 + c 2 4b2 + c 2 + a2 4c 2 + a2 + b2
2
1.24. Let a, b, c be nonnegative real numbers such that a + b + c = 2. Prove that
bc
ca
ab
+ 2
+ 2
≤ 1.
+1 b +1 c +1
a2
1.25. Let a, b, c be nonnegative real numbers such that a + b + c = 1. Prove that
bc
ca
ab
1
+
+
≤ .
a+1 b+1 c+1 4
Symmetric Rational Inequalities
5
1.26. Let a, b, c be positive real numbers such that a + b + c = 1. Prove that
1
1
3
1
+
+
≤
.
2
2
2
a(2a + 1) b(2b + 1) c(2c + 1) 11a bc
1.27. Let a, b, c be positive real numbers such that a + b + c = 3. Prove that
1
1
1
+
+
≤ 1.
a3 + b + c b3 + c + a c 3 + a + b
1.28. Let a, b, c be positive real numbers such that a + b + c = 3. Prove that
a2
b2
c2
+
+
≥ 1.
1 + b3 + c 3 1 + c 3 + a3 1 + a3 + b3
1.29. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that
1
1
1
3
+
+
≤ .
6 − a b 6 − bc 6 − ca
5
1.30. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that
1
1
1
1
+ 2
+ 2
≤ .
+ 7 2b + 7 2c + 7 3
2a2
1.31. Let a, b, c be nonnegative real numbers such that a ≥ b ≥ 1 ≥ c and a + b + c = 3.
Prove that
1
1
1
3
+ 2
+ 2
≤ .
2
a +3 b +3 c +3 4
1.32. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that
1
1
1
3
+
+
≥ .
2a2 + 3 2b2 + 3 2c 2 + 3 5
1.33. Let a, b, c be nonnegative real numbers such that a ≥ 1 ≥ b ≥ c and a + b + c = 3.
Prove that
1
1
1
+ 2
+ 2
≥ 1.
2
a +2 b +2 c +2
6
Vasile Cîrtoaje
1.34. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. Prove that
1
1
a+b+c
3
1
+
+
≥
+
.
a+b b+c c+a
6
a+b+c
1.35. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. Prove that
a2
1
1
1
3
+ 2
+ 2
≥ .
+1 b +1 c +1 2
1.36. Let a, b, c be positive real numbers such that a b + bc + ca = 3. Prove that
a2
b2
c2
+
+
≥ 1.
a2 + b + c b2 + c + a c 2 + a + b
1.37. Let a, b, c be positive real numbers such that a b + bc + ca = 3. Prove that
bc + 4 ca + 4 a b + 4
bc + 2 ca + 2 a b + 2
+ 2
+ 2
≤3≤ 2
+
+
.
2
a +4 b +4 c +4
a + 2 b2 + 2 c 2 + 2
p
1.38. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. If k ≥ 2 + 3,
then
1
1
1
3
+
+
≤
.
a+k b+k c+k
1+k
1.39. Let a, b, c be nonnegative real numbers such that a2 + b2 + c 2 = 3. Prove that
a(b + c) b(c + a) c(a + b)
+
+
≤ 3.
1 + bc
1 + ca
1 + ab
1.40. Let a, b, c be positive real numbers such that a2 + b2 + c 2 = 3. Prove that
a2 + b2 b2 + c 2 c 2 + a2
+
+
≤ 3.
a+b
b+c
c+a
Symmetric Rational Inequalities
1.41. Let a, b, c be positive real numbers such that a2 + b2 + c 2 = 3. Prove that
ab
bc
ca
7
+
+
+ 2 ≤ (a + b + c).
a+b b+c c+a
6
1.42. Let a, b, c be positive real numbers such that a2 + b2 + c 2 = 3. Prove that
(a)
1
1
3
1
+
+
≤ ;
3 − a b 3 − bc 3 − ca
2
(b)
1
1
1
3
+p
+p
≤p
.
p
6 − ab
6 − bc
6 − ca
6−1
1.43. Let a, b, c be positive real numbers such that a2 + b2 + c 2 = 3. Prove that
1
1
3
1
+
+
≥ .
1 + a5 1 + b5 1 + c 5
2
1.44. Let a, b, c be positive real numbers such that a bc = 1. Prove that
1
1
1
+
+
≥ 1.
a2 + a + 1 b2 + b + 1 c 2 + c + 1
1.45. Let a, b, c be positive real numbers such that a bc = 1. Prove that
a2
1
1
1
+ 2
+ 2
≤ 3.
−a+1 b − b+1 c −c+1
1.46. Let a, b, c be positive real numbers such that a bc = 1. Prove that
3+ b
3+c
3+a
+
+
≥ 3.
2
2
(1 + a)
(1 + b)
(1 + c)2
1.47. Let a, b, c be positive real numbers such that a bc = 1. Prove that
7 − 6a 7 − 6b 7 − 6c
+
+
≥ 1.
2 + a2 2 + b2 2 + c 2
7
8
Vasile Cîrtoaje
1.48. Let a, b, c be positive real numbers such that a bc = 1. Prove that
b6
c6
a6
+
+
≥ 1.
1 + 2a5 1 + 2b5 1 + 2c 5
1.49. Let a, b, c be positive real numbers such that a bc = 1. Prove that
a
b
c
1
+
+
≤ .
a2 + 5 b2 + 5 c 2 + 5 2
1.50. Let a, b, c be positive real numbers such that a bc = 1. Prove that
1
1
1
2
+
+
+
≥ 1.
2
2
2
(1 + a)
(1 + b)
(1 + c)
(1 + a)(1 + b)(1 + c)
1.51. Let a, b, c be nonnegative real numbers such that
1
1
1
3
+
+
= .
a+b b+c c+a
2
Prove that
3
2
1
≥
+ 2
.
a+b+c
a b + bc + ca a + b2 + c 2
1.52. Let a, b, c be nonnegative real numbers such that
7(a2 + b2 + c 2 ) = 11(a b + bc + ca).
Prove that
51
a
b
c
≤
+
+
≤ 2.
28
b+c c+a a+b
1.53. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
a2
1
1
1
10
+ 2
+ 2
≥
.
2
2
2
+b
b +c
c +a
(a + b + c)2
1.54. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
a2
1
1
1
3
+ 2
+ 2
≥
.
2
2
2
− ab + b
b − bc + c
c − ca + a
max{a b, bc, ca}
Symmetric Rational Inequalities
1.55. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
a(2a + b + c) b(2b + c + a) c(2c + a + b)
+
+
≥ 6.
b2 + c 2
c 2 + a2
a2 + b2
1.56. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
a2 (b + c)2 b2 (c + a)2 c 2 (a + b)2
+ 2
+
≥ 2(a b + bc + ca).
b2 + c 2
c + a2
a2 + b2
1.57. If a, b, c are real numbers such that a bc > 0, then

‹

‹
X
a
b
c
1 1 1
a
+5
+
+
≥8
+ +
.
3
b2 − bc + c 2
bc ca a b
a b c
1.58. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
‹

1
1
1
+
+
+ a2 + b2 + c 2 ≥ 2(a b + bc + ca);
(a)
2a bc
a+b b+c c+a
(b)
a2
b2
c2
3(a2 + b2 + c 2 )
+
+
≤
.
a+b b+c c+a
2(a + b + c)
1.59. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
(a)
a2 − bc b2 − ca c 2 − a b 3(a b + bc + ca)
+ 2
+
+
≥ 3;
b2 + c 2
c + a2 a2 + b2
a2 + b2 + c 2
(b)
a2
b2
c2
a b + bc + ca
5
+
+
+ 2
≥ ;
b2 + c 2 c 2 + a2 a2 + b2
a + b2 + c 2
2
(c)
a2 + bc b2 + ca c 2 + a b
a b + bc + ca
+ 2
+ 2
≥ 2
+ 2.
2
2
2
2
b +c
c +a
a +b
a + b2 + c 2
1.60. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
a2
b2
c2
(a + b + c)2
+
+
≥
.
b2 + c 2 c 2 + a2 a2 + b2
2(a b + bc + ca)
9
10
Vasile Cîrtoaje
1.61. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
5
2a b
2bc
2ca
a2 + b2 + c 2
≥ .
+
+
+
(a + b)2 (b + c)2 (c + a)2 a b + bc + ca
2
1.62. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
bc
ca
1
a b + bc + ca
ab
+
+
+ ≥ 2
.
(a + b)2 (b + c)2 (c + a)2 4
a + b2 + c 2
1.63. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
3a b
3bc
3ca
a b + bc + ca 5
+
+
≤ 2
+ .
(a + b)2 (b + c)2 (c + a)2
a + b2 + c 2
4
1.64. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
(a)
a3 + a bc b3 + a bc c 3 + a bc
+
+
≥ a2 + b2 + c 2 ;
b+c
c+a
a+b
(b)
a3 + 2a bc b3 + 2a bc c 3 + 2a bc
1
+
+
≥ (a + b + c)2 ;
b+c
c+a
a+b
2
(c)
a3 + 3a bc b3 + 3a bc c 3 + 3a bc
+
+
≥ 2(a b + bc + ca).
b+c
c+a
a+b
1.65. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
a3 + 3a bc b3 + 3a bc c 3 + 3a bc
+
+
≥ a + b + c.
(b + c)2
(c + a)2
(a + b)2
1.66. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
(a)
a3 + 3a bc b3 + 3a bc c 3 + 3a bc
3
+
+
≥ ;
3
3
3
(b + c)
(c + a)
(a + b)
2
(b)
3a3 + 13a bc 3b3 + 13a bc 3c 3 + 13a bc
+
+
≥ 6.
(b + c)3
(c + a)3
(a + b)3
Symmetric Rational Inequalities
11
1.67. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
(a)
b3
c3
3
a3
+
+
+ a b + bc + ca ≥ (a2 + b2 + c 2 );
b+c c+a a+b
2
(b)
2a2 + bc 2b2 + ca 2c 2 + a b
9(a2 + b2 + c 2 )
+
+
≥
.
b+c
c+a
a+b
2(a + b + c)
1.68. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
a(b + c)
b(c + a)
c(a + b)
+ 2
+ 2
≥ 2.
2
2
+ bc + c
c + ca + a
a + a b + b2
b2
1.69. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
Y  a − b ‹2
a(b + c)
b(c + a)
c(a + b)
+
+
≥2+4
.
b2 + bc + c 2 c 2 + ca + a2 a2 + a b + b2
a+b
1.70. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
a b − bc + ca bc − ca + a b ca − a b + bc
3
+
+
≥ .
2
2
2
2
2
2
b +c
c +a
a +b
2
1.71. Let a, b, c be nonnegative real numbers, no two of which are zero. If k > −2, then
X a b + (k − 1)bc + ca
b2
+ k bc
+ c2
≥
3(k + 1)
.
k+2
1.72. Let a, b, c be nonnegative real numbers, no two of which are zero. If k > −2, then
X 3bc − a(b + c)
b2
+ k bc
+ c2
≤
3
.
k+2
1.73. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. Prove that
ab + 1
bc + 1
ca + 1
4
+ 2
+ 2
≥ .
2
2
2
2
a +b
b +c
c +a
3
12
Vasile Cîrtoaje
1.74. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. Prove that
5a b + 1 5bc + 1 5ca + 1
+
+
≥ 2.
(a + b)2 (b + c)2 (c + a)2
1.75. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
b2 − ca
c2 − a b
a2 − bc
+
+
≥ 0.
2b2 − 3bc + 2c 2 2c 2 − 3ca + 2a2 2a2 − 3a b + 2b2
1.76. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
2a2 − bc
2b2 − ca
2c 2 − a b
+
+
≥ 3.
b2 − bc + c 2 c 2 − ca + a2 a2 − a b + b2
1.77. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
a2
b2
c2
+
+
≥ 1.
2b2 − bc + 2c 2 2c 2 − ca + 2a2 2a2 − a b + 2b2
1.78. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
4b2
1
1
9
1
+ 2
+ 2
≥
.
2
2
2
2
− bc + 4c
4c − ca + 4a
4a − a b + 4b
7(a + b2 + c 2 )
1.79. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
2a2 + bc 2b2 + ca 2c 2 + a b
9
+ 2
+ 2
≥ .
2
2
2
2
b +c
c +a
a +b
2
1.80. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
2b2 + 3ca
2c 2 + 3a b
2a2 + 3bc
+
+
≥ 5.
b2 + bc + c 2 c 2 + ca + a2 a2 + a b + b2
Symmetric Rational Inequalities
13
1.81. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
21
2a2 + 5bc 2b2 + 5ca 2c 2 + 5a b
+
+
≥
.
(b + c)2
(c + a)2
(a + b)2
4
1.82. Let a, b, c be nonnegative real numbers, no two of which are zero. If k > −2, then
X 2a2 + (2k + 1)bc
b2
+ k bc
+ c2
≥
3(2k + 3)
.
k+2
1.83. Let a, b, c be nonnegative real numbers, no two of which are zero. If k > −2, then
X
3bc − 2a2
3
≤
.
2
2
b + k bc + c
k+2
1.84. If a, b, c are nonnegative real numbers, no two of which are zero, then
a2 + 16bc b2 + 16ca c 2 + 16a b
+ 2
+ 2
≥ 10.
b2 + c 2
c + a2
a + b2
1.85. If a, b, c are nonnegative real numbers, no two of which are zero, then
a2 + 128bc b2 + 128ca c 2 + 128a b
+
+
≥ 46.
b2 + c 2
c 2 + a2
a2 + b2
1.86. If a, b, c are nonnegative real numbers, no two of which are zero, then
a2 + 64bc b2 + 64ca c 2 + 64a b
+
+
≥ 18.
(b + c)2
(c + a)2
(a + b)2
1.87. Let a, b, c be nonnegative real numbers, no two of which are zero. If k ≥ −1, then
X a2 (b + c) + ka bc
b2 + k bc + c 2
≥ a + b + c.
14
Vasile Cîrtoaje
1.88. Let a, b, c be nonnegative real numbers, no two of which are zero. If k ≥
then
X a3 + (k + 1)a bc
b2 + k bc + c 2
−3
,
2
≥ a + b + c.
1.89. Let a, b, c be nonnegative real numbers, no two of which are zero. If k > 0, then
2a k − b k − c k 2b k − c k − a k 2c k − a k − b k
+ 2
+ 2
≥ 0.
b2 − bc + c 2
c − ca + a2
a − a b + b2
1.90. If a, b, c are the lengths of the sides of a triangle, then
(a)
c+a−b
a+b−c
2(a + b + c)
b+c−a
+
+
≥
;
b2 − bc + c 2 c 2 − ca + a2 a2 − a b + b2
a2 + b2 + c 2
(b)
a2 − 2bc
b2 − 2ca
c 2 − 2a b
+
+
≤ 0.
b2 − bc + c 2 c 2 − ca + a2 a2 − a b + b2
1.91. If a, b, c are nonnegative real numbers, then
a2
b2
c2
1
+
+
≤ .
2
2
2
2
2
2
5a + (b + c)
5b + (c + a)
5c + (a + b)
3
1.92. If a, b, c are nonnegative real numbers, then
b2 + c 2 − a2
c 2 + a2 − b2
a2 + b2 − c 2
1
+
+
≥ .
2
2
2
2
2
2
2a + (b + c)
2b + (c + a)
2c + (a + b)
2
1.93. Let a, b, c be positive real numbers. If k > 0, then
3a2 − 2bc
3b2 − 2ca
3c 2 − 2a b
3
+
+
≤ .
2
2
2
2
2
2
ka + (b − c)
k b + (c − a)
kc + (a − b)
k
1.94. Let a, b, c be nonnegative real numbers, no two of which are zero. If k ≥ 3 +
then
(a)
(b)
a2
ka2
a
b
c
9
+ 2
+ 2
≥
;
+ k bc b + kca c + ka b
(1 + k)(a + b + c)
1
1
1
9
+ 2
+ 2
≥
.
+ bc k b + ca kc + a b
(k + 1)(a b + bc + ca)
p
7,
Symmetric Rational Inequalities
15
1.95. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
2a2
1
1
1
6
+ 2
+ 2
≥ 2
.
2
2
+ bc 2b + ca 2c + a b
a + b + c + a b + bc + ca
1.96. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
22a2
1
1
1
1
+
+
≥
.
2
2
+ 5bc 22b + 5ca 22c + 5a b
(a + b + c)2
1.97. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
1
1
1
8
+
+
≥
.
2a2 + bc 2b2 + ca 2c 2 + a b
(a + b + c)2
1.98. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
1
1
12
1
+
+
≥
.
a2 + bc b2 + ca c 2 + a b
(a + b + c)2
1.99. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
(a)
(b)
a2
1
1
1
1
2
+ 2
+ 2
≥ 2
+
;
2
2
+ 2bc b + 2ca c + 2a b
a +b +c
a b + bc + ca
a(b + c)
b(c + a)
c(a + b)
a b + bc + ca
+ 2
+ 2
≥1+ 2
.
2
a + 2bc b + 2ca c + 2a b
a + b2 + c 2
1.100. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
(a)
(b)
a2
a
b
c
a+b+c
+ 2
+ 2
≤
;
+ 2bc b + 2ca c + 2a b
a b + bc + ca
a(b + c)
b(c + a)
c(a + b)
a2 + b2 + c 2
+
+
≤
1
+
.
a2 + 2bc b2 + 2ca c 2 + 2a b
a b + bc + ca
16
Vasile Cîrtoaje
1.101. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
(a)
2a2
b
c
a+b+c
a
+ 2
+ 2
≥ 2
;
+ bc 2b + ca 2c + a b
a + b2 + c 2
b+c
c+a
a+b
6
+ 2
+ 2
≥
.
2
2a + bc 2b + ca 2c + a b
a+b+c
(b)
1.102. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
a(b + c) b(c + a) c(a + b)
(a + b + c)2
+
+
≥
.
a2 + bc
b2 + ca
c2 + a b
a2 + b2 + c 2
1.103. Let a, b, c be nonnegative real numbers, no two of which are zero. If k > 0, then
p
p
p
p
3(2 + 3)
b2 + c 2 + 3bc c 2 + a2 + 3ca a2 + b2 + 3a b
+
+
≥
.
a2 + k bc
b2 + kca
c 2 + ka b
1+k
1.104. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
a2
1
1
1
8
6
+ 2
+ 2
+ 2
≥
.
2
2
2
2
2
+b
b +c
c +a
a +b +c
a b + bc + ca
1.105. If a, b, c are the lengths of the sides of a triangle, then
a(b + c)
b(c + a)
c(a + b)
+ 2
+ 2
≤ 2.
2
a + 2bc b + 2ca c + 2a b
1.106. If a, b, c are real numbers, then
a2 − bc
b2 − ca
c2 − a b
+
+
≥ 0.
2a2 + b2 + c 2 2b2 + c 2 + a2 2c 2 + a2 + b2
1.107. If a, b, c are nonnegative real numbers, then
3a2 − bc
3b2 − ca
3c 2 − a b
3
+
+
≤ .
2
2
2
2
2
2
2
2
2
2a + b + c
2b + c + a
2c + a + b
2
Symmetric Rational Inequalities
1.108. If a, b, c are nonnegative real numbers, then
(b + c)2
(c + a)2
(a + b)2
+
+
≥ 2.
4a2 + b2 + c 2 4b2 + c 2 + a2 4c 2 + a2 + b2
1.109. If a, b, c are positive real numbers, then
P
(a)
(b)
11a2
P
4a2
3
1
≤
;
2
2
+ 2b + 2c
5(a b + bc + ca)
1
1
1
≤
+
.
2
2
2
2
2
+b +c
2(a + b + c ) a b + bc + ca
1.110. If a, b, c are nonnegative real numbers such that a b + bc + ca = 3, then
p
p
p
a
b
c
3
+
+
≥ .
b+c c+a a+b
2
1.111. If a, b, c are nonnegative real numbers such that a b + bc + ca ≥ 3, then
1
1
1
1
1
1
+
+
≥
+
+
.
2+a 2+ b 2+c
1+ b+c 1+c+a 1+a+ b
1.112. If a, b, c are the lengths of the sides of a triangle, then
(a)
a2 − bc
b2 − ca
c2 − a b
+
+
≤ 0;
3a2 + b2 + c 2 3b2 + c 2 + a2 3c 2 + a2 + b2
(b)
a4 − b2 c 2
b4 − c 2 a2
c 4 − a2 b2
+
+
≤ 0.
3a4 + b4 + c 4 3b4 + c 4 + a4 3c 4 + a4 + b4
1.113. If a, b, c are the lengths of the sides of a triangle, then
4a2
bc
ca
ab
1
+ 2
+ 2
≥ .
2
2
2
2
2
2
+b +c
4b + c + a
4c + a + b
2
1.114. If a, b, c are the lengths of the sides of a triangle, then
b2
1
1
1
9
+ 2
+ 2
≤
.
2
2
2
+c
c +a
a +b
2(a b + bc + ca)
17
18
Vasile Cîrtoaje
1.115. If a, b, c are the lengths of the sides of a triangle, then
a + b b + c c + a
(a)
a − b + b − c + c − a > 5;
2
a + b2 b2 + c 2 c 2 + a2 (b)
a2 − b2 + b2 − c 2 + c 2 − a2 ≥ 3.
1.116. If a, b, c are the lengths of the sides of a triangle, then

‹
b+c c+a a+b
a
b
c
+
+
+3≥6
+
+
.
a
b
c
b+c c+a a+b
1.117. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
X 3a(b + c) − 2bc
3
≥ .
(b + c)(2a + b + c) 2
1.118. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
X
a(b + c) − 2bc
≥ 0.
(b + c)(3a + b + c)
1.119. Let a, b, c be positive real numbers such that a2 + b2 + c 2 ≥ 3. Prove that
b5 − b2
c5 − c2
a5 − a2
+
+
≥ 0.
a5 + b2 + c 2 b5 + c 2 + a2 c 5 + a2 + b2
1.120. Let a, b, c be positive real numbers such that a2 + b2 + c 2 = a3 + b3 + c 3 . Prove
that
a2
b2
c2
3
+
+
≥ .
b+c c+a a+b
2
1.121. If a, b, c ∈ [0, 1], then
(a)
a
b
c
+
+
≤ 1;
bc + 2 ca + 2 a b + 2
(b)
ab
bc
ca
+
+
≤ 1.
2bc + 1 2ca + 1 2a b + 1
Symmetric Rational Inequalities
19
1.122. Let a, b, c be positive real numbers such that a + b + c = 2. Prove that

‹
1
1
1
+
+
5(1 − a b − bc − ca)
+ 9 ≥ 0.
1 − a b 1 − bc 1 − ca
1.123. Let a, b, c be nonnegative real numbers such that a + b + c = 2. Prove that
2 − c2
2 − a2 2 − b2
+
+
≤ 3.
2 − bc 2 − ca 2 − a b
1.124. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that
3 + 5a2 3 + 5b2 3 + 5c 2
+
+
≥ 12.
3 − bc
3 − ca
3 − ab
1.125. Let a, b, c be nonnegative real numbers such that a + b + c = 2. If
7
−1
≤m≤ ,
7
8
then
a2 + m
b2 + m
c2 + m
3(4 + 9m)
+
+
≥
.
3 − 2bc 3 − 2ca 3 − 2a b
19
1.126. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that
47 − 7a2 47 − 7b2 47 − 7c 2
+
+
≥ 60.
1 + bc
1 + ca
1 + ab
1.127. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that
26 − 7a2 26 − 7b2 26 − 7c 2
57
+
+
≤
.
1 + bc
1 + ca
1 + ab
2
1.128. Let a, b, c be nonnegative real numbers, no all are zero. Prove that
X 5a(b + c) − 6bc
≤ 3.
a2 + b2 + c 2 + bc
20
Vasile Cîrtoaje
1.129. Let a, b, c be nonnegative real numbers, no two of which are zero, and let
x=
a2 + b2 + c 2
.
a b + bc + ca
Prove that
a
b
c
1
1
+
+
+ ≥x+ ;
b+c c+a a+b 2
x
‹

b
c
4
a
+
+
≥ 5x + ;
6
b+c c+a a+b
x
(a)
(b)

‹
a
b
c
3 1
1
+
+
− ≥
x−
.
b+c c+a a+b 2 3
x
(c)
1.130. If a, b, c are real numbers, then
a2
1
1
1
9
+ 2
+ 2
≤
.
2
2
2
2
2
2
+ 7(b + c ) b + 7(c + a ) c + 7(a + b ) 5(a + b + c)2
1.131. If a, b, c are real numbers, then
ca
ab
3
bc
+
+
≤ .
3a2 + b2 + c 2 3b2 + c 2 + a2 3c 2 + a2 + b2
5
1.132. If a, b, c are real numbers such that a + b + c = 3, then
(a)
1
1
1
3
+
+
≤ ;
2
2
2
2
2
2
2+ b +c
2+c +a
2+a + b
4
(b)
1
1
1
1
+
+
≤ .
2
2
2
2
2
2
8 + 5(b + c ) 8 + 5(c + a ) 8 + 5(a + b ) 6
1.133. If a, b, c are real numbers, then
(a + b)(a + c)
(b + c)(b + a)
(c + a)(c + b)
4
+ 2
+ 2
≤ .
2
2
2
2
2
2
2
a + 4(b + c ) b + 4(c + a ) c + 4(a + b ) 3
Symmetric Rational Inequalities
21
1.134. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
X
1
1
≤
.
(b + c)(7a + b + c) 2(a b + bc + ca)
1.135. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
X
b2
+ c2
9
1
≤
.
+ 4a(b + c) 10(a b + bc + ca)
1.136. If a, b, c are nonnegative real numbers such that a + b + c = 3, then
1
1
1
9
+
+
≤
.
3 − a b 3 − bc 3 − ca
2(a b + bc + ca)
1.137. If a, b, c are nonnegative real numbers such that a + b + c = 3, then
a2
bc
ca
ab
3
+ 2
+ 2
≤ .
+a+6 b + b+6 c +c+6 8
1.138. If a, b, c are nonnegative real numbers such that a b + bc + ca = 3, then
1
1
1
1
+
+
≥ .
8a2 − 2bc + 21 8b2 − 2ca + 21 8c 2 − 2a b + 21 9
1.139. Let a, b, c be real numbers, no two of which are zero. Prove that
(a)
a2 + bc b2 + ca c 2 + a b
(a + b + c)2
+
+
≥
;
b2 + c 2
c 2 + a2 a2 + b2
a2 + b2 + c 2
(b)
a2 + 3bc b2 + 3ca c 2 + 3a b
6(a b + bc + ca)
+ 2
+ 2
≥
.
b2 + c 2
c + a2
a + b2
a2 + b2 + c 2
1.140. Let a, b, c be real numbers such that a b + bc + ca ≥ 0 and no two of which are
zero. Prove that
a(b + c) b(c + a) c(a + b)
3
+ 2
+ 2
≥
.
2
2
2
2
b +c
c +a
a +b
10
22
Vasile Cîrtoaje
1.141. If a, b, c are positive real numbers such that a bc > 1, then
1
4
1
+
≥
.
a + b + c − 3 a bc − 1
a b + bc + ca − 3
1.142. Let a, b, c be positive real numbers, no two of which are zero. Prove that
X (4b2 − ac)(4c 2 − a b)
b+c
≤
27
a bc.
2
1.143. Let a, b, c be nonnegative real numbers, no two of which are zero, such that
a + b + c = 3.
Prove that
b
c
2
a
+
+
≥ .
3a + bc 3b + ca 3c + a b
3
1.144. Let a, b, c be positive real numbers such that

‹
1 1 1
(a + b + c)
+ +
= 10.
a b c
Prove that
19
a
b
c
5
≤
+
+
≤ .
12
b+c c+a a+b
3
1.145. Let a, b, c be nonnegative real numbers, no two of which are zero, such that
a + b + c = 3. Prove that
9
a
b
c
<
+
+
≤ 1.
10 2a + bc 2b + ca 2c + a b
1.146. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
a3
b3
c3
a3 + b3 + c 3
+
+
≤
.
2a2 + bc 2b2 + ca 2c 2 + a b
a2 + b2 + c 2
Symmetric Rational Inequalities
23
1.147. Let a, b, c be positive real numbers, no two of which are zero. Prove that
b3
c3
a+b+c
a3
+
+
≥
.
4a2 + bc 4b2 + ca 4c 2 + a b
5
1.148. If a, b, c are positive real numbers, then
1
1
3
1
+
+
≥
.
(2 + a)2 (2 + b)2 (2 + c)2
6 + a b + bc + ca
1.149. If a, b, c are positive real numbers, then
1
1
1
3
+
+
≥
.
1 + 3a 1 + 3b 1 + 3c
3 + a bc
1.150. Let a, b, c be real numbers, no two of which are zero. If 1 ≤ k ≤ 3, then

‹
‹
‹
2a b
2bc
2ca
k+ 2
k+ 2
k+ 2
≥ (k − 1)(k2 − 1).
a + b2
b + c2
c + a2
1.151. If a, b, c are non-zero and distinct real numbers, then
•
˜

‹
1
1
1
1
1
1
1
1
1
+
+
+3
+
+
≥4
+
+
.
a2 b2 c 2
(a − b)2 (b − c)2 (c − a)2
a b bc ca
1.152. Let a, b, c be positive real numbers, and let
A=
a b
+ + k,
b a
B=
b c
+ + k,
c
b
C=
c a
+ + k,
a b
where −2 < k ≤ 4. Prove that
1 1 1
1
4
+ + ≤
+
.
A B C
k + 2 A + B + C − (k + 2)
1.153. If a, b, c are nonnegative real numbers, no two of which are zero, then
b2
1
1
1
1
1
1
+ 2
+ 2
≥ 2
+ 2
+ 2
.
2
2
2
+ bc + c
c + ca + a
a + ab + b
2a + bc 2b + ca 2c + a b
24
Vasile Cîrtoaje
1.154. If a, b, c are nonnegative real numbers such that a + b + c = 3, then
1
1
1
1
1
1
+
+
≥ 2
+ 2
+ 2
.
2a b + 1 2bc + 1 2ca + 1
a +2 b +2 c +2
1.155. If a, b, c are nonnegative real numbers such that a + b + c = 4, then
1
1
1
1
1
1
+
+
≥ 2
+ 2
+ 2
.
a b + 2 bc + 2 ca + 2
a +2 b +2 c +2
1.156. If a, b, c are nonnegative real numbers, no two of which are zero, then
a b + bc + ca
(a − b)2 (b − c)2 (c − a)2
+
≤ 1;
a2 + b2 + c 2
(a2 + b2 )(b2 + c 2 )(c 2 + a2 )
(a)
a b + bc + ca
(a − b)2 (b − c)2 (c − a)2
+
≤ 1.
a2 + b2 + c 2
(a2 − a b + b2 )(b2 − bc + c 2 )(c 2 − ca + a2 )
(b)
1.157. If a, b, c are nonnegative real numbers, no two of which are zero, then
a2
1
1
45
1
+ 2
+ 2
≥
.
2
2
2
2
2
2
+b
b +c
c +a
8(a + b + c ) + 2(a b + bc + ca)
1.158. If a, b, c are real numbers, no two of which are zero, then
a2 − 7bc b2 − 7ca c 2 − 7a b 9(a b + bc + ca)
+ 2
+ 2
+
≥ 0.
b2 + c 2
a + b2
a + b2
a2 + b2 + c 2
1.159. If a, b, c are real numbers such that a bc 6= 0, then
(b + c)2 (c + a)2 (a + b)2
10(a + b + c)2
+
+
≥
2
+
.
a2
b2
c2
3(a2 + b2 + c 2 )
1.160. If a, b, c are nonnegative real numbers, no two of which are zero, then
a2 − 4bc b2 − 4ca c 2 − 4a b 9(a b + bc + ca) 9
+ 2
+ 2
+
≥ .
b2 + c 2
a + b2
a + b2
a2 + b2 + c 2
2
Symmetric Rational Inequalities
1.161. If a, b, c are nonnegative real numbers, no two of which are zero, then
9(a − b)2 (b − c)2 (c − a)2
a2 + b2 + c 2
≥1+
.
a b + bc + ca
(a + b)2 (b + c)2 (c + a)2
1.162. If a, b, c are nonnegative real numbers, no two of which are zero, then
p
a2 + b2 + c 2
(a − b)2 (b − c)2 (c − a)2
≥ 1 + (1 + 2)2 2
.
a b + bc + ca
(a + b2 )(b2 + c 2 )(c 2 + a2 )
1.163. If a, b, c are nonnegative real numbers, no two of which are zero, then
2
2
2
5
5
5
+
+
≥
+
+
.
a+b b+c c+a
3a + b + c 3b + c + a 3c + a + b
1.164. If a, b, c are real numbers, no two of which are zero, then
(a)
8a2 + 3bc
8b2 + 3ca
8c 2 + 3a b
+
+
≥ 11;
b2 + bc + c 2 c 2 + ca + a2 a2 + a b + b2
(b)
8a2 − 5bc
8b2 − 5ca
8c 2 − 5a b
+
+
≥ 9.
b2 − bc + c 2 c 2 − ca + a2 a2 − a b + b2
1.165. If a, b, c are real numbers, no two of which are zero, then
4b2 + ca
4c 2 + a b
4a2 + bc
+
+
≥ 1.
4b2 + 7bc + 4c 2 4c 2 + 7ca + 4a2 4a2 + 7a b + 4b2
1.166. If a, b, c are real numbers, no two of which are equal, then
1
1
1
27
+
+
≥
.
2
2
2
2
2
2
(a − b)
(b − c)
(c − a)
4(a + b + c − a b − bc − ca)
1.167. If a, b, c are real numbers, no two of which are zero, then
1
1
1
14
+
+
≥
.
a2 − a b + b2 b2 − bc + c 2 c 2 − ca + a2
3(a2 + b2 + c 2 )
25
26
Vasile Cîrtoaje
1.168. Let a, b, c be real numbers such that a b + bc + ca ≥ 0 and no two of which are
zero. Prove that
b
c
3
a
+
+
≥ ;
b+c c+a a+b
2
(a)
(b) i f a b ≤ 0, then
a
b
c
+
+
≥ 2.
b+c c+a a+b
1.169. If a, b, c are nonnegative real numbers, then
a
b
c
a b + bc + ca
+
+
≥
.
7a + b + c 7b + c + a 7c + a + b
(a + b + c)2
1.170. If a, b, c are the lengths of the sides of a triangle, then
a2
b2
c2
1
+
+
≥ .
2
2
2
4a + 5bc 4b + 5ca 4c + 5a b
3
1.171. If a, b, c are the lengths of the sides of a triangle, then
1
1
1
3
+
+
≥
.
7a2 + b2 + c 2 7b2 + c 2 + a2 7c 2 + a2 + b2
(a + b + c)2
1.172. Let a, b, c be the lengths of the sides of a triangle. If k > −2, then
X a(b + c) + (k + 1)bc
b2
+ k bc
+ c2
≤
3(k + 3)
.
k+2
1.173. Let a, b, c be the lengths of the sides of a triangle. If k > −2, then
X 2a2 + (4k + 9)bc
b2
+ k bc
+ c2
≤
3(4k + 11)
.
k+2
1.174. If a ≥ b ≥ c ≥ d such that a bcd = 1, then
1
1
1
3
+
+
≥
.
p
3
1+a 1+ b 1+c
1 + a bc
Symmetric Rational Inequalities
27
1.175. Let a, b, c, d be positive real numbers such that a bcd = 1. Prove that
X
1
≤ 1.
1 + a b + bc + ca
1.176. Let a, b, c, d be positive real numbers such that a bcd = 1. Prove that
1
1
1
1
+
+
+
≥ 1.
2
2
2
(1 + a)
(1 + b)
(1 + c)
(1 + d)2
1.177. Let a, b, c, d 6=
1
be positive real numbers such that a bcd = 1. Prove that
3
1
1
1
1
+
+
+
≥ 1.
2
2
2
(3a − 1)
(3b − 1)
(3c − 1)
(3d − 1)2
1.178. Let a, b, c, d be positive real numbers such that a bcd = 1. Prove that
1
1
1
1
+
+
+
≥ 1.
1 + a + a2 + a3 1 + b + b2 + b3 1 + c + c 2 + c 3 1 + d + d 2 + d 3
1.179. Let a, b, c, d be positive real numbers such that a bcd = 1. Prove that
1
1
1
1
+
+
+
≥ 1.
2
2
2
1 + a + 2a
1 + b + 2b
1 + c + 2c
1 + d + 2d 2
1.180. Let a, b, c, d be positive real numbers such that a bcd = 1. Prove that
1 1 1 1
9
25
+ + + +
≥
.
a b c d a+b+c+d
4
1.181. If a, b, c, d are real numbers such that a + b + c + d = 0, then
(a − 1)2 (b − 1)2 (c − 1)2 (d − 1)2
+
+ 2
+
≤ 4.
3a2 + 1
3b2 + 1
3c + 1
3d 2 + 1
28
Vasile Cîrtoaje
1.182. If a, b, c, d ≥ −5 such that a + b + c + d = 4, then
1− b
1−c
1−d
1−a
+
+
+
≥ 0.
2
2
2
(1 + a)
(1 + b)
(1 + c)
(1 + d)2
1.183. Let a1 , a2 , . . . , an be positive real numbers such that a1 + a2 + · · · + an = n. Prove
that
X
1
1
≤ .
2
2
2
(n + 1)a1 + a2 + · · · + an2
1.184. Let a1 , a2 , . . . , an be real numbers such that a1 + a2 + · · · + an = 0. Prove that
(a1 + 1)2
a12 + n − 1
+
(a2 + 1)2
a22 + n − 1
+ ··· +
(an + 1)2
n
≥
.
2
an + n − 1
n−1
1.185. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an = 1. Prove that
1
1
1
+
+ ··· +
≥ 1.
1 + (n − 1)a1 1 + (n − 1)a2
1 + (n − 1)an
1.186. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an = 1. Prove that
1
1 − a1 + na12
+
1
1 − a2 + na22
+ ··· +
1
≥ 1.
1 − an + nan2
1.187. Let a1 , a2 , . . . , an be positive real numbers such that
a1 , a2 , . . . , an ≥
k(n − k − 1)
,
kn − k − 1
k>1
and
a1 a2 · · · an = 1.
Prove that
1
1
1
n
+
+ ··· +
≤
.
a1 + k a2 + k
an + k
1+k
Symmetric Rational Inequalities
29
1.188. Let a1 , a2 , . . . , an be positive real numbers such that
a1 ≥ 1 ≥ a2 ≥ · · · ≥ an ,
Prove that
1 − a1
3 + a12
+
1 − a2
3 + a22
a1 a2 · · · an = 1.
+ ··· +
1 − an
≥ 0.
3 + an2
1.189. If a1 , a2 , . . . , an ≥ 0, then
1
1
1
n
+
+ ··· +
≥
.
1 + na1 1 + na2
1 + nan
n + a1 a2 · · · an
1.190. If a1 , a2 , . . . , an are positive real numbers, then
bn
an
a1 a2
b1 b2
+
+ ··· +
≥
+
+ ··· + ,
a1 a2
an
b1 b2
bn
where
bi =
1 X
aj,
n − 1 j6=i
i = 1, 2, · · · , n.
1.191. If a1 , a2 , . . . , an are positive real numbers such that
a1 + a2 + · · · + an =
1
1
1
+
+ ··· + ,
a1 a2
an
then
(a)
1
1
1
+
+ ··· +
≥ 1;
1 + (n − 1)a1 1 + (n − 1)a2
1 + (n − 1)an
(b)
1
1
1
+
+ ··· +
≤ 1.
n − 1 + a1 n − 1 + a2
n − 1 + an
30
Vasile Cîrtoaje
Symmetric Rational Inequalities
1.2
31
Solutions
P 1.1. If a, b are nonnegative real numbers, then
1
1
1
+
≥
.
2
2
(1 + a)
(1 + b)
1 + ab
(Vasile Cîrtoaje, 1994)
First Solution. Use the Cauchy-Schwarz inequality as follows:
1
1
1
(b + a)2
1
+
−
≥
−
2
2
2
2
2
2
(1 + a)
(1 + b)
1 + ab
b (1 + a) + a (1 + b)
1 + ab
a b[a2 + b2 − 2(a + b) + 2]
=
(1 + a b)[b2 (1 + a)2 + a2 (1 + b)2 ]
a b[(a − 1)2 + (b − 1)2 ]
=
≥ 0.
(1 + a b)[b2 (1 + a)2 + a2 (1 + b)2 ]
The equality holds for a = b = 1.
Second Solution. By the Cauchy-Schwarz inequality, we have
‹

1
≥ (a + 1)2 ,
(a + b) a +
b
(a + b)

‹
1
+ b ≥ (1 + b)2 ,
a
hence
1
1
1
1
1
+
≥
+
=
.
2
2
(1 + a)
(1 + b)
(a + b)(a + 1/b) (a + b)(1/a + b) 1 + a b
Third Solution. The desired inequality follows from the identity
1
1
a b(a − b)2 + (1 − a b)2
1
+
−
=
.
(1 + a)2 (1 + b)2 1 + a b
(1 + a)2 (1 + b)2 (1 + a b)
P 1.2. If a, b, c are nonnegative real numbers, then
a2 − bc
b2 − ca
c2 − a b
+
+
≥ 0.
3a + b + c 3b + c + a 3c + a + b
32
Vasile Cîrtoaje
Solution. We use the SOS method. Without loss of generality, assume that a ≥ b ≥ c.
We have
2
X (a − b)(a + c) + (a − c)(a + b)
X a2 − bc
=
3a + b + c
3a + b + c
X (a − b)(a + c) X (b − a)(b + c)
+
=
3a + b + c
3b + c + a
X (a − b)2 (a + b − c)
=
(3a + b + c)(3b + c + a)
Since a + b − c ≥ 0, it suffices to show that
(b − c)2 (b + c − a)(3a + b + c) + (c − a)2 (c + a − b)(3b + c + a) ≥ 0;
that is,
(a − c)2 (c + a − b)(3b + c + a) ≥ (b − c)2 (a − b − c)(3a + b + c).
This inequality is trivial for a ≤ b + c. Otherwise, we can get it by multiplying the
obvious inequalities
c + a − b ≥ a − b − c,
b2 (a − c)2 ≥ a2 (b − c)2 ,
a(3b + c + a) ≥ b(3a + b + c),
a ≥ b.
The equality holds for a = b = c, and for a = 0 and b = c (or any cyclic permutation).
P 1.3. If a, b, c are positive real numbers, then
4a2 − b2 − c 2 4b2 − c 2 − a2 4c 2 − a2 − b2
+
+
≤ 3.
a(b + c)
b(c + a)
c(a + b)
(Vasile Cîrtoaje, 2006)
Solution. We use the SOS method. Write the inequality as follows:
X
4a2 − b2 − c 2
1−
≥ 0,
a(b + c)
X b2 + c 2 − 4a2 + a(b + c)
≥ 0,
a(b + c)
Symmetric Rational Inequalities
33
X (b2 − a2 ) + a(b − a) + (c 2 − a2 ) + a(c − a)
a(b + c)
X (b − a)(2a + b) + (c − a)(2a + c)
a(b + c)
X (b − a)(2a + b)
a(b + c)
X
+
≥ 0,
≥ 0,
X (a − b)(2b + a)
b(c + a)
≥ 0,
c(a + b)(a − b)2 (bc + ca − a b) ≥ 0.
Without loss of generality, assume that a ≥ b ≥ c. Since ca + a b − bc > 0, it suffices to
show that
b(c + a)(c − a)2 (a b + bc − ca) + c(a + b)(a − b)2 (bc + ca − a b) ≥ 0,
that is,
b(c + a)(a − c)2 (a b + bc − ca) ≥ c(a + b)(a − b)2 (a b − bc − ca).
For the nontrivial case a b − bc − ca > 0, this inequality follows by multiplying the
inequalities
a b + bc − ca > a b − bc − ca,
(a − c)2 ≥ (a − b)2 ,
b(c + a) ≥ c(a + b).
The equality holds for a = b = c
P 1.4. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
(a)
(b)
(c)
a2
2a2
a2
1
1
1
3
+ 2
+ 2
≥
;
+ bc b + ca c + a b
a b + bc + ca
1
1
1
2
+ 2
+ 2
≥
.
+ bc 2b + ca 2c + a b
a b + bc + ca
1
1
1
2
+ 2
+ 2
>
.
+ 2bc b + 2ca c + 2a b
a b + bc + ca
(Vasile Cîrtoaje, 2005)
34
Vasile Cîrtoaje
Solution. (a) Since
a(b + c − a)
a b + bc + ca
=1+
,
2
a + bc
a2 + bc
we can write the inequality as
a(b + c − a) b(c + a − b) c(a + b − c)
+
+
≥ 0.
a2 + bc
b2 + ca
c2 + a b
Without loss of generality, assume that a = min{a, b, c}. Since b + c − a > 0, it suffices
to show that
b(c + a − b) c(a + b − c)
+
≥ 0.
b2 + ca
c2 + a b
This is equivalent to each of the following inequalities
(b2 + c 2 )a2 − (b + c)(b2 − 3bc + c 2 )a + bc(b − c)2 ≥ 0,
(b − c)2 a2 − (b + c)(b − c)2 a + bc(b − c)2 + a bc(2a + b + c) ≥ 0,
(b − c)2 (a − b)(a − c) + a bc(2a + b + c) ≥ 0.
The last inequality is obviously true. The equality holds for a = 0 and b = c (or any
cyclic permutation thereof).
(b) According to the identities
2a2 + bc = a(2a − b − c) + a b + bc + ca,
2b2 + ca = b(2b − c − a) + a b + bc + ca,
2c 2 + a b = c(2c − a − b) + a b + bc + ca,
we can write the inequality as
1
1
1
+
+
≥ 2,
1+ x 1+ y 1+z
where
x=
a(2a − b − c)
,
a b + bc + ca
y=
b(2b − c − a)
c(2c − a − b)
, z=
.
a b + bc + ca
a b + bc + ca
Without loss of generality, assume that a = min{a, b, c}. Since x ≤ 0 and
suffices to show that
1
1
+
≥ 1.
1+ y 1+z
This is equivalent to each of the following inequalities
1 ≥ yz,
1
≥ 1, it
1+ x
Symmetric Rational Inequalities
35
(a b + bc + ca)2 ≥ bc(2b − c − a)(2c − a − b),
a2 (b2 + bc + c 2 ) + 3a bc(b + c) + 2bc(b − c)2 ≥ 0.
The last inequality is obviously true. The equality holds for a = 0 and b = c (or any
cyclic permutation thereof).
(c) According to the identities
a2 + 2bc = (a − b)(a − c) + a b + bc + ca,
b2 + 2ca = (b − c)(b − a) + a b + bc + ca,
c 2 + 2a b = (c − a)(c − b) + a b + bc + ca,
we can write the inequality as
1
1
1
+
+
> 2,
1+ x 1+ y 1+z
where
x=
(a − b)(a − c)
,
a b + bc + ca
y=
(b − c)(b − a)
(c − a)(c − b)
, z=
.
a b + bc + ca
a b + bc + ca
Since
x y + yz + z x = 0
and
x yz =
we have
−(a − b)2 (b − c)2 (c − a)2
≤ 0,
(a b + bc + ca)3
1 − 2x yz
1
1
1
+
+
−2=
> 0.
1+ x 1+ y 1+z
(1 + x)(1 + y)(1 + z)
P 1.5. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
a(b + c) b(c + a) c(a + b)
+ 2
+ 2
≥ 2.
a2 + bc
b + ca
c + ab
(Pham Kim Hung, 2006)
Solution. Without loss of generality, assume that a ≥ b ≥ c and write the inequality as
b(c + a) (a − b)(a − c) (a − c)(b − c)
≥
+
.
b2 + ca
a2 + bc
c2 + a b
36
Vasile Cîrtoaje
Since
a−b
(a − b)(a − c) (a − b)a
≤ 2
≤
2
a + bc
a + bc
a
and
(a − c)(b − c)
a(b − c)
b−c
≤ 2
≤
,
2
c + ab
c + ab
b
it suffices to show that
b(c + a)
a−b b−c
≥
+
.
b2 + ca
a
b
This inequality is equivalent to
b2 (a − b)2 − 2a bc(a − b) + a2 c 2 + a b2 c ≥ 0,
(a b − b2 − ac)2 + a b2 c ≥ 0.
The equality holds for for a = 0 and b = c (or any cyclic permutation).
P 1.6. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
a2
b2
c2
a
b
c
+
+
≥
+
+
.
2
2
2
2
2
2
b +c
c +a
a +b
b+c c+a a+b
(Vasile Cîrtoaje, 2002)
Solution. We have
X
a2
a
−
b2 + c 2 b + c
=
X a b(a − b) + ac(a − c)
(b2 + c 2 )(b + c)
X
a b(a − b)
ba(b − a)
+
2
2
2
(b + c )(b + c)
(c + a2 )(c + a)
X
a b(a − b)2
= (a2 + b2 + c 2 + a b + bc + ca)
≥ 0.
(b2 + c 2 )(c 2 + a2 )(b + c)(c + a)
=
X
The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).
P 1.7. Let a, b, c be positive real numbers. Prove that
1
1
1
a
b
c
+
+
≥ 2
+ 2
+ 2
.
b+c c+a a+b
a + bc b + ca c + a b
Symmetric Rational Inequalities
First Solution. Without loss of generality, assume that a = min{a, b, c}. Since
‹
X
X 1
X 1
a
a
−
=
−
b+c
a2 + bc
b + c a2 + bc
X (a − b)(a − c)
=
(b + c)(a2 + bc)
and (a − b)(a − c) ≥ 0, it suffices to show that
(b − c)(b − a)
(c − a)(c − b)
+
≥ 0.
2
(c + a)(b + ca) (a + b)(c 2 + a b)
This inequality is equivalent to
(b − c)[(b2 − a2 )(c 2 + a b) + (a2 − c 2 )(b2 + ca)] ≥ 0,
a(b − c)2 (b2 + c 2 − a2 + a b + bc + ca) ≥ 0,
and is clearly true for a = min{a, b, c}. The equality holds for a = b = c.
Second Solution. Since
X
˜ X •
˜
X•
b
c
1
1
1
=
+
=
a
+
,
b+c
(b + c)2 (b + c)2
(a + b)2 (a + c)2
we can write the inequality as
˜
X •
1
1
1
a
+
−
≥ 0.
(a + b)2 (a + c)2 a2 + bc
This is true, since
1
1
1
bc(b − c)2 + (a2 − bc)2
+
−
=
≥ 0.
(a + b)2 (a + c)2 a2 + bc
(a + b)2 (a + c)2 (a2 + bc)
We can also prove this inequality using the Cauchy-Schwarz inequality, as follows
1
1
1
(c + b)2
1
+
−
≥
−
(a + b)2 (a + c)2 a2 + bc
c 2 (a + b)2 + b2 (a + c)2 a2 + bc
bc[2a2 − 2a(b + c) + b2 + c 2 ]
= 2
(a + bc)[c 2 (a + b)2 + b2 (a + c)2 ]
bc[(2a − b − c)2 + (b − c)2 ]
=
≥ 0.
2(a2 + bc)[c 2 (a + b)2 + b2 (a + c)2 ]
37
38
Vasile Cîrtoaje
P 1.8. Let a, b, c be positive real numbers. Prove that
1
1
2a
2b
2c
1
+
+
≥ 2
+ 2
+ 2
.
b+c c+a a+b
3a + bc 3b + ca 3c + a b
(Vasile Cîrtoaje, 2005)
Solution. Since
X
‹
X
X 1
1
2a
2a
−
=
−
b+c
3a2 + bc
b + c 3a2 + bc
X (a − b)(a − c) + a(2a − b − c)
=
,
(b + c)(3a2 + bc)
it suffices to show that
X
(a − b)(a − c)
≥0
(b + c)(3a2 + bc)
X
a(2a − b − c)
≥ 0.
(b + c)(3a2 + bc)
and
In order to prove the first inequality, assume that a = min{a, b, c}. Since
(a − b)(a − c) ≥ 0,
it is enough to show that
(b − c)(b − a)
(c − a)(c − b)
+
≥ 0.
(c + a)(3b2 + ca) (a + b)(3c 2 + a b)
This is equivalent to the obvious inequality
a(b − c)2 (b2 + c 2 − a2 + 3a b + bc + 3ca) ≥ 0.
The second inequality is also true, since
X
X a(a − b) + a(a − c)
a(2a − b − c)
=
(b + c)(3a2 + bc)
(b + c)(3a2 + bc)
X
X
a(a − b)
b(b − a)
+
=
2
(b + c)(3a + bc)
(c + a)(3b2 + ca)
•
˜
X
a
b
=
(a − b)
−
(b + c)(3a2 + bc) (c + a)(3b2 + ca)
X c(a − b)2 [(a − b)2 + c(a + b)]
=
≥ 0.
(b + c)(c + a)(3a2 + bc)(3b2 + ca)
The equality holds for a = b = c.
Symmetric Rational Inequalities
39
P 1.9. Let a, b, c be positive real numbers. Prove that
(a)
(b)
b
c
13 2(a b + bc + ca)
a
+
+
≥
−
;
b+c c+a a+b
6
3(a2 + b2 + c 2 )

‹
p
a
b
c
3
a b + bc + ca
+
+
− ≥ ( 3 − 1) 1 − 2
.
b+c c+a a+b 2
a + b2 + c 2
(Vasile Cîrtoaje, 2006)
Solution. (a) We use the SOS method. Rewrite the inequality as

‹
a
a b + bc + ca
b
c
3 2
1− 2
.
+
+
− ≥
b+c c+a a+b 2 3
a + b2 + c 2
Since
‹ X
X a
(a − b) + (a − c)
1
−
=
b+c 2
2(b + c)
X a−b
X b−a
=
+
2(b + c)
2(c + a)
‹
X a−b  1
1
=
−
2
b+c c+a
X
(a − b)2
=
2(b + c)(c + a)
and

‹ X
(a − b)2
2
a b + bc + ca
1− 2
=
,
3
a + b2 + c 2
3(a2 + b2 + c 2 )
the inequality can be restated as
•
˜
X
1
1
−
≥ 0.
(a − b)2
2(b + c)(c + a) 3(a2 + b2 + c 2 )
This is true, since
3(a2 + b2 + c 2 ) − 2(b + c)(c + a) = (a + b − c)2 + 2(a − b)2 ≥ 0.
The equality holds for a = b = c.
(b) Let
p = a + b + c,
q = a b + bc + ca,
r = a bc.
We have
X
X a
X 1
a
=
+1 −3= p
−3
b+c
b+c
b+c
p(p2 + q)
=
− 3.
pq − r
40
Vasile Cîrtoaje
According to P 3.57-(a) in Volume 1, for fixed p and q, the product r is minimal when
a = 0 or b = c. Therefore, it suffices to prove the inequality for a = 0 and for b = c = 1.
Case 1: a = 0. The original inequality can be written as

‹
p
bc
b c 3
+ − ≥ ( 3 − 1) 1 − 2
.
c
b 2
b + c2
It suffices to show that
bc
b c 3
+ − ≥1− 2
.
c
b 2
b + c2
Denoting
t=
b2 + c 2
,
bc
t ≥ 2,
this inequality becomes
t−
3
1
≥1− ,
2
t
(t − 2)(2t − 1) ≥ 0.
Case 2: b = c = 1. The original inequality becomes as follows:

‹
p
a
2
3
2a + 1
+
− ≥ ( 3 − 1) 1 − 2
,
2 a+1 2
a +2
p
(a − 1)2
( 3 − 1)(a − 1)2
≥
,
2(a + 1)
a2 + 2
p
(a − 1)2 (a − 3 + 1)2 ≥ 0.
a
The equality holds for a = b = c, and for p
= b = c (or any cyclic permutation).
3−1
P 1.10. Let a, b, c be positive real numbers. Prove that

‹2
1
1
a+b+c
1
.
+
+
≤
a2 + 2bc b2 + 2ca c 2 + 2a b
a b + bc + ca
(Vasile Cîrtoaje, 2006)
First Solution. Write the inequality as
‹
X  a b + bc + ca
(a + b + c)2
−3≥
−
1
,
a b + bc + ca
a2 + 2bc
Symmetric Rational Inequalities
41
(a − b)2 + (b − c)2 + (a − b)(b − c) X (a − b)(a − c)
+
≥ 0.
a b + bc + ca
a2 + 2bc
Without loss of generality, assume that a ≥ b ≥ c. Since (a − b)(a − c) ≥ 0 and (c −
a)(c − b) ≥ 0, it suffices to show that
(a − b)2 + (b − c)2 + (a − b)(b − c) −
(a b + bc + ca)(a − b)(b − c)
≥ 0.
b2 + 2ca
This inequality is equivalent to
(a − b)2 + (b − c)2 −
(a − b)2 (b − c)2
≥ 0,
b2 + 2ca
or
c(a − b)2 (2a + 2b − c)
≥ 0,
b2 + 2ca
which is clearly true. The equality holds for a = b = c.
Second Solution. Assume that a ≥ b ≥ c and write the desired inequality as
‹
X  a b + bc + ca
(a + b + c)2
−3≥
−1 ,
a b + bc + ca
a2 + 2bc
X
X (a − b)(a − c)
1
(a − b)(a − c) +
≥ 0,
a b + bc + ca
a2 + 2bc

‹
X
a b + bc + ca
1+
(a − b)(a − c) ≥ 0.
a2 + 2bc
Since (c − a)(c − b) ≥ 0 and a − b ≥ 0, it suffices to prove that
‹

‹

a b + bc + ca
a b + bc + ca
(a − c) + 1 +
(c − b) ≥ 0.
1+
a2 + 2bc
b2 + 2ca
(b − c)2 +
Write this inequality as
‹
a−c
c−b
a − b + (a b + bc + ca) 2
+
≥ 0,
a + 2bc b2 + 2ca
(a b + bc + ca)(3ac + 3bc − a b − 2c 2
(a − b) 1 +
≥ 0.
(a2 + 2bc)(b2 + 2ca)

Since a − b ≥ 0 and 2ac + 3bc − 2c 2 > 0, it is enough to show that
1+
(a b + bc + ca)(ac − a b)
≥ 0.
(a2 + 2bc)(b2 + 2ca)
We have
1+
(a b + bc + ca)(ac − a b)
(a b + bc + ca)(ac − a b)
≥1+
(a2 + 2bc)(b2 + 2ca)
a2 (b2 + ca)
2
(a + b)c + (a2 − b2 )c
=
> 0.
a(b2 + ca)
42
Vasile Cîrtoaje
P 1.11. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
a2 (b + c) b2 (c + a) c 2 (a + b)
+ 2
+ 2
≥ a + b + c.
b2 + c 2
c + a2
a + b2
(Darij Grinberg, 2004)
First Solution. We use the SOS method. We have
X a2 (b + c) X
X a2 (b + c)
−
a=
−a
b2 + c 2
b2 + c 2
X a b(a − b) + ac(a − c)
=
b2 + c 2
X a b(a − b) X ba(b − a)
=
+
b2 + c 2
c 2 + a2
X a b(a + b)(a − b)2
=
≥ 0.
(b2 + c 2 )(c 2 + a2 )
The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).
Second Solution. By virtue of the Cauchy-Schwarz inequality, we have
P
X a2 (b + c)
[ a2 (b + c)]2
≥P
.
b2 + c 2
a2 (b + c)(b2 + c 2 )
Then, it suffices to show that
X
X
X
[
a2 (b + c)]2 ≥ (
a)[
a2 (b + c)(b2 + c 2 )].
Let p = a + b + c and q = a b + bc + ca. Since
X
[
a2 (b + c)]2 = (pq − 3a bc)2
= p2 q2 − 6a bc pq + 9a2 b2 c 2
and
X
X
(b + c)[(a2 b2 + b2 c 2 + c 2 a2 ) − b2 c 2 ]
X
= 2p(a2 b2 + b2 c 2 + c 2 a2 ) −
b2 c 2 (p − a)
a2 (b + c)(b2 + c 2 ) =
= p(a2 b2 + b2 c 2 + c 2 a2 ) + a bcq = p(q2 − 2a bc p) + a bcq,
the inequality can be written as
a bc(2p3 + 9a bc − 7pq) ≥ 0.
Symmetric Rational Inequalities
43
Using Schur’s inequality
p3 + 9a bc − 4pq ≥ 0,
we have
2p3 + 9a bc − 7pq ≥ p(p2 − 3q) ≥ 0.
P 1.12. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
3(a2 + b2 + c 2 )
a2 + b2 b2 + c 2 c 2 + a2
+
+
≤
.
a+b
b+c
c+a
a + b + c)
Solution. We use the SOS method.
First Solution. Multiplying by 2(a + b + c), the inequality successively becomes
X
a 2
1+
(b + c 2 ) ≤ 3(a2 + b2 + c 2 ),
b+c
X
X a
(b2 + c 2 ) ≤
a2 ,
b+c
X b2 + c 2
a a−
≥ 0,
b+c
X a b(a − b) − ac(c − a)
≥ 0,
b+c
X a b(a − b) X ba(a − b)
−
≥ 0,
b+c
c+a
X a b(a − b)2
≥ 0.
(b + c)(c + a)
The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).
Second Solution. Subtracting a + b + c from the both sides, the desired inequality
becomes as follows:
X a2 + b2 a + b 3(a2 + b2 + c 2 )
− (a + b + c) ≥
−
,
a+b+c
a+b
2
X (a − b)2
X (a − b)2
≥
,
a+b+c
2(a + b)
X (a + b − c)(a − b)2
≥ 0.
a+b
44
Vasile Cîrtoaje
Without loss of generality, assume that a ≥ b ≥ c. Since a + b − c ≥ 0, it suffices to
prove that
(a − b − c)(b − c)2
(a + c − b)(a − c)2
≥
.
a+c
b+c
a−c
b−c
This inequality is true because a + c − b ≥ a − b − c, a − c ≥ b − c and
≥
.
a+c
b+c
The last inequality reduces to c(a − b) ≥ 0.
Third Solution. Write the inequality as follows
X 3(a2 + b2 )
a2 + b2
−
≥ 0,
2(a + b + c)
a+b
X (a2 + b2 )(a + b − 2c)
≥ 0,
a+b
X (a2 + b2 )(a − c) X (a2 + b2 )(b − c)
+
≥ 0,
a+b
a+b
X (a2 + b2 )(a − c) X (b2 + c 2 )(c − a)
+
≥ 0,
a+b
b+c
X (a − c)2 (a b + bc + ca − b2 )
≥ 0.
(a + b)(b + c)
It suffices to prove that
X (a − c)2 (a b + bc − ca − b2 )
(a + b)(b + c)
≥ 0.
Since a b + bc − ca − b2 = (a − b)(b − c), this inequality is equivalent to
(a − b)(b − c)(c − a)
which is true because
X
X
c−a
≥ 0,
(a + b)(b + c)
c−a
= 0.
(a + b)(b + c)
P 1.13. Let a, b, c be positive real numbers. Prove that
a2
1
1
1
9
+ 2
+ 2
≥
.
2
2
2
+ ab + b
b + bc + c
c + ca + a
(a + b + c)2
(Vasile Cîrtoaje, 2000)
Symmetric Rational Inequalities
45
First Solution. Due to homogeneity, we may assume that a + b + c = 1. Let q = a b +
bc + ca. Since
b2 + bc + c 2 = (a + b + c)2 − a(a + b + c) − (a b + bc + ca) = 1 − a − q,
we can write the inequality as
X
1
≥ 9,
1−a−q
9q3 − 6q2 − 3q + 1 + 9a bc ≥ 0.
From Schur’s inequality
(a + b + c)3 + 9a bc ≥ 4(a + b + c)(a b + bc + ca),
we get 1 + 9a bc − 4q ≥ 0. Therefore,
9q3 − 6q2 − 3q + 1 + 9a bc = (1 + 9a bc − 4q) + q(3q − 1)2 ≥ 0.
The equality holds for a = b = c.
Second Solution. Multiplying by a2 + b2 +c 2 +a b+ bc +ca, the inequality can be written
as
X
a
9(a b + bc + ca)
(a + b + c)
+
≥ 6.
2
2
b + bc + c
(a + b + c)2
By the Cauchy-Schwarz inequality, we have
X
a
(a + b + c)2
a+b+c
P
≥
=
.
2
2
2
2
b + bc + c
a b + bc + ca
a(b + bc + c )
Then, it suffices to show that
(a + b + c)2
9(a b + bc + ca)
+
≥ 6.
a b + bc + ca
(a + b + c)2
This follows immediately from the AM-GM inequality.
P 1.14. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
a2
b2
c2
1
+
+
≤ .
(2a + b)(2a + c) (2b + c)(2b + a) (2c + a)(2c + b) 3
(Tigran Sloyan, 2005)
46
Vasile Cîrtoaje
First Solution. The inequality is equivalent to each of the inequalities
X
a
a2
−
≤ 0,
(2a + b)(2a + c) 3(a + b + c)
X a(a − b)(a − c)
≥ 0.
(2a + b)(2a + c)
Due to symmetry, we may consider that a ≥ b ≥ c. Since c(c − a)(c − b) ≥ 0, it suffices
to prove that
b(b − c)(b − a)
a(a − b)(a − c)
+
≥ 0.
(2a + b)(2a + c) (2b + c)(2b + a)
This is equivalent to the obvious inequality
(a − b)2 [(a + b)(2a b − c 2 ) + c(a2 + b2 + 5a b)] ≥ 0.
The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).
Second Solution (by Vo Quoc Ba Can). Apply the Cauchy-Schwarz inequality in the
following manner
9a2
(2a + a)2
2a
a2
=
≤
+
.
(2a + b)(2a + c) 2a(a + b + c) + (2a2 + bc)
a + b + c 2a2 + bc
Then,
X
X
a2
9a2
≤2+
,
(2a + b)(2a + c)
2a2 + bc
and from the known inequality
X
a2
≤ 1,
2a2 + bc
the conclusion follows. The last inequality is equivalent to
X
bc
≥ 1,
2a2 + bc
and can be obtained using the Cauchy-Schwarz inequality, as follows
P
X
( bc)2
bc
≥P
= 1.
2a2 + bc
bc(2a2 + bc)
Remark. From the inequality in P 1.14 and Hölder’s inequality
X
”X Æ
—2
a2
a(2a + b)(2a + c) ≥ (a + b + c)3 ,
(2a + b)(2a + c)
Symmetric Rational Inequalities
47
we get the following result:
• If a, b, c are nonnegative real numbers such that a + b + c = 3, then
Æ
Æ
Æ
a(2a + b)(2a + c) + b(2b + c)(2b + a) + c(2c + a)(2c + bc) ≥ 9,
3 3
with equality for a = b = c = 1, and for (a, b, c) = (0, , ) or any cyclic permutation.
2 2
P 1.15. Let a, b, c be positive real numbers. Prove that
(a)
(b)
P
P
a
1
≤
;
(2a + b)(2a + c)
a+b+c
1
a3
≤
.
(2a2 + b2 )(2a2 + c 2 )
a+b+c
(Vasile Cîrtoaje, 2005)
Solution. (a) Write the inequality as
˜
X•1
a(a + b + c)
−
≥ 0,
3 (2a + b)(2a + c)
X (a − b)(a − c)
≥ 0.
(2a + b)(2a + c)
Assume that a ≥ b ≥ c. Since (a − b)(a − c) ≥ 0, it suffices to prove that
(b − c)(b − a)
(a − c)(b − c)
+
≥ 0.
(2b + c)(2b + a) (2c + a)(2c + b)
Since b − c ≥ 0 and a − c ≥ a − b ≥ 0, it is enough to show that
1
1
≥
.
(2c + a)(2c + b) (2b + c)(2b + a)
This is equivalent to the obvious inequality
(b − c)(a + 4b + 4c) ≥ 0.
The equality holds for a = b = c.
(b) We obtain the desired inequality by summing the inequalities
a3
a
≥
,
2
2
2
2
(2a + b )(2a + c ) (a + b + c)2
48
Vasile Cîrtoaje
b
b3
≥
,
2
2
2
2
(2b + c )(2b + a ) (a + b + c)2
c
c3
≥
,
2
2
2
2
(2c + a )(2c + b ) (a + b + c)2
which are consequences of the Cauchy-Schwarz inequality. For example, from
(a2 + a2 + b2 )(c 2 + a2 + a2 ) ≥ (ac + a2 + ba)2 ,
the first inequality follows. The equality holds for a = b = c.
P 1.16. If a, b, c are positive real numbers, then
X
1
1
2
≥
+
.
(a + 2b)(a + 2c) (a + b + c)2 3(a b + bc + ca)
Solution. Write the inequality as follows
˜
X•
1
1
2
2
−
≥
−
,
2
(a + 2b)(a + 2c) (a + b + c)
3(a b + bc + ca) (a + b + c)2
X
(b − c)2
(b − c)2
≥
,
(a + 2b)(a + 2c)
3(a b + bc + ca)
X
b−c
(a − b)(b − c)(c − a)
≥ 0.
(a + 2b)(a + 2c)
X
Since
X
˜
X•
b−c
b−c
b−c
=
−
(a + 2b)(a + 2c)
(a + 2b)(a + 2c) 3(a b + bc + ca)
(a − b)(b − c)(c − a) X
1
=
,
3(a b + bc + ca)
(a + 2b)(a + 2c)
the desired inequality is equivalent to the obvious inequality
(a − b)2 (b − c)2 (c − a)2
X
1
≥ 0.
(a + 2b)(a + 2c)
The equality holds for a = b, or b = c, or c = a.
Symmetric Rational Inequalities
49
P 1.17. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
1
1
4
1
+
+
≥
;
2
2
2
(a − b)
(b − c)
(c − a)
a b + bc + ca
(a)
(b)
1
1
1
3
+
+
≥
;
a2 − a b + b2 b2 − bc + c 2 c 2 − ca + a2
a b + bc + ca
(c)
a2
1
1
1
5
+ 2
+ 2
≥
.
2
2
2
+b
b +c
c +a
2(a b + bc + ca)
Solution. Let
Ek (a, b, c) =
a b + bc + ca
a b + bc + ca
a b + bc + ca
+ 2
+ 2
.
2
2
2
a − ka b + b
b − k bc + c
c − kca + a2
We will prove that
Ek (a, b, c) ≥ αk ,
where

5 − 2k

, 0≤k≤1

2−k
αk =
.

1≤k≤2
 2 + k,
To show this, assume that a ≤ b ≤ c and prove that
Ek (a, b, c) ≥ Ek (0, b, c) ≥ αk .
For the nontrivial case a > 0, the left inequality is true because
Ek (a, b, c) − Ek (0, b, c)
=
a
b2 + (1 + k)bc − ac
c 2 + (1 + k)bc − a b
b+c
=
+
+
b(a2 − ka b + b2 )
b2 − k bc + c 2
c(c 2 − kca + a2 )
bc − ac
b+c
bc − a b
>
+ 2
+
> 0.
2
2
2
2
b(a − ka b + b ) b − k bc + c
c(c − kca + a2 )
In order to prove the right inequality Ek (0, b, c) ≥ αk , where
Ek (0, b, c) =
b2
bc
b c
+ + ,
2
− k bc + c
c
b
by virtue of the AM-GM inequality, we have
Ek (0, b, c) =
bc
b2 − k bc + c 2
+
+ k ≥ 2 + k.
b2 − k bc + c 2
bc
50
Vasile Cîrtoaje
Thus, it remains to consider the case 0 ≤ k ≤ 1. We have
Ek (0, b, c) =
b2 − k bc + c 2
bc
+
b2 − k bc + c 2
(2 − k)2 bc
•
˜
‹
1
b c
k
+ 1−
+
+
2
(2 − k)
c
b
(2 − k)2
•
˜
2
1
k
5 − 2k
≥
+2 1−
+
=
.
2−k
(2 − k)2
(2 − k)2
2−k
b
c
+
= 1 + k (or any cyclic
c
b
permutation). For 0 ≤ k ≤ 1, the equality holds when a = 0 and b = c (or any cyclic
permutation).
For 1 ≤ k ≤ 2, the equality holds when a = 0 and
P 1.18. Let a, b, c be positive real numbers, no two of which are zero. Prove that
(a2 + b2 )(a2 + c 2 ) (b2 + c 2 )(b2 + a2 ) (c 2 + a2 )(c 2 + b2 )
+
+
≥ a2 + b2 + c 2 .
(a + b)(a + c)
(b + c)(b + a)
(c + a)(c + b)
(Vasile Cîrtoaje, 2011)
Solution. Using the identity
(a2 + b2 )(a2 + c 2 ) = a2 (a2 + b2 + c 2 ) + b2 c 2 ,
we can write the inequality as follows
X
X
b2 c 2
a2
2
2
2
≥ (a + b + c ) 1 −
,
(a + b)(a + c)
(a + b)(a + c)
X
b2 c 2 (b + c) ≥ 2a bc(a2 + b2 + c 2 ),
X
X
a3 (b2 + c 2 ) ≥ 2
a3 bc,
X
a3 (b − c)2 ≥ 0.
Since the last form is obvious, the proof is completed. The equality holds for a = b = c.
P 1.19. Let a, b, c be positive real numbers such that a + b + c = 3. Prove that
a2
1
1
1
+ 2
+ 2
≤ 1.
+b+c b +c+a c +a+b
Symmetric Rational Inequalities
51
First Solution. By virtue of the Cauchy-Schwarz inequality, we have
(a2 + b + c)(1 + b + c) ≥ (a + b + c)2 .
Therefore,
X
X 1+ b+c
1
3 + 2(a + b + c)
≤
=
= 1.
2
2
a +b+c
(a + b + c)
(a + b + c)2
The equality occurs for a = b = c = 1.
Second Solution. Rewrite the inequality as
1
1
1
+
+
≤ 1.
a2 − a + 3 b2 − b + 3 c 2 − c + 3
We see that the equality holds for a = b = c = 1. Thus, if there exists a real number k
such that

‹
1
1
≤k+
−k a
a2 − a + 3
3
for all a ∈ [0, 3], then
X

‹ ˜

‹X
X•
1
1
1
≤
k+
− k a = 3k +
−k
a = 1.
a2 − a + 3
3
3
We have
‹
1
(a − 1)[(1 − 3k)a2 + 3ka + 3(1 − 3k)]
1
−k a− 2
=
.
k+
3
a −a+3
3(a2 − a + 3)

Setting k = 4/9, we get
k+

‹
1
(a − 1)2 (3 − a)
1
−k a− 2
=
≥ 0.
3
a −a+3
9(a2 − a + 3)
P 1.20. Let a, b, c be real numbers such that a + b + c = 3. Prove that
a2 − bc b2 − ca c 2 − a b
+ 2
+ 2
≥ 0.
a2 + 3
b +3
c +3
(Vasile Cîrtoaje, 2005)
52
Vasile Cîrtoaje
Solution. Apply the SOS method. We have
2
X a2 − bc
a2 + 3
=
X (a − b)(a + c) + (a − c)(a + b)
a2 + 3
X (a − b)(a + c) X (b − a)(b + c)
=
+
a2 +
3
b2 + 3

‹
X
a+c
b+c
− 2
=
(a − b) 2
a +3 b +3
X
(a − b)2
= (3 − a b − bc − ca)
≥ 0.
(a2 + 3)(b2 + 3)
Thus, it suffices to show that 3 − a b − bc − ca ≥ 0. This follows immediately from the
known inequality (a + b + c)2 ≥ 3(a b + bc + ca), which is equivalent to (a − b)2 + (b −
c)2 + (c − a)2 ≥ 0. The equality holds for a = b = c = 1.
P 1.21. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that
1 − bc 1 − ca 1 − a b
+
+
≥ 0.
5 + 2a 5 + 2b 5 + 2c
Solution. Since
9(1 − bc) = (a + b + c)2 − 9bc,
we can write the inequality as
X a2 + b2 + c 2 + 2a(b + c) − 7bc
5 + 2a
≥ 0.
From
(a − b)(a + k b + mc) + (a − c)(a + kc + mb) =
= 2a2 − k(b2 + c 2 ) + (k + m − 1)a(b + c) − 2mbc,
choosing k = −2 and m = 7, we get
(a − b)(a − 2b + 7c) + (a − c)(a − 2c + 7b) = 2[a2 + b2 + c 2 + 2a(b + c) − 7bc].
Therefore, the desired inequality becomes as follows:
X (a − b)(a − 2b + 7c)
5 + 2a
X (a − b)(a − 2b + 7c)
5 + 2a
+
+
X (a − c)(a − 2c + 7b)
5 + 2a
X (b − a)(b − 2a + 7c)
5 + 2b
≥ 0,
≥ 0,
Symmetric Rational Inequalities
53
X
(a − b)(5 + 2c)[(5 + 2b)(a − 2b + 7c) − (5 + 2a)(b − 2a + 7c)] ≥ 0,
X
(a − b)2 (5 + 2c)(15 + 4a + 4b − 14c) ≥ 0,
X
(a − b)2 (5 + 2c)(a + b − c) ≥ 0.
Without loss of generality, assume that a ≥ b ≥ c. Clearly, it suffices to show that
(a − c)2 (5 + 2b)(a + c − b) ≥ (b − c)2 (5 + 2a)(a − b − c).
Since a − c ≥ b − c ≥ 0 and a + c − b ≥ a − b − c, it suffices to prove that
(a − c)(5 + 2b) ≥ (b − c)(5 + 2a).
Indeed,
(a − c)(5 + 2b) − (b − c)(5 + 2a) = (a − b)(5 + 2c) ≥ 0.
The equality holds for a = b = c = 1, and for c = 0 and a = b = 3/2 (or any cyclic
permutation).
P 1.22. Let a, b, c be positive real numbers such that a + b + c = 3. Prove that
a2
1
1
1
3
+ 2
+ 2
≤ .
2
2
2
+ b +2 b +c +2 c +a +2 4
(Vasile Cîrtoaje, 2006)
Solution. Since
1
a2 + b2
1
=
−
,
a2 + b2 + 2 2 a2 + b2 + 2
we write the inequality as
a2 + b2
b2 + c 2
c 2 + a2
3
+
+
≥ .
2
2
2
2
2
2
a + b +2 b +c +2 c +a +2 2
By the Cauchy-Schwarz inequality, we have
Pp
P
Pp
X a2 + b2
(
a2 + b2 )2
2 a2 + 2
(a2 + b2 )(a2 + c 2 )
P
P
≥
=
a2 + b2 + 2
(a2 + b2 + 2)
2 a2 + 6
P 2
P 2
P 2
2 a + 2 (a + bc) 3 a + 9 3
P
≥
= P
= .
2 a2 + 6
2 a2 + 6 2
The equality holds for a = b = c = 1
54
Vasile Cîrtoaje
P 1.23. Let a, b, c be positive real numbers such that a + b + c = 3. Prove that
4a2
1
1
1
1
+ 2
+ 2
≤ .
2
2
2
2
2
2
+b +c
4b + c + a
4c + a + b
2
(Vasile Cîrtoaje, 2007)
Solution. According to the Cauchy-Schwarz inequality, we have
9
(a + b + c)2
=
4a2 + b2 + c 2
2a2 + (a2 + b2 ) + (a2 + c 2 )
1
b2
c2
≤ + 2
+
.
2 a + b2 a2 + c 2
Therefore,
X
b2
9
3 X
c2
≤ +
+
4a2 + b2 + c 2
2
a2 + b2 a2 + c 2
b2
3 X
a2
3
9
= +
+
= +3= .
2
a2 + b2 b2 + a2
2
2
The equality holds for a = b = c = 1.
P 1.24. Let a, b, c be nonnegative real numbers such that a + b + c = 2. Prove that
ca
ab
bc
+ 2
+ 2
≤ 1.
+1 b +1 c +1
a2
(Pham Kim Hung, 2005)
Solution. Let p = a + b + c and q = a b + bc + ca, q ≤ p2 /3 = 4/3. If one of a, b, c is
zero, then the inequality is true. Otherwise, write the inequality as
1
1
≤
,
+ 1)
a bc
‹
X1
a
1
− 2
≤
,
a a +1
a bc
X a
1 1 1
1
≥ + + −
,
2
a +1
a b c a bc
X a
q−1
≥
,
2
a +1
a bc
X
a(a2
Symmetric Rational Inequalities
55
Using the inequality
a2
2
≥ 2 − a,
+1
which is equivalent to a(a − 1)2 ≥ 0, we get
X
X a(2 − a) X a(b + c)
a
≥
=
= q.
a2 + 1
2
2
Therefore, it suffices to prove that
1 + a bcq ≥ q.
Since
a4 + b4 + c 4 = (a2 + b2 + c 2 )2 − 2(a2 b2 + b2 c 2 + c 2 a2 )
= (p2 − 2q)2 − 2(q2 − 2a bc p) = p4 − 4p2 q + 2q2 + 4a bc p
by Schur’s inequality of degree four
a4 + b4 + c 4 + 2a bc(a + b + c) ≥ (a b + bc + ca)(a2 + b2 + c 2 ),
we get
a bc ≥
(p2 − q)(4q − p2 )
,
6p
a bc ≥
(4 − q)(q − 1)
.
3
Thus
1 + a bcq − q ≥ 1 +
q(4 − q)(q − 1)
(3 − q)(q − 1)2
−q =
≥ 0.
3
3
The equality holds if a = 0 and b = c = 1 (or any cyclic permutation).
P 1.25. Let a, b, c be nonnegative real numbers such that a + b + c = 1. Prove that
bc
ca
ab
1
+
+
≤ .
a+1 b+1 c+1 4
(Vasile Cîrtoaje, 2009)
56
Vasile Cîrtoaje
First Solution. We have
X
X bc
bc
=
a+1
(a + b) + (c + a)

‹
1X
1
1
≤
bc
+
4
a+b c+a
1 X bc
1 X bc
+
=
4
a+b 4
c+a
1 X bc
1 X ca
=
+
4
a+b 4
a+b
1X
1 X bc + ca
1
=
=
c= .
4
a+b
4
4
The equality holds for a = b = c = 1/3, and for a = 0 and b = c = 1/2 (or any cyclic
permutation).
Second Solution. It is easy to check that the inequality is true if one of a, b, c is zero.
Otherwise, write the inequality as
1
1
1
1
+
+
≤
.
a(a + 1) b(b + 1) c(c + 1) 4a bc
Since
1
1
1
= −
,
a(a + 1)
a a+1
we can write the required inequality as
1
1
1
1 1 1
1
+
+
≥ + + −
.
a+1 b+1 c+1
a b c 4a bc
In virtue of the Cauchy-Schwarz inequality,
1
1
1
9
9
+
+
≥
= .
a + 1 b + 1 c + 1 (a + 1) + (b + 1) + (c + 1) 4
Therefore, it suffices to prove that
9 1 1 1
1
≥ + + −
.
4
a b c 4a bc
This is equivalent to Schur’s inequality
(a + b + c)3 + 9a bc ≥ 4(a + b + c)(a b + bc + ca).
Symmetric Rational Inequalities
57
P 1.26. Let a, b, c be positive real numbers such that a + b + c = 1. Prove that
1
1
3
1
+
+
≤
.
2
2
2
a(2a + 1) b(2b + 1) c(2c + 1) 11a bc
(Vasile Cîrtoaje, 2009)
Solution. Since
1
2a
1
= − 2
,
2
a(2a + 1)
a 2a + 1
we can write the inequality as
X 2a
1 1 1
3
≥ + + −
.
2a2 + 1
a b c 11a bc
By the Cauchy-Schwarz inequality, we have
P
X 2a
2( a)2
2
≥P
=
.
2
3
3
2
2a + 1
a(2a + 1) 2(a + b + c 3 ) + 1
Therefore, it suffices to show that
2(a3
where q = a b + bc + ca, q ≤
11q − 3
2
≥
,
3
+ +c )+1
11a bc
b3
1
1
(a + b + c)2 = . Since
3
3
a3 + b3 + c 3 = 3a bc + (a + b + c)3 − 3(a + b + c)(a b + bc + ca) = 3a bc + 1 − 3q,
we need to prove that
22a bc ≥ (11q − 3)(6a bc + 3 − 6q),
or, equivalently,
2(20 − 33q)a bc ≥ 3(11q − 3)(1 − 2q).
From Schur’s inequality
(a + b + c)3 + 9a bc ≥ 4(a + b + c)(a b + bc + ca),
we get
9a bc ≥ 4q − 1.
Thus,
2(20 − 33q)a bc − 3(11q − 3)(1 − 2q) ≥
2(20 − 33q)(4q − 1)
≥
− 3(11q − 3)(1 − 2q)
9
330q2 − 233q + 41 (1 − 3q)(41 − 110q)
=
=
≥ 0.
9
9
This completes the proof. The equality holds for a = b = c = 1/3.
58
Vasile Cîrtoaje
P 1.27. Let a, b, c be positive real numbers such that a + b + c = 3. Prove that
a3
1
1
1
+ 3
+ 3
≤ 1.
+b+c b +c+a c +a+b
(Vasile Cîrtoaje, 2009)
Solution. Write the inequality in the form
1
1
1
+
+
≤ 1.
a3 − a + 3 b3 − b + 3 c 3 − c + 3
Assume that a ≥ b ≥ c. There are two cases to consider.
Case 1: c ≤ b ≤ a ≤ 2. The desired inequality follows by adding the inequalities
a3
1
5 − 2a
1
5 − 2b
1
5 − 2c
≤
, 3
≤
, 3
≤
.
−a+3
9
b − b+3
9
c −c+3
9
Indeed, we have
1
5 − 2a
(a − 1)2 (a − 2)(2a + 3)
−
=
≤ 0.
a3 − a + 3
9
9(a3 − a + 3)
Case 2: a > 2. From a + b + c = 3, we get b + c < 1. Since
X
a3
1
1
1
1
1
1
1
< 3
+
+
< +
+
,
−a+3
a −a+3 3− b 3−c
9 3− b 3−c
it suffices to prove that
1
1
8
+
≤ .
3− b 3−c
9
We have
1
1
8 −3 − 15(1 − b − c) − 8bc
+
− =
< 0.
3− b 3−c 9
9(3 − b)(3 − c)
The equality holds for a = b = c = 1.
P 1.28. Let a, b, c be positive real numbers such that a + b + c = 3. Prove that
a2
b2
c2
+
+
≥ 1.
1 + b3 + c 3 1 + c 3 + a3 1 + a3 + b3
Symmetric Rational Inequalities
59
Solution. Using the Cauchy-Schwarz inequality, we have
P
X
( a2 )2
a2
≥P
,
1 + b3 + c 3
a2 (1 + b3 + c 3 )
and it remains to show that
(a2 + b2 + c 2 )2 ≥ (a2 + b2 + c 2 ) +
X
a2 b2 (a + b).
Let p = a + b + c and q = a b + bc + ca, q ≤ 3. Since a2 + b2 + c 2 = 9 − 2q and
X
X
X
a2 b2 (a + b) =
a2 b2 (3 − c) = 3
a2 b2 − qa bc = 3q2 − (q + 18)a bc,
the desired inequality can be written as
q2 − 34q + 72 + (q + 18)a bc ≥ 0.
This inequality is clearly true for q ≤ 2. Consider further that 2 < q ≤ 3. Since
a4 + b4 + c 4 = (a2 + b2 + c 2 )2 − 2(a2 b2 + b2 c 2 + c 2 a2 )
= (p2 − 2q)2 − 2(q2 − 2a bc p) = p4 − 4p2 q + 2q2 + 4a bc p
by Schur’s inequality of degree four
a4 + b4 + c 4 + 2a bc(a + b + c) ≥ (a b + bc + ca)(a2 + b2 + c 2 ),
we get
a bc ≥
(p2 − q)(4q − p2 ) (9 − q)(4q − 9)
=
.
6p
18
Therefore
(q + 18)(9 − q)(4q − 9)
18
(3 − q)(4q2 + 21q − 54)
=
≥ 0.
18
q2 − 34q + 72 + (q + 18)a bc ≥ q2 − 34q + 72 +
The equality holds for a = b = c = 1.
P 1.29. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that
1
1
1
3
+
+
≤ .
6 − a b 6 − bc 6 − ca
5
60
Vasile Cîrtoaje
Solution. Rewrite the inequality as
108 − 48(a b + bc + ca) + 13a bc(a + b + c) − 3a2 b2 c 2 ≥ 0,
4[9 − 4(a b + bc + ca) + 3a bc] + a bc(1 − a bc) ≥ 0.
By the AM-GM inequality,
a+b+c
3
Consequently, it suffices to show that
1=

‹3
≥ a bc.
9 − 4(a b + bc + ca) + 3a bc ≥ 0.
We see that the homogeneous form of this inequality is just Schur’s inequality of third
degree
(a + b + c)3 + 9a bc ≥ 4(a + b + c)(a b + bc + ca).
The equality holds for a = b = c = 1, as well as for a = 0 and b = c = 3/2 (or any cyclic
permutation).
P 1.30. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that
1
1
1
1
+ 2
+ 2
≤ .
+ 7 2b + 7 2c + 7 3
2a2
(Vasile Cîrtoaje, 2005)
Solution. Assume that a = max{a, b, c} and prove that
E(a, b, c) ≤ E(a, s, s) ≤
where
s=
and
E(a, b, c) =
1
,
3
b+c
, 0 ≤ s ≤ 1,
2
1
1
1
+
+
.
2a2 + 7 2b2 + 7 2c 2 + 7
We have
‹ 
‹
1
1
1
1
−
+
−
2s2 + 7 2b2 + 7
2s2 + 7 2c 2 + 7
•
˜
1
(b − c)(b + s) (c − b)(c + s)
= 2
+
2s + 7
2b2 + 7
2c 2 + 7
2
2
(b − c) (7 − 4s − 2bc)
=
.
(2s2 + 7)(2b2 + 7)(2c 2 + 7)
E(a, s, s) − E(a, b, c) =

Symmetric Rational Inequalities
61
Since bc ≤ s2 ≤ 1, it follows that 7 − 4s2 − 2bc > 0, and hence E(a, s, s) ≥ E(a, b, c).
Also,
1
1
4(s − 1)2 (2s − 1)2
− E(a, s, s) = − E(3 − 2s, s, s) =
≥ 0.
3
3
3(2a2 + 7)(2s2 + 7)
The equality holds for a = b = c = 1, as well as for a = 2 and b = c = 1/2 (or any cyclic
permutation).
P 1.31. Let a, b, c be nonnegative real numbers such that a ≥ b ≥ 1 ≥ c and a + b + c = 3.
Prove that
1
1
3
1
+
+
≤ .
a2 + 3 b2 + 3 c 2 + 3 4
(Vasile Cîrtoaje, 2005)
First Solution (by Nguyen Van Quy). Write the inequality as follows:

‹ 
‹ 
‹
1
3−a
1
3− b
3−c
1
−
+
−
≤
− 2
,
a2 + 3
8
b2 + 3
8
8
c +3
(a − 1)3 (b − 1)3
(1 − c)3
+
≤
.
a2 + 3
b2 + 3
c2 + 3
Indeed, we have
(1 − c)3
(a − 1 + b − 1)3
(a − 1)3 + (b − 1)3
(a − 1)3 (b − 1)3
=
≥
≥
+ 2
.
c2 + 3
c2 + 3
c2 + 3
a2 + 3
b +3
The proof is completed. The equality holds for a = b = c = 1.
Second Solution. Let d be a positive number such that
c + d = 2.
We have
a + b = 1 + d,
In addition, we claim that
c2
Indeed,
d ≥ a ≥ b ≥ 1.
1
1
1
+ 2
≤ .
+3 d +3 2
1
1
1
(cd − 1)2
− 2
− 2
=
≥ 0.
2 c + 3 d + 3 2(c 2 + 3)(d 2 + 3)
Thus, it suffices to show that
1
1
1
1
+
≤ 2
+ .
a2 + 3 b2 + 3
d +3 4
62
Vasile Cîrtoaje
Since
a2
1
1
(d − a)(d + a)
− 2
= 2
,
+ 3 d + 3 (a + 3)(d 2 + 3)
we need to prove that
(a2
1
1
(b − 1)(b + 1)
− 2
=
,
4 b +3
4(b2 + 3)
d+a
b+1
≤
.
2
+ 3)(d + 3) 4(b2 + 3)
We can get this inequality by multiplying the inequalities
a+1
d+a
≤
,
2
d +3
4
a+1
b+1
≤ 2
.
2
a +3
b +3
We have
a+1
d+a
(d − 1)(ad + a + d − 3)
− 2
=
≥ 0,
4
d +3
4(d 2 + 3)
b+1
a+1
(a − b)(a b + a + b − 3)
− 2
=
≥ 0.
2
b +3 a +3
(a2 + 3)(b2 + 3)
P 1.32. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that
1
1
1
3
+ 2
+ 2
≥ .
+ 3 2b + 3 2c + 3 5
2a2
(Vasile Cîrtoaje, 2005)
First Solution (by Nguyen Van Quy). Write the inequality as
‹
X1
1
2
− 2
≤ ,
3 2a + 3
5
a2
3
≤ .
2
2a + 5 5
Using the Cauchy-Schwarz inequality gives
X
25
25
=
3(2a2 + 3) 6a2 + (a + b + c)2
(2 + 2 + 1)2
=
2(2a2 + bc) + 2a(a + b + c) + a2 + b2 + c 2
22
22
1
≤
+
+ 2
,
2
2(2a + bc) 2a(a + b + c) a + b2 + c 2
Symmetric Rational Inequalities
63
hence
X
X 2a2
X
X
25a2
2a
a2
≤
+
+
3(2a2 + 3)
2a2 + bc
a+b+c
a2 + b2 + c 2
X 2a2
+ 3.
=
2a2 + bc
Therefore, it suffices to show that
X
a2
≤ 1,
2a2 + bc
which is equivalent to
a2
1
− 2
≥ ,
2 2a + bc
2
X
bc
≥ 1.
2
2a + bc
Using again the Cauchy-Schwarz inequality, we get
P
X
( bc)2
bc
= 1.
≥P
2a2 + bc
bc(2a2 + bc)
X1
The equality holds for a = b = c = 1, as well as for a = 0 and b = c = 3/2 (or any cyclic
permutation).
Second Solution. First, we can check that the desired inequality becomes an equality
for a = b = c = 1, and for a = 0 and b = c = 3/2. Consider then the inequality
f (x) ≥ 0, where
1
f (x) =
− A − B x.
2
2x + 3
We have
−4x
− B.
f 0 (x) =
(2x 2 + 3)2
From the conditions f (1) = 0 and f 0 (1) = 0, we get A = 9/25 and B = −4/25. Also,
from the conditions f (3/2) = 0 and f 0 (3/2) = 0, we get A = 22/75 and B = −8/75.
From these values of A and B, we obtain the identities
1
9 − 4x
2(x − 1)2 (4x − 1)
−
=
,
2x 2 + 3
25
25(2x 2 + 3)
1
22 − 8x
(2x − 3)2 (4x + 1)
−
=
,
2x 2 + 3
75
75(2x 2 + 3)
and the inequalities
1
9 − 4x
≥
,
2x 2 + 3
25
x≥
1
,
4
64
Vasile Cîrtoaje
22 − 8x
1
≥
,
2x 2 + 3
75
x ≥ 0.
Without loss of generality, assume that a ≥ b ≥ c.
1
. By summing the inequalities
4
Case 1: a ≥ b ≥ c ≥
9 − 4a
1
≥
,
2a2 + 3
25
we get
1
9 − 4b
≥
,
2b2 + 3
25
1
9 − 4c
≥
,
2c 2 + 3
25
1
1
27 − 4(a + b + c) 3
1
+ 2
+ 2
≥
= .
+ 3 2b + 3 2c + 3
25
5
2a2
Case 2: a ≥ b ≥
X
1
≥ c. We have
4
1
22 − 8a 22 − 8b
1
≥
+
+ 2
+3
75
75
2c + 3
44 − 8(a + b)
1
20 + 8c
1
=
+ 2
=
+ 2
.
75
2c + 3
75
2c + 3
2a2
Therefore, it suffices to show that
1
3
20 + 8c
+ 2
≥ ,
75
2c + 3 5
which is equivalent to the obvious inequality
c(8c 2 − 25c + 12) ≥ 0.
Case 3: a ≥
1
≥ b ≥ c. We have
4
X
1
2a2
+3
>
1
1
+ 2
≥
+ 3 2c + 3
2b2
2
1
8
+3
>
3
.
5
P 1.33. Let a, b, c be nonnegative real numbers such that a ≥ 1 ≥ b ≥ c and a + b + c = 3.
Prove that
1
1
1
+ 2
+ 2
≥ 1.
2
a +2 b +2 c +2
(Vasile Cîrtoaje, 2005)
Symmetric Rational Inequalities
65
First Solution. Let b1 and c1 be positive numbers such that
b + b1 = 2,
c + c1 = 2.
We have
b1 + c1 = 1 + a, 1 ≤ b1 ≤ c1 ≤ a.
In addition, we claim that
b2
1
1
2
+ 2
≥ ,
+ 2 b1 + 2 3
c2
1
1
2
+ 2
≥ .
+ 2 c1 + 2 3
Indeed,
2b b1 (1 − b b1 )
b b1 (b − b1 )2
1
1
2
+
=
−
=
≥ 0,
b2 + 2 b12 + 2 3 3(b2 + 2)(b12 + 2) 6(b2 + 2)(b12 + 2)
cc1 (c − c1 )2
1
2
1
+
−
=
≥ 0.
c 2 + 2 c12 + 2 3 6(c 2 + 2)(c12 + 2)
Using these inequalities, it suffices to show that
a2
1
1
1
1
+ ≥ 2
+ 2
.
+2 3
b1 + 2 c1 + 2
Since
(b1 − 1)(b1 + 1)
1
1
=
,
− 2
3 b1 + 2
3(b12 + 2)
we need to prove that
b1 + 1
3(b12
+ 2)
≥
1
c12
+2
−
a2
(a − c1 )(a + c1 )
1
,
=
+ 2 (a2 + 2)(c12 + 2)
a + c1
(a2
+ 2)(c12 + 2)
.
We can get this inequality by multiplying the inequalities
b1 + 1
b12
+2
≥
c1 + 1
c12 + 2
,
c1 + 1
a + c1
≥ 2
.
3
a +2
We have
b1 + 1
b12
+2
−
c1 + 1
c12
+2
=
(c1 − b1 )(b1 c1 + b1 + c1 − 2)
(b12 + 2)(c12 + 2)
≥ 0,
c1 + 1 a + c1
(a − 1)(ac1 + a + c1 − 2)
− 2
=
≥ 0.
3
a +2
3(a2 + 2)
66
Vasile Cîrtoaje
The proof is completed. The equality holds for a = b = c = 1, as well as for a = 2, b = 1
and c = 0.
Second Solution. First, we can check that the desired inequality becomes an equality
for a = b = c = 1, and also for a = 2, b = 1, c = 0. Consider then the inequality
f (x) ≥ 0, where
1
f (x) = 2
− A − B x.
x +2
We have
−2x
− B.
f 0 (x) = 2
(x + 2)2
From the conditions f (1) = 0 and f 0 (1) = 0, we get A = 5/9 and B = −2/9. Also, from
the conditions f 2) = 0 and f 0 (2) = 0, we get A = 7/18 and B = −1/9. From these
values of A and B, we obtain the identities
1
5 − 2x
(x − 1)2 (2x − 1)
−
=
,
x2 + 2
9
9(x 2 + 2)
7 − 2x
(x − 2)2 (2x + 1)
1
−
=
,
x2 + 2
18
18(x 2 + 2)
and the inequalities
1
5 − 2x
≥
,
x2 + 2
9
1
7 − 2x
≥
,
2
x +2
18
Let us define
g(x) =
x≥
1
,
2
x ≥ 0.
1
.
x2 + 2
p
Notice that for d ∈ (0, 2] and x ∈ [0, d], we have
g(x) ≥ g(0) +
because
g(x) − g(0) −
g(d) − g(0)
x,
d
g(d) − g(0)
x(d − x)(2 − d x)
x=
≥ 0.
d
2(d 2 + 2)(x 2 + 2)
For d = 1/2 and d = 1, we get the inequalities
x2
1
9 − 2x
≥
,
+2
18
0≤ x ≤
1
,
2
1
3− x
≥
, 0 ≤ x ≤ 1.
x2 + 2
6
Consider further two cases: c ≥ 1/2 and c ≤ 1/2.
Symmetric Rational Inequalities
Case 1: c ≥
67
1
. By summing the inequalities
2
1
5 − 2a
≥
,
a2 + 2
9
1
5 − 2b
≥
,
b2 + 2
9
1
5 − 2c
≥
,
c2 + 2
9
we get
a2
Case 2: c ≤
1
1
15 − 2(a + b + c)
1
+ 2
+ 2
≥
= 1.
+2 b +2 c +2
9
1
. We have
2
a2
1
7 − 2a
≥
,
+2
18
1
3− b
8 − 2b
≥
≥
,
b2 + 2
6
18
9 − 2c
1
≥
.
c2 + 2
18
Therefore,
a2
1
1
1
7 − 2a 8 − 2b 9 − 2c
+ 2
+ 2
≥
+
+
= 1.
+2 b +2 c +2
18
18
18
P 1.34. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. Prove that
1
1
a+b+c
3
1
+
+
≥
+
.
a+b b+c c+a
6
a+b+c
(Vasile Cîrtoaje, 2007)
First Solution. Denoting x = a + b + c, we have
1
1
1
(a + b + c)2 + a b + bc + ca
x2 + 3
+
+
=
=
.
a+b b+c c+a
(a + b + c)(a b + bc + ca) − a bc
3x − a bc
Then, the inequality becomes
x2 + 3
x 3
≥ + ,
3x − a bc
6 x
or
3(x 3 + 9a bc − 12x) + a bc(x 2 − 9) ≥ 0.
68
Vasile Cîrtoaje
This inequality is true since from (a + b + c)2 ≥ 3(a b + bc + ca), we get x 2 − 9 ≥ 0, and
from Schur’s inequality of degree three
(a + b + c)3 + 9a bc ≥ 4(a + b + c)(a b + bc + ca),
3
we get xp
+ 9a bc − 12 ≥ 0. The equality holds for a = b = c = 1, and for a = 0 and
b = c = 3 (or any cyclic permutation).
Second Solution. We apply the SOS method. Write the inequality as follows:
1
1
1
a+b+c
3
+
+
≥
+
,
a+b b+c c+a
2(a b + bc + ca) a + b + c
‹
1
1
(a + b + c)2
1
+
+
≥
+ 6,
a+b b+c c+a
a b + bc + ca

‹
1
1
1
(a + b + c)2
[(a + b) + (b + c) + (c + a)]
+
+
−9≥
− 3,
a+b b+c c+a
a b + bc + ca
X
X
1
(b − c)2
≥
(b − c)2 ,
(a + b)(c + a) 2(a b + bc + ca)
X a b + bc + ca − a2
(b − c)2 ≥ 0.
(a + b)(c + a)
2(a + b + c)

Without loss of generality, assume that a ≥ b ≥ c. Since a b + bc + ca − c 2 ≥ 0, it suffices
to show that
a b + bc + ca − b2
a b + bc + ca − a2
(b − c)2 +
(c − a)2 ≥ 0.
(a + b)(c + a)
(b + c)(a + b)
Since a b + bc + ca − b2 ≥ b(a − b) ≥ 0 and (c − a)2 ≥ (b − c)2 , it is enough to prove that
a b + bc + ca − a2
a b + bc + ca − b2
(b − c)2 +
(b − c)2 ≥ 0.
(a + b)(c + a)
(b + c)(a + b)
This is true if
a b + bc + ca − a2 a b + bc + ca − b2
+
≥ 0, .
(a + b)(c + a)
(b + c)(a + b)
which is equivalent to
3 − a2 3 − b2
+
≥ 0,
3 + a2 3 + b2
Indeed,
2(9 − a2 b2 )
2c(a + b)(3 + a b)
3 − a2 3 − b2
+
=
=
≥ 0.
2
2
2
2
3+a
3+ b
(3 + a )(3 + b )
(3 + a2 )(3 + b2 )
Symmetric Rational Inequalities
69
P 1.35. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. Prove that
a2
1
1
1
3
+ 2
+ 2
≥ .
+1 b +1 c +1 2
(Vasile Cîrtoaje, 2005)
First Solution. After expanding, the inequality can be restated as
a2 + b2 + c 2 + 3 ≥ a2 b2 + b2 c 2 + c 2 a2 + 3a2 b2 c 2 .
From
(a + b + c)(a b + bc + ca) − 9a bc = a(b − c)2 + b(c − a)2 + c(a − b)2 ≥ 0,
we get
a + b + c ≥ 3a bc.
So, it suffices to show that
a2 + b2 + c 2 + 3 ≥ a2 b2 + b2 c 2 + c 2 a2 + a bc(a + b + c).
This is equivalent to the homogeneous inequalities
(a b + bc + ca)(a2 + b2 + c 2 ) + (a b + bc + ca)2 ≥ 3(a2 b2 + b2 c 2 + c 2 a2 ) + 3a bc(a + b + c),
a b(a2 + b2 ) + bc(b2 + c 2 ) + ca(c 2 + a2 ) ≥ 2(a2 b2 + b2 c 2 + c 2 a2 ),
a b(a − b)2 + bc(b − c)2 + ca(c − a)2 ≥ 0.
The equality holds for a = b = c = 1, and for a = 0 and b = c =
permutation).
p
3 (or any cyclic
Second Solution. Without loss of generality, assume that a = min{a, b, c}. From a b +
bc + ca = 3, we get bc ≥ 1. Also, from
(a + b + c)(a b + bc + ca) − 9a bc = a(b − c)2 + b(c − a)2 + c(a − b)2 ≥ 0,
we get
a + b + c ≥ 3a bc.
The desired inequality follows by summing the inequalities
b2
1
1
2
+ 2
≥
,
+1 c +1
bc + 1
a2
1
2
3
+
≥ .
+ 1 bc + 1 2
70
Vasile Cîrtoaje
We have
b2
1
2
b(c − b)
c(b − c)
1
+ 2
−
= 2
+ 2
+ 1 c + 1 bc + 1 (b + 1)(bc + 1) (c + 1)(bc + 1)
(b − c)2 (bc − 1)
= 2
≥0
(b + 1)(c 2 + 1)(bc + 1)
and
2
3
a2 − bc + 3 − 3a2 bc
a(a + b + c − 3a bc)
1
+
−
=
=
≥ 0.
2
2
a + 1 bc + 1 2
2(a + 1)(bc + 1)
2(a2 + 1)(bc + 1)
Third Solution. Since
1
a2
1
b2
1
c2
=
1
−
,
=
1
−
,
=
1
−
,
a2 + 1
a2 + 1 b2 + 1
b2 + 1 c 2 + 1
c2 + 1
we can rewrite the inequality as
a2
b2
c2
3
+
+
≤ ,
2
2
2
a +1 b +1 c +1 2
or, in the homogeneous form
X
a2
1
≤ .
2
3a + a b + bc + ca
2
According to the Cauchy-Schwarz inequality, we have
4a2
(a + a)2
a
a2
=
≤
+
.
3a2 + a b + bc + ca
a(a + b + c) + (2a2 + bc)
a + b + c 2a2 + bc
Then,
X
X
4a2
a2
≤
1
+
,
3a2 + a b + bc + ca
2a2 + bc
and it suffices to show that
X
a2
≤ 1,
2a2 + bc
or, equivalently,
X
bc
≥ 1.
+ bc
2a2
This follows from the Cauchy-Schwarz inequality as follows:
P
P 2 2
P
X
b c + 2a bc a
( bc)2
bc
P
P
≥P
=
= 1.
2a2 + bc
bc(2a2 + bc) 2a bc a + b2 c 2
Symmetric Rational Inequalities
71
Remark. We can write the inequality in P 1.35 in the homogeneous form
1
1+
2
3a
a b + bc + ca
Substituting a, b, c by
1
+
1+
2
3b
a b + bc + ca
1
+
1+
2
3c
a b + bc + ca
≥
3
.
2
1 1 1
, , , respectively, we get
x y z
y
x
z
3
+
+
≥ .
3 yz
3z x
3x y
2
y+
x+
z+
x + y +z
x + y +z
x + y +z
So, we find the following result.
• If x, y, z are positive real numbers such that x + y + z = 3, then
y
x
z
3
+
+
≥ .
x + yz
y + zx z + x y
2
P 1.36. Let a, b, c be positive real numbers such that a b + bc + ca = 3. Prove that
b2
c2
a2
+
+
≥ 1.
a2 + b + c b2 + c + a c 2 + a + b
(Vasile Cîrtoaje, 2005)
Solution. We apply the Cauchy-Schwarz inequality in the following way
2
P 3
P
X
a3/2 + b3/2 + c 3/2
a + 2 (a b)3/2
a2
P
≥ P
=
.
a2 + b + c
a(a2 + b + c)
a3 + 6
Then, we still have to show that
(a b)3/2 + (bc)3/2 + (ca)3/2 ≥ 3.
By the AM-GM inequality,
(a b)3/2 =
and hence
(a b)3/2 + (a b)3/2 + 1 1 3a b 1
− ≥
− ,
2
2
2
2
X
3X
3
(a b)3/2 ≥
a b − = 3.
2
2
The equality holds for a = b = c = 1.
72
Vasile Cîrtoaje
P 1.37. Let a, b, c be positive real numbers such that a b + bc + ca = 3. Prove that
bc + 2 ca + 2 a b + 2
bc + 4 ca + 4 a b + 4
+ 2
+ 2
≤3≤ 2
+
+
.
2
a +4 b +4 c +4
a + 2 b2 + 2 c 2 + 2
(Vasile Cîrtoaje, 2007)
Solution. More general, using the SOS method, we will show that
‹

bc + k ca + k a b + k
+
+ 2
−3 ≤0
(k − 3) 2
a + k b2 + k
c +k
for k > 0. This inequality is equivalent to
(k − 3)
X a2 − bc
a2 + k
≥ 0.
Since
2
X a2 − bc
a2 + k
we have
2(k − 3)
=
X (a − b)(a + c) + (a − c)(a + b)
a2 + k
X (a − b)(a + c) X (b − a)(b + c)
=
+
a2 + k
b2 + k
X
(a − b)2
= (k − a b − bc − ca)
(a2 + p)(b2 + p)
2
X
(a − b)
= (k − 3)
,
2
(a + p)(b2 + p)
X a2 − bc
a2 + k
= (k − 3)2
X
(a − b)2
≥ 0.
(a2 + k)(b2 + k)
Equality in both inequalities holds for a = b = c = 1.
p
P 1.38. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. If k ≥ 2 + 3,
then
1
1
1
3
+
+
≤
.
a+k b+k c+k
1+k
(Vasile Cîrtoaje, 2007)
Solution. Let us denote p = a + b + c, p ≥ 3. By expanding, the inequality becomes
k(k − 2)p + 3a bc ≥ 3(k − 1)2 .
Symmetric Rational Inequalities
73
Since this inequality is true for p ≥ 3(k − 1)2 /(k2 − 2k), consider further that
p≤
3(k − 1)2
.
k(k − 2)
From Schur’s inequality
(a + b + c)3 + 9a bc ≥ 4(a b + bc + ca)(a + b + c),
we get 9a bc ≥ 12p − p3 . Therefore, it suffices to prove that
3k(k − 2)p + 12p − p3 ≥ 9(k − 1)2 ,
or, equivalently,
(p − 3)[(3(k − 1)2 − p2 − 3p] ≥ 0.
Thus, it remains to prove that
3(k − 1)2 − p2 − 3p ≥ 0.
p
Since p ≤ 3(k − 1)2 /(k2 − 2k) and k ≥ 2 + 3, we have
9(k − 1)4
9(k − 1)2
−
k2 (k − 2)2
k(k − 2)
2 2
2
3(k − 1) (k − 3)(k − 4k + 1)
=
≥ 0.
k2 (k − 2)2
p
The equality holds for p
a = b = c = 1. In the case k = 2 + 3, the equality holds again
for a = 0 and b = c = 3 (or any cyclic permutation).
3(k − 1)2 − p2 − 3p ≥ 3(k − 1)2 −
P 1.39. Let a, b, c be nonnegative real numbers such that a2 + b2 + c 2 = 3. Prove that
a(b + c) b(c + a) c(a + b)
+
+
≤ 3.
1 + bc
1 + ca
1 + ab
(Vasile Cîrtoaje, 2010)
Solution. Write the inequality in the homogeneous form
X
a2
or
X•
a(b + c)
≤ 1,
+ b2 + c 2 + 3bc
˜
a(b + c)
a
−
≤ 0,
a2 + b2 + c 2 + 3bc a + b + c
74
Vasile Cîrtoaje
X a(a − b)(a − c)
≥ 0.
a2 + b2 + c 2 + 3bc
Without loss of generality, assume that a ≥ b ≥ c. Then, it suffices to prove that
b(b − c)(b − a)
a(a − b)(a − c)
+ 2
≥ 0.
2
2
+ b + c + 3bc a + b2 + c 2 + 3ca
a2
This is true if
a(a − c)
b(b − c)
≥ 2
.
2
2
+ b + c + 3bc
a + b2 + c 2 + 3ca
Since a(a − c) ≥ b(b − c) and
a2
a2
+
b2
1
1
≥ 2
,
2
2
+ c + 3bc
a + b + c 2 + 3ca
the conclusion
follows. The equality holds for a = b = c = 1, and for a = 0 and
p
b = c = 3/2 (or any cyclic permutation).
P 1.40. Let a, b, c be positive real numbers such that a2 + b2 + c 2 = 3. Prove that
a2 + b2 b2 + c 2 c 2 + a2
+
+
≤ 3.
a+b
b+c
c+a
(Cezar Lupu, 2005)
First Solution. We apply the SOS method. Write the inequality in the homogeneous
form
X b2 + c 2 b + c Æ
−
≥ 3(a2 + b2 + c 2 ) − a − b − c,
b+c
2
or
P
X (b − c)2
(b − c)2
≥p
.
2(b + c)
3(a2 + b2 + c 2 ) + a + b + c
p
Since 3(a2 + b2 + c 2 ) + a + b + c ≥ 2(a + b + c) > 2(b + c), the conclusion follows.
The equality holds for a = b = c = 1.
Second Solution. By virtue of the Cauchy-Schwarz inequality, we have
Pp
P
Pp
X a2 + b2
(
a2 + b2 )2
2 a2 + 2
(a2 + b2 )(a2 + c 2 )
P
P
≥
=
a+b
(a + b)
2 a
P 2
P 2
P 2
P
2 a + 2 (a + bc) 3 a + ( a)2
P
P
≥
=
2 a
2 a
P
P
9 + ( a)2
( a − 3)2
P
P
=
=3+
≥ 3.
2 a
2 a
Symmetric Rational Inequalities
75
P 1.41. Let a, b, c be positive real numbers such that a2 + b2 + c 2 = 3. Prove that
bc
ca
7
ab
+
+
+ 2 ≤ (a + b + c).
a+b b+c c+a
6
(Vasile Cîrtoaje, 2011)
Solution. We apply the SOS method. Write the inequality as
‹
X
4bc
b+c−
≥ 8(3 − a − b − c).
3
b+c
Since
b+c−
4bc
(b − c)2
=
b+c
b+c
and
3−a− b−c =
3(a2 + b2 + c 2 ) − (a + b + c)2
9 − (a + b + c)2
=
3+a+ b+c
3+a+ b+c
X
1
=
(b − c)2 ,
3+a+ b+c
we can write the inequality as
Sa (b − c)2 + S b (c − a)2 + Sc (a − b)2 ≥ 0,
where
8
3
−
.
b+c 3+a+ b+c
Without loss of generality, assume that a ≥ b ≥ c. Since Sa ≥ S b ≥ Sc , it suffices to show
that S b + Sc ≥ 0. Indeed, if this condition is fulfilled, then Sa ≥ S b ≥ (S b + Sc )/2 ≥ 0,
and hence
Sa =
Sa (b − c)2 + S b (c − a)2 + Sc (a − b)2 ≥ S b (c − a)2 + Sc (a − b)2
≥ S b (a − b)2 + Sc (a − b)2 = (a − b)2 (S b + Sc ) ≥ 0.
Since
S b + Sc =
4(9 − 5a − b − c)
,
(a + b + 2c)(3 + a + b + c)
we need to show that 9 ≥ 5a+ b+c. This follows immediately from the Cauchy-Schwarz
inequality
(25 + 1 + 1)(a2 + b2 + c 2 ) ≥ (5a + b + c)2 .
Thus, the proof is completed. The equality holds for a = b = c = 1, and for a = 5/3
and b = c = 1/3 (or any cyclic permutation).
76
Vasile Cîrtoaje
P 1.42. Let a, b, c be positive real numbers such that a2 + b2 + c 2 = 3. Prove that
(a)
1
1
3
1
+
+
≤ ;
3 − a b 3 − bc 3 − ca
2
(b)
1
1
1
3
+p
+p
≤p
.
p
6 − ab
6 − bc
6 − ca
6−1
(Vasile Cîrtoaje, 2005)
Solution. (a) Since
ab
2a b
3
=1+
=1+ 2
2
3 − ab
3 − ab
a + b + 2c 2 + (a − b)2
2a b
(a + b)2
≤1+ 2
≤
1
+
,
a + b2 + 2c 2
2(a2 + b2 + 2c 2 )
it suffices to prove that
(b + c)2
(c + a)2
(a + b)2
+
+
≤ 3.
a2 + b2 + 2c 2 b2 + c 2 + 2a2 c 2 + a2 + 2b2
By the Cauchy-Schwarz inequality, we have
(a + b)2
(a + b)2
a2
b2
=
≤
+
.
a2 + b2 + 2c 2
(a2 + c 2 ) + (b2 + c 2 )
a2 + c 2 b2 + c 2
Thus,
X
X a2
X b2
X a2
X c2
(a + b)2
≤
+
=
+
= 3.
a2 + b2 + 2c 2
a2 + c 2
b2 + c 2
a2 + c 2
c 2 + a2
The equality holds for a = b = c.
(b) According to P 1.29, the following inequality holds
1
1
1
3
+
+
≤ .
2
2
2
2
2
2
6−a b
6− b c
6−c a
5
Since
p
2 6
1
1
=p
+p
,
2
2
6−a b
6 − ab
6 + ab
this inequality becomes
X
1
+
p
6 − ab
X
p
6 6
.
≤
p
5
6 + ab
1
Symmetric Rational Inequalities
77
Thus, it suffices to show that
X
3
1
≥p
.
p
6 + ab
6+1
Since a b + bc + ca ≤ a2 + b2 + c 2 = 3, by the Cauchy-Schwarz inequality, we have
X
1
9
9
3
≥Pp
= p
≥p
.
p
6 + ab
( 6 + a b) 3 6 + a b + bc + ca
6+1
The equality holds for a = b = c = 1.
P 1.43. Let a, b, c be positive real numbers such that a2 + b2 + c 2 = 3. Prove that
1
1
1
3
+
+
≥ .
5
5
5
1+a
1+ b
1+c
2
(Vasile Cîrtoaje, 2007)
Solution. Let a = min{a, b, c}. There are two cases to consider
Case 1: a ≥ 1/2. The desired inequality follows by summing the inequalities
8
≥ 9 − 5a2 ,
1 + a5
8
≥ 9 − 5b2 ,
1 + b5
8
≥ 9 − 5c 2 ;
1 + c5
8
≥ p + q x 2 , whose
1 + x5
coefficients p and q will be determined such that the polynomial P(x) = 8−(1+ x 5 )(p +
q x 2 ) divides by (x − 1)2 . It is easy to check that P(1) = 0 involves p + q = 4, when
To obtain these inequalities, we start from the inequality
P(x) = 4(2 − x 2 − x 7 ) − p(1 − x 2 + x 5 − x 7 ) = (1 − x)Q(x),
where
Q(x) = 4(2 + 2x + x 2 + x 3 + x 4 + x 5 + x 6 ) − p(1 + x + x 5 + x 6 ).
In addition, Q(1) = 0 involves p = 9. In this case,
P(x) = (1 − x)2 (5x 5 + 10x 4 + 6x 3 + 2x 2 − 2x − 1)
= (1 − x)2 [x 5 + (2x − 1)(2x 4 + 6x 3 + 6x 2 + 4x + 1)] ≥ 0.
Case 2: a ≥ 1/2. Write the inequality as
1
1
b5 c 5 − 1
−
≥
.
1 + a5 2 (1 + b5 )(1 + c 5 )
78
Vasile Cîrtoaje
Since
1
1 32 1 31
− ≥
− =
5
1+a
2 33 2 66
and
(1 + b5 )(1 + c 5 ) ≥ (1 +
it suffices to show that
31(1 +
p
p
b5 c 5 )2 ,
b5 c 5 )2 ≥ 66(b5 c 5 − 1),
which is equivalent to bc ≤ (97/35)2/5 . Indeed, from
3 = a2 + b2 + c 2 > b2 + c 2 ≥ 2bc,
we get bc < 3/2 < (97/35)2/5 . This completes the proof. The equality holds for a =
b = c = 1.
P 1.44. Let a, b, c be positive real numbers such that a bc = 1. Prove that
1
1
1
+
+
≥ 1.
a2 + a + 1 b2 + b + 1 c 2 + c + 1
First Solution. Using the substitutions a = yz/x 2 , b = z x/ y 2 , c = x y/z 2 , where x, y, z
are positive real numbers, the inequality becomes
X
x4
≥ 1.
x 4 + x 2 yz + y 2 z 2
By the Cauchy-Schwarz inequality, we have
P
P 4
P
X
( x 2 )2
x + 2 y 2z2
x4
P
P
≥P
=P
.
x 4 + x 2 yz + y 2 z 2
(x 4 + x 2 yz + y 2 z 2 )
x 4 + x yz x + y 2 z 2
Therefore, it suffices to show that
X
which is equivalent to
P
y 2 z 2 ≥ x yz
X
x,
x 2 ( y − z)2 ≥ 0. The equality holds for a = b = c = 1.
Second Solution. Using the substitutions a = y/x, b = z/ y, c = x/z, where x, y, z > 0,
we need to prove that
y2
x2
z2
+
+
≥ 1.
x2 + x y + y2
y 2 + yz + z 2 z 2 + z x + z 2
Symmetric Rational Inequalities
Since
79
x 2 z(x + y + z)
x 2 (x 2 + y 2 + z 2 + x y + yz + z x)
2
=
x
+
,
x2 + x y + y2
x2 + x y + y2
multiplying by x 2 + y 2 + z 2 + x y + yz + z x, the inequality can be written as
X
x y + yz + z x
x 2z
≥
.
x2 + x y + y2
x + y +z
By the Cauchy-Schwarz inequality, we have
P
X
( xz)2
x y + yz + z x
x 2z
≥P
.
=
2
2
2
2
x +xy+ y
x + y +z
z(x + x y + y )
Remark. The inequality in P 1.44 is a particular case of the following more general
inequality: (Vasile Cîrtoaje, 2009).
• Let a1 , a2 , . . . , an (n ≥ 3) be positive real numbers such that a1 a2 · · · an = 1. If p and
q are nonnegative real numbers satisfying p + q = n − 1, then
i=n
X
1
i=1
1 + pai + qai2
≥ 1.
P 1.45. Let a, b, c be positive real numbers such that a bc = 1. Prove that
a2
1
1
1
+ 2
+ 2
≤ 3.
−a+1 b − b+1 c −c+1
First Solution. Since
1
1
2(a2 + 1)
2a4
+
=
=
2
−
,
a2 − a + 1 a2 + a + 1
a4 + a2 + 1
a4 + a2 + 1
we can rewrite the inequality as
X
X
1
a4
+
2
≥ 3.
a2 + a + 1
a4 + a2 + 1
Thus, it suffices to show that
X
and
X
1
≥1
a2 + a + 1
a4
≥ 1.
a4 + a2 + 1
80
Vasile Cîrtoaje
The first inequality is just the inequality in P 1.44, while the second follows from the
first by substituting a, b, c with a−2 , b−2 , c −2 , respectively. The equality holds for a =
b = c = 1.
Second Solution. Write the inequality as
‹
X4
1
− 2
≥ 1,
3 a −a+1
X (2a − 1)2
≥ 3.
a2 − a + 1
Let p = a + b + c and q = a b + bc + ca. By the Cauchy-Schwarz inequality, we have
P
X (2a − 1)2
(2 a − 3)2
(2p − 3)2
P
≥
.
=
a2 − a + 1
p2 − 2q − p + 3
(a2 − a + 1)
Thus, it suffices to show that
(2p − 3)2 ≥ 3(p2 − 2q − p + 3),
which is equivalent to
p2 + 6q − 9p ≥ 0.
From the known inequality
(a b + bc + ca)2 ≥ 3a bc(a + b + c),
we get q2 ≥ 3p. Using this inequality and the AM-GM inequality, we get
Æ
Æ
3
3
p2 + 6q = p2 + 3q + 3q ≥ 3 9p2 q2 ≥ 3 9p2 (3p) = 9p.
P 1.46. Let a, b, c be positive real numbers such that a bc = 1. Prove that
3+a
3+ b
3+c
+
+
≥ 3.
2
2
(1 + a)
(1 + b)
(1 + c)2
Solution. Using the inequality in P 1.1, we have
X 3+a
X
X 1
2
=
+
(1 + a)2
(1 + a)2
1+a
˜ X
X•
1
1
1
=
+
+
2
2
(1 + a)
(1 + b)
1+c
X 1
X ab
≥
+
= 3.
1 + ab
1 + ab
The equality holds for a = b = c = 1.
Symmetric Rational Inequalities
81
P 1.47. Let a, b, c be positive real numbers such that a bc = 1. Prove that
7 − 6a 7 − 6b 7 − 6c
+
+
≥ 1.
2 + a2 2 + b2 2 + c 2
(Vasile Cîrtoaje, 2008)
Solution. Write the inequality as

‹ 
‹ 
‹
7 − 6a
7 − 6b
7 − 6c
+1 +
+1 +
+ 1 ≥ 4,
2 + a2
2 + b2
2 + c2
(3 − a)2 (3 − b)2 (3 − c)2
+
+
≥ 4.
2 + a2
2 + b2
2 + c2
Substituting a, b, c by 1/a, 1/b, 1/c, respectively, we need to prove that a bc = 1 involves
(3a − 1)2 (3b − 1)2 (3c − 1)2
+
+
≥ 4.
2a2 + 1
2b2 + 1
2c 2 + 1
By the Cauchy-Schwarz inequality, we have
P
P
P
P
X (3a − 1)2
(3 a − 3)2
9 a2 + 18 a b − 18 a + 9
P
≥ P
=
.
2a2 + 1
(2a2 + 1)
2 a2 + 3
Thus, it suffices to prove that
f (a) + f (b) + f (c) ≥ 3,
where
‹
1
f (x) = x + 18
−x .
x
In order to do this, we use the mixing method. Without loss of generality, assume that
a = max{a, b, c}, a ≥ 1, bc ≤ 1. Since
‹

p
p
p
1
f (b) + f (c) − 2 f ( bc) = (b − c)2 + 18( b − c)2
− 1 ≥ 0,
bc
2
it suffices to show that

p
f (a) + 2 f ( bc) ≥ 3.
Write this inequality as
 ‹
1
f (x ) + 2 f
≥ 3,
x
p
where x = a. It is equivalent to
2
x 6 − 18x 4 + 36x 3 − 3x 2 − 36x + 20 ≥ 0,
(x − 1)2 (x − 2)2 (x + 1)(x + 5) ≥ 0.
Since the last inequality is true, the proof is completed. The equality holds for a = b =
c = 1, and also for a = 1/4 and b = c = 2 (or any cyclic permutation).
82
Vasile Cîrtoaje
P 1.48. Let a, b, c be positive real numbers such that a bc = 1. Prove that
b6
c6
a6
+
+
≥ 1.
1 + 2a5 1 + 2b5 1 + 2c 5
(Vasile Cîrtoaje, 2008)
Solution. Using the substitutions
a=
v
t
2
3 x
yz
,
b=
v
t
2
3 y
zx
,
v
t
2
3 z
c=
,
xy
the inequality becomes
X
x4
≥ 1.
p
y 2 z 2 + 2x 3 3 x yz
By the Cauchy-Schwarz inequality, we have
P
P
X
( x 2 )2
( x 2 )2
x4
P .
≥P
=P
p
p
p
y 2 z 2 + 2x 3 3 x yz
( y 2 z 2 + 2x 3 3 x yz)
x 2 y 2 + 2 3 x yz x 3
Therefore, we need to show that
X
X
X
p
(
x 2 )2 ≥
x 2 y 2 + 2 3 x yz
x 3.
p
Since x + y + z ≥ 3 3 x yz, it suffices to prove that
X
X
X
X
3(
x 2 )2 ≥ 3
x 2 y 2 + 2(
x)(
x 3 );
that is,
X
x4 + 3
X
x2 y2 ≥ 2
X
x y(x 2 + y 2 ),
or, equivalently,
X
(x − y)4 ≥ 0.
The equality holds for a = b = c = 1.
P 1.49. Let a, b, c be positive real numbers such that a bc = 1. Prove that
a
b
c
1
+
+
≤ .
a2 + 5 b2 + 5 c 2 + 5 2
(Vasile Cîrtoaje, 2008)
Symmetric Rational Inequalities
83
Solution. Let
F (a, b, c) =
a2
a
b
c
+ 2
+ 2
.
+5 b +5 c +5
Without loss of generality, assume that a = min{a, b, c}.
Case 1: a ≤ 1/5. We have
a
1
b
c
1
1
+ p
+ p
≤
+p < .
5 2 5b2 2 5c 2
25
5 2
p
p
Case 2: a > 1/5. Let x = bc, a = 1/x 2 , x < 5. We will show that
F (a, b, c) <
F (a, b, c) ≤ F (a, x, x) ≤
1
.
2
The left inequality, F (a, b, c) ≤ F (a, x, x), is equivalent to
p
p
( b − c)2 [10x(b + c) + 10x 2 − 25 − x 4 ] ≥ 0.
This is true since
10x(b + c) + 10x 2 − 25 − x 4 ≥ 20x 2 + 10x 2 − 25x 2 − x 4 = x 2 (5 − x 2 ) > 0.
The right inequality, F (a, x, x) ≤ 1/2, is equivalent to
(x − 1)2 (5x 4 − 10x 3 − 2x 2 + 6x + 5) ≥ 0.
It is also true since
5x 4 − 10x 3 − 2x 2 + 6x + 5 = 5(x − 1)4 + 2x(5x 2 − 16x + 13)
and
5x 2 + 13 ≥ 2
p
65x 2 > 16x.
The equality holds for a = b = c = 1.
P 1.50. Let a, b, c be positive real numbers such that a bc = 1. Prove that
1
1
1
2
+
+
+
≥ 1.
2
2
2
(1 + a)
(1 + b)
(1 + c)
(1 + a)(1 + b)(1 + c)
(Pham Van Thuan, 2006)
84
Vasile Cîrtoaje
First Solution. There are two of a, b, c either greater than or equal to 1, or less than or
equal to 1. Let b and c be these numbers; that is, (1 − b)(1 − c) ≥ 0. Since
1
1
1
+
≥
2
2
(1 + b)
(1 + c)
1 + bc
(see P 1.1), it suffices to show that
1
1
2
+
+
≥ 1.
2
(1 + a)
1 + bc (1 + a)(1 + b)(1 + c)
This inequality is equivalent to
b2 c 2
1
2bc
+
+
≥ 1,
2
(1 + bc)
1 + bc (1 + bc)(1 + b)(1 + c)
which can be written in the obvious form
bc(1 − b)(1 − c)
≥ 0.
(1 + bc)(1 + b)(1 + c)
The equality holds for a = b = c = 1.
Second Solution. Setting a = yz/x 2 , b = z x/ y 2 , c = x y/z 2 , where x, y, z > 0, the
inequality becomes
X
2x 2 y 2 z 2
x4
+
≥ 1.
(x 2 + yz)2 (x 2 + yz)( y 2 + z x)(z 2 + x y)
Since (x 2 + yz)2 ≤ (x 2 + y 2 )(x 2 + z 2 ), we have
X
X
2x 2 y 2 z 2
x4
x4
≥
=
1
−
.
(x 2 + yz)2
(x 2 + y 2 )(x 2 + z 2 )
(x 2 + y 2 )( y 2 + z 2 )(z 2 + x 2 )
Then, it suffices to show that
(x 2 + y 2 )( y 2 + z 2 )(z 2 + x 2 ) ≥ (x 2 + yz)( y 2 + z x)(z 2 + x y).
This inequality follows by multiplying the inequalities
(x 2 + y 2 )(x 2 + z 2 ) ≥ (x 2 + yz)2 ,
( y 2 + z 2 )( y 2 + x 2 ) ≥ ( y 2 + z x)2 ,
(z 2 + x 2 )(z 2 + y 2 ) ≥ (z 2 + x y)2 .
Third Solution. We make the substitutions
1+ y
1
1+ x
1
1
1+z
=
,
=
,
=
;
1+a
2
1+ b
2
1+c
2
Symmetric Rational Inequalities
that is,
a=
85
1− x
,
1+ x
b=
1− y
,
1+ y
c=
1−z
,
1+z
where −1 < x, y, z < 1. Since a bc = 1 involves x + y + z + x yz = 0, we need to prove
that
x + y + z + x yz = 0
implies
(1 + x)2 + (1 + y)2 + (1 + z)2 + (1 + x)(1 + y)(1 + z) ≥ 4.
This inequality is equivalent to
x 2 + y 2 + z 2 + (x + y + z)2 + 4(x + y + z) ≥ 0.
By virtue of the AM-GM inequality, we have
x 2 + y 2 + z 2 + (x + y + z)2 + 4(x + y + z) = x 2 + y 2 + z 2 + x 2 y 2 z 2 − 4x yz
Æ
4
≥ 4 x 4 y 4 z 4 − 4x yz = 4|x yz| − 4x yz ≥ 0.
P 1.51. Let a, b, c be nonnegative real numbers such that
1
1
1
3
+
+
= .
a+b b+c c+a
2
Prove that
3
2
1
≥
+ 2
.
a+b+c
a b + bc + ca a + b2 + c 2
Solution. Write the inequality in the homogeneous form

‹
1
2
1
1
2
1
+
+
≥
+ 2
.
a+b+c a+b b+c c+a
a b + bc + ca a + b2 + c 2
Denote q = a b + bc + ca and assume that a + b + c = 1. From the known inequality
(a + b + c)2 ≥ 3(a b + bc + ca), we get 1 − 3q ≥ 0. Rewrite the desired inequality as
follows

‹
1
1
1
2
1
2
+
+
≥ +
,
1−c 1−a 1− b
q 1 − 2q
2(q + 1)
2 − 3q
≥
,
q − a bc
q(1 − 2q)
q2 (1 − 4q) + (2 − 3q)a bc ≥ 0.
86
Vasile Cîrtoaje
By Schur’s inequality, we have
(a + b + c)3 + 9a bc ≥ 4(a + b + c)(a b + bc + ca),
1 − 4q ≥ −9a bc.
Then,
q2 (1 − 4q) + (2 − 3q)a bc ≥ −9q2 a bc + (2 − 3q)a bc
= (1 − 3q)(2 + 3q)a bc ≥ 0.
The equality holds for a = b = c = 1, and for a = 0 and b = c = 5/3 (or any cyclic
permutation).
P 1.52. Let a, b, c be nonnegative real numbers such that
7(a2 + b2 + c 2 ) = 11(a b + bc + ca).
Prove that
a
b
c
51
≤
+
+
≤ 2.
28
b+c c+a a+b
Solution. Due to homogeneity, we may assume that b + c = 2. Let us denote x = bc,
0 ≤ x ≤ 1. By the hypothesis 7(a2 + b2 + c 2 ) = 11(a b + bc + ca), we get
x=
7a2 − 22a + 28
.
25
Then, x ≤ 1 involves 1/7 ≤ a ≤ 3. Since
a
b
c
a
a(b + c) + (b + c)2 − 2bc
+
+
=
+
b+c c+a a+b
b+c
a2 + (b + c)a + bc
a 2(a + 2 − x) 4a3 + 27a + 11
= + 2
=
,
2
a + 2a + x
8a2 + 7a + 7
the required inequalities become
51 4a3 + 27a + 11
≤
≤ 2.
28
8a2 + 7a + 7
We have
4a3 + 27a + 11 51 (7a − 1)(4a − 7)2
−
=
≥0
8a2 + 7a + 7
28
28(8a2 + 7a + 7)
Symmetric Rational Inequalities
87
and
4a3 + 27a + 11 (3 − a)(2a − 1)2
=
≥ 0.
8a2 + 7a + 7
8a2 + 7a + 7
This completes the proof. The left inequality becomes an equality for 7a = b = c (or
any cyclic permutation), while the right inequality is an equality for a/3 = b = c (or
any cyclic permutation).
2−
P 1.53. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
a2
1
1
1
10
+ 2
+ 2
≥
.
2
2
2
+b
b +c
c +a
(a + b + c)2
Solution. Assume that a = min{a, b, c}, and denote
a
x = b+ ,
2
a
y =c+ .
2
Since
a2 + b2 ≤ x 2 , b2 + c 2 ≤ x 2 + y 2 , c 2 + a2 ≤ y 2 ,
(a + b + c)2 = (x + y)2 ≥ 4x y,
it suffices to show that
1
1
1
5
+ 2
+ 2≥
.
2
2
x
x +y
y
2x y
We have
1
1
1
5
+ 2
+ 2−
=
2
2
x
x +y
y
2x y
‹ 
‹
1
1
2
1
1
+ 2−
+
−
x2
y
xy
x 2 + y 2 2x y
(x − y)2
(x − y)2
−
=
x2 y2
2x y(x 2 + y 2 )
2
(x − y) (2x 2 − x y + 2 y 2 )
=
≥ 0.
2x 2 y 2 (x 2 + y 2 )

The equality holds for a = 0 and b = c (or any cyclic permutation).
P 1.54. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
a2
1
1
1
3
+ 2
+ 2
≥
.
2
2
2
− ab + b
b − bc + c
c − ca + a
max{a b, bc, ca}
88
Vasile Cîrtoaje
Solution. Assume that a = min{a, b, c}, hence bc = max{a b, bc, ca}. Since
a2
1
1
1
1
1
1
+ 2
+ 2
≥ 2+ 2
+ 2,
2
2
2
2
− ab + b
b − bc + c
c − ca + a
b
b − bc + c
c
it suffices to show that
We have
1
1
3
1
.
+
+
≥
b2 b2 − bc + c 2 c 2
bc
1
1
1
3
(b − c)4
+
+
−
=
≥ 0.
b2 b2 − bc + c 2 c 2 bc
b2 c 2 (b2 − bc + c 2 )
The equality holds for a = b = c, and also a = 0 and b = c (or any cyclic permutation).
P 1.55. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
a(2a + b + c) b(2b + c + a) c(2c + a + b)
+
+
≥ 6.
b2 + c 2
c 2 + a2
a2 + b2
Solution. By the Cauchy-Schwarz inequality, we have
P
X a(2a + b + c)
[ a(2a + b + c)]2
≥P
.
b2 + c 2
a(2a + b + c)(b2 + c 2 )
Thus, we still need to show that
X
X
X
2(
a2 +
a b)2 ≥ 3
a(2a + b + c)(b2 + c 2 ),
which is equivalent to
X
X
X
X
a+
a b(a2 + b2 ) ≥ 6
a2 b2 .
2
a4 + 2a bc
We can obtain this inequality by adding Schur’s inequality of degree four
X
X
X
a4 + a bc
a≥
a b(a2 + b2 )
and
X
a b(a2 + b2 ) ≥ 2
X
a2 b2 ,
multiplied by 2 and 3, respectively. The equality occurs for a = b = c, and for a = 0
and b = c (or any cyclic permutation).
Symmetric Rational Inequalities
89
P 1.56. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
a2 (b + c)2 b2 (c + a)2 c 2 (a + b)2
+ 2
+
≥ 2(a b + bc + ca).
b2 + c 2
c + a2
a2 + b2
Solution. We apply the SOS method. Since
a2 (b + c)2 )
2a2 bc
2
=
a
+
,
b2 + c 2
b2 + c 2
we can write the inequality as
X
X
X 
2(
a2 −
a b) −
a2 1 −
2bc
b2 + c 2
‹
≥ 0,
X
X a2 (b − c)2
≥ 0,
(b − c)2 −
b2 + c 2
X
a2
1− 2
(b − c)2 ≥ 0.
b + c2
Without loss of generality, assume that a ≥ b ≥ c. Since 1 −
prove that
c2
> 0, it suffices to
a2 + b2
a2
b2
2
1− 2
(b − c) + 1 − 2
(a − c)2 ≥ 0,
b + c2
c + a2
which is equivalent to
(a2 − b2 + c 2 )(a − c)2
(a2 − b2 − c 2 )(b − c)2
≥
.
a2 + c 2
b2 + c 2
This inequality follows by multiplying the inequalities
a2 − b2 + c 2 ≥ a2 − b2 − c 2 ,
(a − c)2
(b − c)2
≥
.
a2 + c 2
b2 + c 2
The latter inequality is true since
(a − c)2 (b − c)2
2bc
2ac
2c(a − b)(a b − c 2 )
−
=
−
=
≥ 0.
a2 + c 2
b2 + c 2
b2 + c 2 a2 + c 2
(b2 + c 2 )(a2 + c 2 )
The equality occurs for a = b = c, and for a = 0 and b = c (or any cyclic permutation).
90
Vasile Cîrtoaje
P 1.57. If a, b, c are real numbers such that a bc > 0, then

‹

‹
X
a
b
c
a
1 1 1
3
+
5
+
+
≥
8
+
+
.
b2 − bc + c 2
bc ca a b
a b c
(Vasile Cîrtoaje, 2011)
Solution. In order to apply the SOS method, we multiply the inequality by a bc and
write it as follows:
‹
X
X
X 
bc
2
2
≥ 0,
8(
a −
bc) − 3
a 1− 2
b − bc + c 2
X
X a2 (b − c)2
(b − c)2 − 3
≥ 0,
b2 − bc + c 2
X (b − c)2 (4b2 − 4bc + 4c 2 − 3a2 )
≥ 0.
b2 − bc + c 2
Without loss of generality, assume that a ≥ b ≥ c. Since
4
4a2 − 4a b + 4b2 − 3c 2 = (2a − b)2 + 3(b2 − c 2 ) ≥ 0,
it suffices to prove that
(a − c)2 (4a2 − 4ac + 4c 2 − 3b2 ) (b − c)2 (3a2 − 4b2 + 4bc − 4c 2 )
≥
.
a2 − ac + c 2
b2 − bc + c 2
Notice that
4a2 − 4ac + 4c 2 − 3b2 = (a − 2c)2 + 3(a2 − b2 ) ≥ 0.
Thus, the desired inequality follows by multiplying the inequalities
4a2 − 4ac + 4c 2 − 3b2 ≥ 3a2 − 4b2 + 4bc − 4c 2
and
(a − c)2
(b − c)2
≥
.
a2 − ac + c 2
b2 − bc + c 2
The first inequality is equivalent to
(a − 2c)2 + (b − 2c)2 ≥ 0.
Also, we have
(a − c)2
(b − c)2
bc
ac
−
= 2
− 2
2
2
2
2
2
a − ac + c
b − bc + c
b − bc + c
a − ac + c 2
c(a − b)(a b − c 2 )
= 2
≥ 0.
(b − bc + c 2 )(a2 − ac + c 2 )
The equality occurs for a = b = c, and for 2a = b = c (or any cyclic permutation).
Symmetric Rational Inequalities
91
P 1.58. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that

‹
1
1
1
(a) 2a bc
+
+
+ a2 + b2 + c 2 ≥ 2(a b + bc + ca);
a+b b+c c+a
b2
c2
3(a2 + b2 + c 2 )
a2
+
+
≤
.
a+b b+c c+a
2(a + b + c)
(b)
Solution. (a) First Solution. We have
2a bc
X
X
X a(2bc + a b + ac)
1
+
a2 =
b+c
b+c
X a b(a + c) X ac(a + b)
=
+
b+c
b+c
X a b(a + c) X ba(b + c)
+
=
b+ c
‹c + a X
X
a+c b+c
=
ab
+
≥2
a b.
b+c a+c
The equality occurs for a = b = c, and for a = 0 and b = c (or any cyclic permutation).
Second Solution. Write the inequality as
‹
X  2a bc
2
+ a − a b − ac ≥ 0.
b+c
We have
X  2a bc
b+c
2
‹
+ a − a b − ac =
X a b(a − b) + ac(a − c)
b+c
X a b(a − b) X ba(b − a)
=
+
b+c
c+a
X a b(a − b)2
=
≥ 0.
(b + c)(c + a)
(b) Since
‹
X a2
X
X ab
ab
=
a−
=a+b+c−
,
a+b
a+b
a+b
we can write the desired inequality as
X ab
3(a2 + b2 + c 2 )
+
≥ a + b + c.
a+b
2(a + b + c)
Multiplying by 2(a + b + c), the inequality can be written as
X
a 2
1+
bc + 3(a2 + b2 + c 2 ) ≥ 2(a + b + c)2 ,
b+c
92
Vasile Cîrtoaje
or
1
+ a2 + b2 + c 2 ≥ 2(a b + bc + ca),
b+c
which is just the inequality in (a).
2a bc
X
P 1.59. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
(a)
a2 − bc b2 − ca c 2 − a b 3(a b + bc + ca)
+ 2
+
+
≥ 3;
b2 + c 2
c + a2 a2 + b2
a2 + b2 + c 2
(b)
a2
b2
c2
a b + bc + ca
5
+
+
+ 2
≥ ;
2
2
2
2
2
2
2
2
b +c
c +a
a +b
a +b +c
2
(c)
a b + bc + ca
a2 + bc b2 + ca c 2 + a b
+ 2
+ 2
≥ 2
+ 2.
2
2
2
2
b +c
c +a
a +b
a + b2 + c 2
(Vasile Cîrtoaje, 2014)
Solution. (a) Write the inequality as follows:
X
‹

‹
X 2a2
2bc
a b + bc + ca
−
1
+
1
−
−
6
1
−
≥ 0,
b2 + c 2
b2 + c 2
a2 + b2 + c 2
X 2a2 − b2 − c 2
b2 + c 2
+
X (b − c)2
b2 + c 2
−3
X
(b − c)2
≥ 0.
a2 + b2 + c 2
Since
X 2a2 − b2 − c 2
b2 + c 2
=
=
X a2 − b2
X
+
X a2 − c 2
=
X a2 − b2
+
X b2 − a2
c 2 + a2
b2 + c 2
b2 + c 2
b2 + c 2
2
2 2
X
(a − b )
(b2 − c 2 )2
=
,
(b2 + c 2 )(c 2 + a2 )
(a2 + b2 )(a2 + c 2 )
we can write the inequality as
X
(b − c)2 Sa ≥ 0,
where
Sa =
(b + c)2
1
3
+ 2
− 2
.
2
2
2
2
2
(a + b )(a + c ) b + c
a + b2 + c 2
It suffices to show that Sa ≥ 0 for all nonnegative real numbers a, b, c, no two of which
are zero. Denoting x 2 = b2 + c 2 , we have
Sa =
x 2 + 2bc
1
3
+ 2− 2
,
4
2
2
2
2
a +a x +b c
x
a + x2
Symmetric Rational Inequalities
93
and the inequality Sa ≥ 0 becomes
(a2 − 2x 2 )b2 c 2 + 2x 2 (a2 + x 2 )bc + (a2 + x 2 )(a2 − x 2 )2 ≥ 0.
Clearly, this is true if
−2x 2 b2 c 2 + 2x 4 bc ≥ 0.
Indeed,
−2x 2 b2 c 2 + 2x 4 bc = 2x 2 bc(x 2 − bc) = 2bc(b2 + c 2 )(b2 + c 2 − bc) ≥ 0.
The equality occurs for a = b = c, and for a = 0 and b = c (or any cyclic permutation).
(b) First Solution. We get the desired inequality by summing the inequality in (a)
and the inequality
b2
ca
ab
1 2(a b + bc + ca)
bc
+ 2
+ 2
+ ≥
.
2
2
2
+c
c +a
a +b
2
a2 + b2 + c 2
This inequality is equivalent to
‹
X  2bc
4(a b + bc + ca)
+
1
≥
+ 2,
b2 + c 2
a2 + b2 + c 2
X (b + c)2
≥
b2 + c 2
By the Cauchy-Schwarz inequality, we have
X (b + c)2
b2 + c 2
2(a + b + c)2
.
a2 + b2 + c 2
P
2
(b + c)
2(a + b + c)2
≥ P
.
= 2
a + b2 + c 2
(b2 + c 2 )
The equality occurs for a = b = c, and for a = 0 and b = c (or any cyclic permutation).
Second Solution. Let
p = a + b + c,
q = a b + bc + ca,
r = a bc.
By the Cauchy-Schwarz inequality, we have
X
P 2 2
a
(p2 − 2q)2
a2
P
≥
.
=
b2 + c 2
a2 (b2 + c 2 ) 2(q2 − 2pr)
Therefore, it suffices to show that
(p2 − 2q)2
2q
+ 2
≥ 5.
2
q − 2pr
p − 2q
(*)
94
Vasile Cîrtoaje
Consider the following cases: p2 ≥ 4q and 3q ≤ p2 < 4q.
Case 1: p2 ≥ 4q. The inequality (*) is true if
2q
(p2 − 2q)2
+ 2
≥ 5,
2
q
p − 2q
which is equivalent to the obvious inequality
(p2 − 4q) (p2 − q)2 − 2q2 ≥ 0.
Case 2: 3q ≤ p2 < 4q. Using Schur’s inequality of degree four
6pr ≥ (p2 − q)(4q − p2 ),
the inequality (*) is true if
3(p2 − 2q)2
2q
+
≥ 5,
3q2 − (p2 − q)(4q − p2 ) p2 − 2q
which is equivalent to the obvious inequality
(p2 − 3q)(p2 − 4q)(2p2 − 5q) ≤ 0.
Third Solution (by Nguyen Van Quy). Write the inequality (*) from the preceding solution as follows:
(a2 + b2 + c 2 )2
2(a b + bc + ca)
+
≥ 5,
2
2
2
2
2
2
a b +b c +c a
a2 + b2 + c 2
(a2 + b2 + c 2 )2
2(a b + bc + ca)
−3≥2−
,
2
2
2
2
2
2
a b +b c +c a
a2 + b2 + c 2
a4 + b4 + c 4 − a2 b2 − b2 c 2 − c 2 a2
2(a2 + b2 + c 2 − a b − bc − ca)
≥
.
a2 b2 + b2 c 2 + c 2 a2
a2 + b2 + c 2
Since
2(a2 b2 + b2 c 2 + c 2 a2 ) ≤
X
a b(a2 + b2 ) ≤ (a b + bc + ca)(a2 + b2 + c 2 ),
it suffices to prove that
a4 + b4 + c 4 − a2 b2 − b2 c 2 − c 2 a2
≥ a2 + b2 + c 2 − a b − bc − ca,
a b + bc + ca
which is just Schur’s inequality of degree four
a4 + b4 + c 4 + a bc(a + b + c) ≥ a b(a2 + b2 ) + bc(b2 + c 2 ) + ca(c 2 + a2 ).
Symmetric Rational Inequalities
95
(c) We get the desired inequality by summing the inequality in (a) and the inequality
2ca
2a b
4(a b + bc + ca)
2bc
+ 2
+ 2
+1≥
,
2
2
2
+c
c +a
a +b
a2 + b2 + c 2
which is proved at (b). The equality occurs for a = b = c, and for a = 0 and b = c (or
any cyclic permutation).
b2
P 1.60. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
a2
b2
c2
(a + b + c)2
+
+
≥
.
b2 + c 2 c 2 + a2 a2 + b2
2(a b + bc + ca)
Solution. Applying the Cauchy-Schwarz inequality, we have
X
P 2 2
a
(a2 + b2 + c 2 )2
a2
P
≥
=
.
b2 + c 2
a2 (b2 + c 2 ) 2(a2 b2 + b2 c 2 + c 2 a2 )
Therefore, it suffices to show that
(a2 + b2 + c 2 )2
(a + b + c)2
≥
,
2(a2 b2 + b2 c 2 + c 2 a2 ) 2(a b + bc + ca)
which is equivalent to
(a2 + b2 + c 2 )2
(a + b + c)2
−
3
≥
− 3,
a2 b2 + b2 c 2 + c 2 a2
a b + bc + ca
a4 + b4 + c 4 − a2 b2 − b2 c 2 − c 2 a2
a2 + b2 + c 2 − a b − bc − ca
≥
.
a2 b2 + b2 c 2 + c 2 a2
a b + bc + ca
Since a2 b2 + b2 c 2 + c 2 a2 ≤ (a b + bc + ca)2 , it suffices to show that
a4 + b4 + c 4 − a2 b2 − b2 c 2 − c 2 a2 ≥ (a2 + b2 + c 2 − a b − bc − ca)(a b + bc + ca),
which is just Schur’s inequality of degree four
a4 + b4 + c 4 + a bc(a + b + c) ≥ a b(a2 + b2 ) + bc(b2 + c 2 ) + ca(c 2 + a2 ).
The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).
96
Vasile Cîrtoaje
P 1.61. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
2bc
2ca
a2 + b2 + c 2
5
2a b
+
+
+
≥ .
(a + b)2 (b + c)2 (c + a)2 a b + bc + ca
2
(Vasile Cîrtoaje, 2006)
First Solution. We use the SOS method. Write the inequality as follows
‹
X1
a2 + b2 + c 2
2bc
−1≥
−
,
a b + bc + ca
2 (b + c)2
X
X (b − c)2
(b − c)2
≥
,
ab + bc + ca
(b + c)2
(b − c)2 Sa + (c − a)2 S b + (a − b)2 Sc ≥ 0,
where
Sa = 1 −
a b + bc + ca
a b + bc + ca
a b + bc + ca
, Sb = 1 −
, Sc = 1 −
.
2
2
(b + c)
(c + a)
(a + b)2
Without loss of generality, assume that a ≥ b ≥ c. We have Sc > 0 and
Sb ≥ 1 −
(c + a)(c + b)
a−b
=
≥ 0.
2
(c + a)
c+a
If b2 Sa + a2 S b ≥ 0, then
X
a2
(b − c)2 Sa ≥ (b − c)2 Sa + (c − a)2 S b ≥ (b − c)2 Sa + 2 (b − c)2 S b
b
2 2
2
(b − c) (b Sa + a S b )
≥ 0.
=
b2
We have
2
2
2

‹2
a 2 b Sa + a S b = a + b − (a b + bc + ca)
+
c+a

‹2 b
a 2
2
2
≥ a + b − (b + c)(c + a))
+
b+c
c+a
‹

c+a
b+c
= a2 1 −
+ b2 1 −
c+a
b+c
2
(a − b) (a b + bc + ca)
=
≥ 0.
(b + c)(c + a)
2
b
b+c
The equality occurs for a = b = c, and for a = 0 and b = c (or any cyclic permutation).
Symmetric Rational Inequalities
97
Second Solution. Multiplying by a b + bc + ca, the inequality becomes
X 1
X 2a2 b2
5
+
2a
bc
+ a2 + b2 + c 2 ≥ (a b + bc + ca),
2
(a + b)
a+b
2
X 1
X1 •
X 4a b ˜
2a bc
+ a2 + b2 + c 2 − 2(a b + bc + ca) −
ab 1 −
≥ 0.
a+b
2
(a + b)2
According to the second solution of P 1.58-(a), we can write the inequality as follows
X a b(a − b)2
X a b(a − b)2
−
≥ 0,
(b + c)(c + a)
2(a + b)2
(b − c)2 Sa + (c − a)2 S b + (a − b)2 Sc ≥ 0,
where
bc
[2(b + c)2 − (a + b)(a + c)].
b+c
Without loss of generality, assume that a ≥ b ≥ c. We have Sc > 0 and
Sa =
ac
ac
[2(a + c)2 − (a + b)(b + c)] ≥
[2(a + c)2 − (2a)(a + c)]
a+c
a+c
2ac 2 (a + c)
=
≥ 0.
a+c
Sb =
If Sa + S b ≥ 0, then
X
(b − c)2 Sa ≥ (b − c)2 Sa + (a − c)2 S b ≥ (b − c)2 (Sa + S b ) ≥ 0.
The inequality Sa + S b ≥ 0 is true if
bc
ac
[2(a + c)2 − (a + b)(b + c)] ≥
[(a + b)(a + c) − 2(b + c)2 ].
a+c
b+c
Since
ac
bc
≥
,
a+c
b+c
it suffices to show that
2(a + c)2 − (a + b)(b + c) ≥ (a + b)(a + c) − 2(b + c)2 .
This is true since is equivalent to
(a − b)2 + 2c(a + b) + 4c 2 ≥ 0.
98
Vasile Cîrtoaje
P 1.62. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
bc
ca
1
a b + bc + ca
ab
+
+
+ ≥ 2
.
2
2
2
(a + b)
(b + c)
(c + a)
4
a + b2 + c 2
(Vasile Cîrtoaje, 2011)
First Solution. We use the SOS method. Write the inequality as follows
1−
•
˜
bc
a b + bc + ca X 1
−
≥
,
a2 + b2 + c 2
4 (b + c)2
X (b − c)2
(b − c)2
≥
,
a2 + b2 + c 2
(b + c)2
X
a2 + b2 + c 2
2
(b − c) 2 −
≥ 0.
(b + c)2
2
Since
2−
X
a 2
2bc − a2
a2 + b2 + c 2
=
1
+
≥
1
−
,
(b + c)2
(b + c)2
b+c
it suffices to show that
(b − c)2 Sa + (c − a)2 S b + (a − b)2 Sc ≥ 0,
where

‹2
c 2
a 2
b
, Sb = 1 −
, Sc = 1 −
.
Sa = 1 −
b+c
c+a
a+b
Without loss of generality, assume that a ≥ b ≥ c. Since S b ≥ 0 and Sc > 0, if b2 Sa +
a2 S b ≥ 0, then
X
a2
(b − c)2 Sa ≥ (b − c)2 Sa + (c − a)2 S b ≥ (b − c)2 Sa + 2 (b − c)2 S b
b
(b − c)2 (b2 Sa + a2 S b )
=
≥ 0.
b2
We have
‹

‹
ab 2
ab 2
b Sa + a S b = a + b −
−
b+c
c+a

‹2
•
a 2 ˜
b
= a2 1 −
+ b2 1 −
≥ 0.
b+c
c+a
2
2
2
2

The equality occurs for a = b = c, and for a = 0 and b = c (or any cyclic permutation).
Symmetric Rational Inequalities
99
Second Solution. Since (a + b)2 ≤ 2(a2 + b2 ), it suffices to prove that
ab
1
a b + bc + ca
+ ≥ 2
,
2
+b ) 4
a + b2 + c 2
X
2(a2
which is equivalent to
X 2a b
4(a b + bc + ca)
+1≥
,
2
2
a +b
a2 + b2 + c 2
X (a + b)2
≥2+
+
X (a + b)2
a2
b2
4(a b + bc + ca)
,
a2 + b2 + c 2
2(a + b + c)2
.
a2 + b2
a2 + b2 + c 2
The last inequality follows immediately by the Cauchy-Schwarz inequality
P
X (a + b)2
[ (a + b)]2
≥ P
.
a2 + b2
(a2 + b2 )
≥
Remark. The following generalization of the inequalities in P 1.61 and P 1.62 holds:
• Let a, b, c be nonnegative real numbers, no two of which are zero. If 0 ≤ k ≤ 2, then
X
4a b
a2 + b2 + c 2
a b + bc + ca
+
k
≥ 3k − 1 + 2(2 − k) 2
.
2
(a + b)
a b + bc + ca
a + b2 + c 2
with equality for a = b = c, and for a = 0 and b = c (or any cyclic permutation).
P 1.63. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
3a b
3bc
3ca
a b + bc + ca 5
+
+
≤ 2
+ .
2
2
2
(a + b)
(b + c)
(c + a)
a + b2 + c 2
4
(Vasile Cîrtoaje, 2011)
Solution. We use the SOS method. Write the inequality as follows
˜
X•1
bc
a b + bc + ca
3
−
≥1− 2
,
2
4 (b + c)
a + b2 + c 2
3
X (b − c)2
(b + c)2
≥2
X
(b − c)2
,
a2 + b2 + c 2
(b − c)2 Sa + (c − a)2 S b + (a − b)2 Sc ≥ 0,
100
Vasile Cîrtoaje
where
Sa =
3(a2 + b2 + c 2 )
3(a2 + b2 + c 2 )
3(a2 + b2 + c 2 )
−
2,
S
=
−
2,
S
=
− 2.
b
c
(b + c)2
(c + a)2
(a + b)2
Without loss of generality, assume that a ≥ b ≥ c. Since Sa > 0 and
Sb =
a2 + 3b2 + c 2 − 4ac
(a − 2c)2 + 3(b2 − c 2 )
=
> 0,
(c + a)2
(c + a)2
if S b + Sc ≥ 0, then
X
(b − c)2 Sa ≥ (c − a)2 S b + (a − b)2 Sc ≥ (a − b)2 (S b + Sc ) ≥ 0.
Using the Cauchy-Schwarz Inequality, we have
˜
•
1
1
+
−4
S b + Sc = 3(a2 + b2 + c 2 )
(c + a)2 (a + b)2
12(a2 + b2 + c 2 )
4(a − b − c)2 + 4(b − c)2
≥
−
4
=
≥ 0.
(c + a)2 + (a + b)2
(c + a)2 + (a + b)2
The equality occurs for a = b = c, and for
a
= b = c (or any cyclic permutation).
2
P 1.64. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
(a)
a3 + a bc b3 + a bc c 3 + a bc
+
+
≥ a2 + b2 + c 2 ;
b+c
c+a
a+b
(b)
a3 + 2a bc b3 + 2a bc c 3 + 2a bc
1
+
+
≥ (a + b + c)2 ;
b+c
c+a
a+b
2
(c)
a3 + 3a bc b3 + 3a bc c 3 + 3a bc
+
+
≥ 2(a b + bc + ca).
b+c
c+a
a+b
Solution. (a) First Solution. Write the inequality as
X a3 + a bc
2
− a ≥ 0,
b+c
X a(a − b)(a − c)
b+c
≥ 0.
Symmetric Rational Inequalities
101
Assume that a ≥ b ≥ c. Since (c − a)(c − b) ≥ 0 and
a(a − b)(a − c) b(b − c)(b − a) (a − b)2 (a2 + b2 + c 2 + a b)
+
=
≥ 0,
b+c
b+c
(b + c)(c + a)
the conclusion follows. The equality occurs for a = b = c, and for a = 0 and b = c (or
any cyclic permutation).
(b) Taking into account the inequality in (a), it suffices to show that
a bc
a bc
a bc
1
+
+
+ a2 + b2 + c 2 ≥ (a + b + c)2 ,
b+c c+a a+b
2
which is just the inequality (a) from P 1.58. The equality occurs for a = b = c, and for
a = 0 and b = c (or any cyclic permutation).
(c) The desired inequality follows by adding the inequality in (a) to the inequality (a)
from P 1.58. The equality occurs for a = b = c, and for a = 0 and b = c (or any cyclic
permutation).
P 1.65. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
a3 + 3a bc b3 + 3a bc c 3 + 3a bc
+
+
≥ a + b + c.
(b + c)2
(c + a)2
(a + b)2
(Vasile Cîrtoaje, 2005)
Solution. We use the SOS method. We have
X 3
X a3 + 3a bc X
X a3 + 3a bc
a − a(b2 − bc + c 2 )
−
a
=
−
a
=
(b + c)2
(b + c)2
(b + c)2
=
=
X a3 (b + c) − a(b3 + c 3 )
X a b(a2 − b2 )
+
=
(b + c)3
X ba(b2 − a2 )
X a b(a2 − b2 ) + ac(a2 − c 2 )
=
(b + c)3
X a b(a2 − b2 )[(c + a)3 − (b + c)3 ]
(b + c)3
(c + a)3
(b + c)3 (c + a)3
X a b(a + b)(a − b)2 [(c + a)2 + (c + a)(b + c) + (b + c)2 ]
=
≥ 0.
(b + c)3 (c + a)3
The equality occurs for a = b = c, and for a = 0 and b = c (or any cyclic permutation).
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Vasile Cîrtoaje
P 1.66. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
(a)
a3 + 3a bc b3 + 3a bc c 3 + 3a bc
3
+
+
≥ ;
3
3
3
(b + c)
(c + a)
(a + b)
2
(b)
3a3 + 13a bc 3b3 + 13a bc 3c 3 + 13a bc
+
+
≥ 6.
(b + c)3
(c + a)3
(a + b)3
(Vasile Cîrtoaje and Ji Chen, 2005)
Solution. (a) First Solution. Use the SOS method. We have
X a3 + 3a bc
(b + c)3
=
=
X a(b + c)2 + a(a2 + bc − b2 − c 2 )
3
2
3
=
2
3
=
2
3
=
2
≥
(b + c)3
X a3 − a(b2 − bc + c 2 )
a
+
b+c
(b + c)3
X a3 (b + c) − a(b3 + c 3 )
+
(b + c)4
X a b(a2 − b2 ) + ac(a2 − c 2 )
+
(b + c)4
X a b(a2 − b2 ) X ba(b2 − a2 )
+
+
(b + c)4
(c + a)4
X a b(a + b)(a − b)[(c + a)4 − (b + c)4 ]
+
≥ 0.
(b + c)4 (c + a)4
X
The equality occurs for a = b = c.
Second Solution. Assume that a ≥ b ≥ c. Since
a3 + 3a bc
b3 + 3a bc
c 3 + 3a bc
≥
≥
b+c
c + ac
a+b
and
1
1
1
≥
≥
,
2
2
(b + c)
(c + a)
(a + b)2
by Chebyshev’s inequality, we get
X a3 + 3a bc
(b + c)3
Thus, it suffices to show that
X
1 X a3 + 3a bc X
1
≥
.
3
b+c
(b + c)2
1
9
a3 + 3a bc X
≥ .
2
b+c
(b + c)
2
Symmetric Rational Inequalities
103
We can obtain this inequality by multiplying the known inequality (Iran-1996)
X
9
1
≥
2
(b + c)
4(a b + bc + ca)
and the inequality (c) from P 1.64.
(b) We have
X 3a3 + 13a bc
(b + c)3
=
=
(b + c)3
X 3a
X
X a3 − a(b2 − bc + c 2 )
1
+ 4a bc
+
3
.
b+c
(b + c)3
(b + c)3
Since
X
and
X 3a(b + c)2 + 4a bc + 3a(a2 + bc − b2 − c 2 )
1
3
≥
(b + c)3
(a + b)(b + c)(c + a)
X a3 − a(b2 − bc + c 2 )
(b + c)3
=
X a3 (b + c) − a(b3 + c 3 )
(b + c)4
X a b(a2 − b2 ) + ac(a2 − c 2 ) X a b(a2 − b2 ) X ba(b2 − a2 )
=
=
+
(b + c)4
(b + c)4
(c + a)4
X a b(a + b)(a − b)[(c + a)4 − (b + c)4 ]
≥ 0,
=
(b + c)4 (c + a)4
it suffices to prove that
X 3a
12a bc
+
≥ 6.
b + c (a + b)(b + c)(c + a)
This inequality is equivalent to the third degree Schur’s inequality
X
a3 + b3 + c 3 + 3a bc ≥
a b(a + b).
The equality occurs for a = b = c, and for a = 0 and b = c (or any cyclic permutation).
P 1.67. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
(a)
(b)
a3
b3
c3
3
+
+
+ a b + bc + ca ≥ (a2 + b2 + c 2 );
b+c c+a a+b
2
2a2 + bc 2b2 + ca 2c 2 + a b
9(a2 + b2 + c 2 )
+
+
≥
.
b+c
c+a
a+b
2(a + b + c)
(Vasile Cîrtoaje, 2006)
104
Vasile Cîrtoaje
Solution. (a) We apply the SOS method. Write the inequality as
X
X 2a3
2
(a − b)2 .
−a ≥
b+c
Since
X 2
X 2a3
a (a − b) + a2 (a − c)
2
−a =
b+c
b+c
=
X a2 (a − b)
b+c
+
X b2 (b − a) X (a − b)2 (a2 + b2 + a b + bc + ca)
=
,
c+a
(b + c)(c + a)
we can write the inequality as
(b − c)2 Sa + (c − a)2 S b + (a − b)2 Sc ≥ 0,
where
Sa = (b + c)(b2 + c 2 − a2 ), S b = (c + a)(c 2 + a2 − b2 ), Sc = (a + b)(a2 + b2 − c 2 ).
Without loss of generality, assume that a ≥ b ≥ c. Since S b ≥ 0, Sc ≥ 0, and
Sa + S b = (a + b)(a − b)2 + c 2 (a + b + 2c) ≥ 0,
we have
X
(b − c)2 Sa ≥ (b − c)2 Sa + (a − c)2 S b ≥ (b − c)2 (Sa + S b ) ≥ 0.
The equality occurs for a = b = c, and for a = 0 and b = c (or any cyclic permutation).
(b) Multiplying by a + b + c, the inequality can be written as
X
1+
or
a 9
(2a2 + bc) ≥ (a2 + b2 + c 2 ),
b+c
2
X 2a3 + a bc
5
+ a b + bc + ca ≥ (a2 + b2 + c 2 ).
b+c
2
This inequality follows using the inequality in (a) and the first inequality from P 1.58.
The equality occurs for a = b = c, and for a = 0 and b = c (or any cyclic permutation).
P 1.68. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
a(b + c)
b(c + a)
c(a + b)
+ 2
+ 2
≥ 2.
2
2
+ bc + c
c + ca + a
a + a b + b2
b2
Symmetric Rational Inequalities
105
First Solution. Apply the SOS method. We have
˜ X•
˜
•X
a(b + c)(a + b + c)
a(b + c)
−2 =
− 2a
(a + b + c)
b2 + bc + c 2
b2 + bc + c 2
=
X a(a b + ac − b2 − c 2 )
=
X a b(a − b) − ca(c − a)
b2 + bc + c 2
b2 + bc + c 2
X a b(a − b)
X a b(a − b)
=
−
b2 + bc + c 2
c 2 + ca + a2
X
a b(a − b)2
= (a + b + c)
≥ 0.
(b2 + bc + c 2 )(c 2 + ca + a2 )
The equality occurs for a = b = c, and for a = 0 and b = c (or any cyclic permutation).
Second Solution. By the AM-GM inequality, we have
4(b2 + bc + c 2 )(a b + bc + ca) ≤ (b2 + bc + c 2 + a b + bc + ca)2 = (b + c)2 (a + b + c)2 .
Thus,
X
a(b + c)(a b + bc + ca)
a(b + c)
=
2
2
2
b + bc + c
(b + bc + c 2 )(a b + bc + ca)
X 4a(a b + bc + ca)
4(a b + bc + ca) X a
≥
=
,
(b + c)(a + b + c)2
(a + b + c)2
b+c
X
and it suffices to show that
X
(a + b + c)2
a
≥
.
b+c
2(a b + bc + ca)
This follows immediately from the Cauchy-Schwarz inequality
X
a
(a + b + c)2
≥ P
.
b+c
a(b + c)
Third Solution. By the Cauchy-Schwarz inequality, we have
X
a(b + c)
(a + b + c)2
≥
.
P a(b2 + bc + c 2 )
b2 + bc + c 2
b+c
Thus, it is enough to show that
(a + b + c)2 ≥ 2
Since
X a(b2 + bc + c 2 )
b+c
.

‹
a(b2 + bc + c 2 )
bc
a bc
=a b+c−
= a b + ca −
,
b+c
b+c
b+c
106
Vasile Cîrtoaje
this inequality is equivalent to

‹
1
1
1
2a bc
+
+
+ a2 + b2 + c 2 ≥ 2(a b + bc + ca),
b+c c+a a+b
which is just the inequality (a) from P 1.58.
Fourth Solution. By direct calculation, we can write the inequality as
X
X
a b(a4 + b4 ) ≥
a2 b2 (a2 + b2 ),
which is equivalent to the obvious inequality
X
a b(a − b)(a3 − b3 ) ≥ 0.
P 1.69. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
Y  a − b ‹2
a(b + c)
b(c + a)
c(a + b)
+
+
≥2+4
.
b2 + bc + c 2 c 2 + ca + a2 a2 + a b + b2
a+b
(Vasile Cîrtoaje, 2011)
Solution. For b = c = 1, the inequality reduces to a(a − 1)2 ≥ 0. Assume further that
a < b < c. According to the first solution of P 1.68, we have
X
X
a(b + c)
bc(b − c)2
−
2
=
.
b2 + bc + c 2
(a2 + a b + b2 )(a2 + ac + c 2 )
Therefore, it remains to show that
X
Y  a − b ‹2
bc(b − c)2
≥4
.
(a2 + a b + b2 )(a2 + ac + c 2 )
a+b
Since
(a2 + a b + b2 )(a2 + ac + c 2 ) ≤ (a + b)2 (a + c)2 ,
it suffices to show that
X
Y  a − b ‹2
bc(b − c)2
≥4
,
(a + b)2 (a + c)2
a+b
which is equivalent to
X
bc(b + c)2
≥ 4.
(a − b)2 (a − c)2
Symmetric Rational Inequalities
107
We have
X
bc(b + c)2
bc(b + c)2
≥
(a − b)2 (a − c)2
(a − b)2 (a − c)2
bc(b + c)2
(b + c)2
≥ 4.
≥
=
b2 c 2
bc
The equality occurs for a = b = c, and for a = 0 and b = c (or any cyclic permutation).
P 1.70. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
a b − bc + ca bc − ca + a b ca − a b + bc
3
+
+
≥ .
b2 + c 2
c 2 + a2
a2 + b2
2
Solution. Use the SOS method. We have
X  a b − bc + ca 1 ‹ X (b + c)(2a − b − c)
−
=
b2 + c 2
2
2(b2 + c 2 )
=
=
X (b + c)(a − b)
2(b2 + c 2 )
X (b + c)(a − b)
+
+
X (b + c)(a − c)
2(b2 + c 2 )
X (c + a)(b − a)
2(b2 + c 2 )
2(c 2 + a2 )
X (a − b)2 (a b + bc + ca − c 2 )
=
.
2(b2 + c 2 )(c 2 + a2 )
Since
a b + bc + ca − c 2 = (b − c)(c − a) + 2a b ≥ (b − c)(c − a),
it suffices to show that
X
(a2 + b2 )(a − b)2 (b − c)(c − a) ≥ 0.
This inequality is equivalent to
(a − b)(b − c)(c − a)
X
(a − b)(a2 + b2 ) ≥ 0,
or
(a − b)2 (b − c)2 (c − a)2 ≥ 0.
The equality occurs for a = b = c, and for a = 0 and b = c (or any cyclic permutation).
108
Vasile Cîrtoaje
P 1.71. Let a, b, c be nonnegative real numbers, no two of which are zero. If k > −2, then
X a b + (k − 1)bc + ca
b2
+ k bc
≥
+ c2
3(k + 1)
.
k+2
(Vasile Cîrtoaje, 2005)
First Solution. Apply the SOS method. Write the inequality as
X • a b + (k − 1)bc + ca
b2 + k bc + c 2
X
b2
˜
k+1
−
≥ 0,
k+2
A
≥ 0,
+ k bc + c 2
where
A = (b + c)(2a − b − c) + k(a b + ac − b2 − c 2 ).
Since
A =(b + c)[(a − b) + (a − c)] + k[b(a − b) + c(a − c)]
= (a − b)[(k + 1)b + c] + (a − c)[(k + 1)c + b],
the inequality is equivalent to
X (a − b)[(k + 1)b + c]
b2 + k bc + c 2
X (a − b)[(k + 1)b + c]
+
+
X (a − c)[(k + 1)c + b]
b2 + k bc + c 2
X (b − a)[(k + 1)a + c]
b2 + k bc + c 2
c 2 + kca + a2
X
(b − c)2 R a Sa ≥ 0,
≥ 0,
≥ 0,
where
R a = b2 + k bc + c 2 , Sa = a(b + c − a) + (k + 1)bc.
Without loss of generality, assume that a ≥ b ≥ c.
Case 1: k ≥ −1. Since Sa ≥ a(b + c − a), it suffices to show that
X
a(b + c − a)(b − c)2 R a ≥ 0.
We have
X
a(b + c − a)(b − c)2 R a ≥ a(b + c − a)(b − c)2 R a + b(c + a − b)(c − a)2 R b
≥ (b − c)2 [a(b + c − a)R a + b(c + a − b)R b ].
Symmetric Rational Inequalities
109
Thus, it is enough to prove that
a(b + c − a)R a + b(c + a − b)R b ≥ 0.
Since b + c − a ≥ −(c + a − b), we have
a(b + c − a)R a + b(c + a − b)R b ≥ (c + a − b)(bR b − aR a )
= (c + a − b)(a − b)(a b − c 2 ) ≥ 0.
Case 2: −2 < k ≤ 1. Since
Sa = (a − b)(c − a) + (k + 2)bc ≥ (a − b)(c − a),
we have
X
X
(b − c)2 R a Sa ≥ (a − b)(b − c)(c − a)
(b − c)R a .
From
X
X
(b − c)R a =
(b − c)[b2 + bc + c 2 − (1 − k)bc]
X
X
=
(b3 − c 3 ) − (1 − k)
bc(b − c)
= (1 − k)(a − b)(b − c)(c − a),
we get
(a − b)(b − c)(c − a)
X
(b − c)R a = (1 − k)(a − b)2 (b − c)2 (c − a)2 ≥ 0.
This completes the proof. The equality occurs for a = b = c, and for a = 0 and b = c
(or any cyclic permutation).
Second Solution. Let p = a + b + c, q = a b + bc + ca, r = a bc. Write the inequality in
the form f6 (a, b, c) ≥ 0, where
X
[a(b + c) + (k − 1)bc](a2 + ka b + b2 )(a2 + kac + c 2 )
f6 (a, b, c) = (k + 2)
−3(k + 1)
= (k + 2)
Y
(b2 + k bc + c 2 )
X
[(k − 2)bc + q](ka b − c 2 + p2 − 2q)(kac − b2 + p2 − 2q)
Y
−3(k + 1)
(k bc − a2 + p2 − 2q).
Thus, f6 (a, b, c) has the same highest coefficient A as
(k + 2)(k − 2)P2 (a, b, c) − 3(k + 1)P3 (a, b, c),
where
P2 (a, b, c) =
X
bc(ka b − c 2 )(kac − b2 ),
110
Vasile Cîrtoaje
P3 (a, b, c) =
Y
(k bc − a2 ).
According to Remark 2 from the proof of P 2.75 in Volume 1,
A = (k + 2)(k − 2)P2 (1, 1, 1) − 3(k + 1)P3 (1, 1, 1)
= 3(k + 2)(k − 2)(k − 1)2 − 3(k + 1)(k − 1)3 = −9(k − 1)2 .
Since A ≤ 0, according to P 3.76-(a) in Volume 1, it suffices to prove the original inequality for b = c = 1, and for a = 0. For these cases, this inequality has the obvious
forms
(k + 2)a(a − 1)2 ≥ 0
and
(b − c)2 [(k + 2)(b2 + c 2 ) + (k2 + k + 1)bc] ≥ 0,
respectively.
Remark. For k = 1 and k = 0, from P 1.71, we get the inequalities in P 1.68 and P 1.70,
respectively. Besides, for k = 2, we get the well-known inequality (Iran 1996):
1
1
1
9
+
+
≥
.
2
2
2
(a + b)
(b + c)
(c + a)
4(a b + bc + ca)
P 1.72. Let a, b, c be nonnegative real numbers, no two of which are zero. If k > −2, then
X 3bc − a(b + c)
b2 + k bc + c 2
≤
3
.
k+2
(Vasile Cîrtoaje, 2011)
Solution. Write the inequality in P 1.71 as
˜
X•
a b + (k − 1)bc + ca
3
1−
≤
,
2
2
b + k bc + c
k+2
X b2 + c 2 + bc − a(b + c)
3
≤
.
b2 + k bc + c 2
k+2
Since b2 + c 2 ≥ 2bc, we get
X 3bc − a(b + c)
b2
+ k bc
+ c2
≤
3
,
k+2
which is just the desired inequality. The equality occurs for a = b = c.
Symmetric Rational Inequalities
111
P 1.73. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. Prove that
bc + 1
ca + 1
4
ab + 1
+ 2
+ 2
≥ .
2
2
2
2
a +b
b +c
c +a
3
Solution. Write the inequality in the homogeneous form E(a, b, c) ≥ 4, where
E(a, b, c) =
4a b + bc + ca 4bc + ca + a b 4ca + a b + bc
+
+
.
a2 + b2
b2 + c 2
c 2 + a2
Without loss of generality, assume that a = min{a, b, c}. We will show that
E(a, b, c) ≥ E(0, b, c) ≥ 4.
For a = 0, we have E(a, b, c) = E(0, b, c), and for a > 0, we have
E(a, b, c) − E(0, b, c) 4b2 + c(b − a)
b+c
4c 2 + b(c − a)
=
+
+
> 0.
a
b(a2 + b2 )
b2 + c 2
c(c 2 + a2 )
Also,
b
4bc
c
(b − c)4
+ 2
+
−
4
=
≥ 0.
c
b + c2 b
bc(b2 + c 2 )
p
The equality holds for a = 0 and b = c = 3 (or any cyclic permutation).
E(0, b, c) − 4 =
P 1.74. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. Prove that
5a b + 1 5bc + 1 5ca + 1
+
+
≥ 2.
(a + b)2 (b + c)2 (c + a)2
Solution. Write the inequality as E(a, b, c) ≥ 6, where
E(a, b, c) =
16a b + bc + ca 16bc + ca + a b 16ca + a b + bc
+
+
.
(a + b)2
(b + c)2
(c + a)2
Without loss of generality, assume that a ≤ b ≤ c.
Case 1: 16b2 ≥ c(a + b). We will show that
E(a, b, c) ≥ E(0, b, c) ≥ 6.
For a = 0, we have E(a, b, c) = E(0, b, c), and for a > 0, we have
E(a, b, c) − E(0, b, c) 16b2 − c(a + b)
1
16c 2 − b(a + c)
=
+
+
> 0.
a
b(a + b)2
b+c
c(c + a)2
112
Vasile Cîrtoaje
Also,
E(0, b, c) − 6 =
16bc
c
(b − c)4
b
+
+
−
6
=
≥ 0.
c (b + c)2 b
bc(b + c)2
Case 2: 16b2 < c(a + b). We have
16a b + 16b2
2(5b − 3a)
16a b + bc + ca
−
6
>
−6=
> 0.
2
2
(a + b)
(a + b)
a+b
p
The equality holds for a = 0 and b = c = 3 (or any cyclic permutation).
E(a, b, c) − 6 >
P 1.75. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
a2 − bc
b2 − ca
c2 − a b
+
+
≥ 0.
2b2 − 3bc + 2c 2 2c 2 − 3ca + 2a2 2a2 − 3a b + 2b2
(Vasile Cîrtoaje, 2005)
Solution. The hint is applying the Cauchy-Schwarz inequality after we made the numerators of the fractions to be nonnegative and as small as possible. Thus, we write the
inequality as
X
a2 − bc
+ 1 ≥ 3,
2b2 − 3bc + 2c 2
X a2 + 2(b − c)2
≥ 3.
2b2 − 3bc + 2c 2
Using the Cauchy-Schwarz inequality
P
P
X a2 + 2(b − c)2
(5 a2 − 4 a b)2
≥P
,
2b2 − 3bc + 2c 2
[a2 + 2(b − c)2 ](2b2 − 3bc + 2c 2 )
it suffices to prove that
X
X
X
(5
a2 − 4
a b)2 ≥ 3
[a2 + 2(b − c)2 ](2b2 − 3bc + 2c 2 ).
This inequality is equivalent to
X
X
X
X
a4 + a bc
a+2
a b(a2 + b2 ) ≥ 6
a2 b2 .
We can obtain it by summing Schur’s inequality of degree four
X
X
X
a4 + a bc
a≥
a b(a2 + b2 )
to the obvious inequality
3
X
a b(a2 + b2 ) ≥ 6
X
a2 b2 .
The equality holds for a = b = c, and for a = 0 and b = c (or any cyclic permutation).
Symmetric Rational Inequalities
113
P 1.76. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
2b2 − ca
2c 2 − a b
2a2 − bc
+
+
≥ 3.
b2 − bc + c 2 c 2 − ca + a2 a2 − a b + b2
(Vasile Cîrtoaje, 2005)
Solution. Write the inequality such that the numerators of the fractions are nonnegative
and as small as possible:
X 2a2 − bc
+ 1 ≥ 6,
b2 − bc + c 2
X 2a2 + (b − c)2
≥ 6.
b2 − bc + c 2
Applying the Cauchy-Schwarz inequality, we get
P
P
X 2a2 + (b − c)2
4(2 a2 − a b)2
≥P
.
b2 − bc + c 2
[2a2 + (b − c)2 ](b2 − bc + c 2 )
Thus, we still have to prove that
X
X
X
2(2
a2 −
a b)2 ≥ 3
[2a2 + (b − c)2 ](b2 − bc + c 2 ).
This inequality is equivalent to
X
X
X
X
2
a4 + 2a bc
a+
a b(a2 + b2 ) ≥ 6
a2 b2 .
We can obtain it by summing up Schur’s inequality of degree four
X
X
X
a4 + a bc
a≥
a b(a2 + b2 )
and
X
a b(a2 + b2 ) ≥ 2
X
a2 b2 ,
multiplied by 2 and 3, respectively. The equality holds for a = b = c, and for a = 0 and
b = c (or any cyclic permutation).
P 1.77. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
a2
b2
c2
+
+
≥ 1.
2b2 − bc + 2c 2 2c 2 − ca + 2a2 2a2 − a b + 2b2
(Vasile Cîrtoaje, 2005)
114
Vasile Cîrtoaje
Solution. By the Cauchy-Schwarz inequality, we have
P
X
( a2 )2
a2
≥P
.
2b2 − bc + 2c 2
a2 (2b2 − bc + 2c 2 )
Therefore, it suffices to show that
X
X
(
a2 )2 ≥
a2 (2b2 − bc + 2c 2 ),
which is equivalent to
X
a4 + a bc
X
a≥2
X
a2 b2 .
This inequality follows by adding Schur’s inequality of degree four
X
X
X
a4 + a bc
a≥
a b(a2 + b2 )
and
X
a b(a2 + b2 ) ≥ 2
X
a2 b2 .
The equality holds for a = b = c, and for a = 0 and b = c (or any cyclic permutation).
P 1.78. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
4b2
1
1
1
9
+ 2
+ 2
≥
.
2
2
2
2
− bc + 4c
4c − ca + 4a
4a − a b + 4b
7(a + b2 + c 2 )
(Vasile Cîrtoaje, 2005)
Solution. We use the SOS method. Without loss of generality, assume that a ≥ b ≥ c.
Write the inequality as
X 7(a2 + b2 + c 2 )
− 3 ≥ 0,
4b2 − bc + 4c 2
X 7a2 − 5b2 − 5c 2 + 3bc
≥ 0,
4b2 − bc + 4c 2
X 5(2a2 − b2 − c 2 ) − 3(a2 − bc)
≥ 0.
4b2 − bc + 4c 2
Since
2(a2 − bc) = (a − b)(a + c) + (a − c)(a + b),
we have
10(2a2 − b2 − c 2 ) − 6(a2 − bc) =
= 10(a2 − b2 ) − 3(a − b)(a + c) + 10(a2 − c 2 ) − 3(a − c)(a + b)
= (a − b)(7a + 10b − 3c) + (a − c)(7a + 10c − 3b).
Symmetric Rational Inequalities
115
Thus, we can write the desired inequality as follows
X (a − b)(7a + 10b − 3c)
4b2 − bc + 4c 2
X (a − b)(7a + 10b − 3c)
4b2 − bc + 4c 2
+
+
X (a − c)(7a + 10c − 3b)
4b2 − bc + 4c 2
≥ 0,
X (b − a)(7b + 10a − 3c)
4c 2 − ca + 4a2
≥ 0,
X (a − b)2 (28a2 + 28b2 − 9c 2 + 68a b − 19bc − 19ca)
(4b2 − bc + 4c 2 )(4c 2 − ca + 4a2 )
,
(b − c)2 Sa + (c − a)2 S b + (a − b)2 Sc ≥ 0,
where
Sa = (4b2 − bc + 4c 2 )[(b − a)(28b + 9a) + c(−19a + 68b + 28c)],
S b = (4c 2 − ca + 4a2 )[(a − b)(28a + 9b) + c(−19b + 68a + 28c)],
Sc = (4a2 − a b + 4b2 )[(b − c)(28b + 9c) + a(68b − 19c + 28a)].
Since S b ≥ 0 and Sc > 0, it suffices to show that Sa + S b ≥ 0. We have
Sa ≥ (4b2 − bc + 4c 2 )[(b − a)(28b + 9a) − 19ac],
S b ≥ (4c 2 − ca + 4a2 )[(a − b)(28a + 9b) + 19ac],
19ac[(4c 2 − ca + 4a2 ) − (4b2 − bc + 4c 2 ] = 19ac(a − b)(4a + 4b − c) ≥ 0,
and hence
Sa + S b ≥ (a − b)[−(4b2 − bc + 4c 2 )(28b + 9a) + (4c 2 − ca + 4a2 )(28a + 9b)]
= (a − b)2 [112(a2 + a b + b2 ) + 76c 2 − 28c(a + b)] ≥ 0.
The equality holds for a = b = c, and for a = 0 and b = c (or any cyclic permutation).
P 1.79. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
2a2 + bc 2b2 + ca 2c 2 + a b
9
+
+
≥
.
b2 + c 2
c 2 + a2
a2 + b2
2
(Vasile Cîrtoaje, 2005)
116
Vasile Cîrtoaje
First Solution. We apply the SOS method. Since
X 2a2 − b2 − c 2 X (b − c)2
X 2(2a2 + bc)
−
3
=
2
−
b2 + c 2
b2 + c 2
b2 + c 2
and
X 2a2 − b2 − c 2
b2 + c 2
=
X a2 − b2
b2 + c 2
+
X a2 − c 2
=
X a2 − b2
+
X b2 − a2
c 2 + a2
b2 + c 2
b2 + c 2
‹ X

X
1
(a2 − b2 )2
1
−
=
=
(a2 − b2 ) 2
b + c 2 c 2 + a2
(b2 + c 2 )(c 2 + a2 )
≥
X (a − b)2 (a2 + b2 )
(b2 + c 2 )(c 2 + a2 )
we can write the inequality as
2
X (b − c)2 (b2 + c 2 )
X (b − c)2
≥
,
(c 2 + a2 )(a2 + b2 )
b2 + c 2
or
(b − c)2 Sa + (c − a)2 S b + (c − a)2 Sc ≥ 0,
where
Sa = 2(b2 + c 2 )2 − (c 2 + a2 )(a2 + b2 ).
Assume that a ≥ b ≥ c. We have
S b = 2(c 2 + a2 )2 − (a2 + b2 )(b2 + c 2 )
≥ 2(c 2 + a2 )(c 2 + b2 ) − (a2 + b2 )(b2 + c 2 )
= (b2 + c 2 )(a2 − b2 + 2c 2 ) ≥ 0,
Sc = 2(a2 + b2 )2 − (b2 + c 2 )(c 2 + a2 ) > 0,
Sa + S b = (a2 − b2 )2 + 2c 2 (a2 + b2 + 2c 2 ) ≥ 0.
Therefore,
(b − c)2 Sa + (c − a)2 S b + (c − a)2 Sc ≥ (b − c)2 Sa + (c − a)2 S b
≥ (b − c)2 (Sa + S b ) ≥ 0.
The equality holds for a = b = c, and for a = 0 and b = c (or any cyclic permutation).
Second Solution. Since
bc ≥
2b2 c 2
,
b2 + c 2
Symmetric Rational Inequalities
117
we have
X 2a2 + bc
b2 + c 2
≥
2b2 c 2
b2 +c 2
b2 + c 2
X 2a2 +
= 2(a2 b2 + b2 c 2 + c 2 a2 )
X
1
.
(b2 + c 2 )2
Therefore, it suffices to show that
X
1
9
≥
,
(b2 + c 2 )2
4(a2 b2 + b2 c 2 + c 2 a2 )
which is just the known Iran-1996 inequality (see Remark from P 1.71).
Third Solution. We get the desired inequality by summing the inequality in P 1.59-(a),
namely
2a2 − 2bc 2b2 − 2ca 2c 2 − 2a b 6(a b + bc + ca)
+ 2
+ 2
+
≥ 6,
b2 + c 2
c + a2
a + b2
a2 + b2 + c 2
and the inequality
3bc
3ca
3a b
3 6(a b + bc + ca)
+ 2
+ 2
+ ≥
.
2
2
2
+c
c +a
a +b
2
a2 + b2 + c 2
b2
This inequality is equivalent to
‹
X  2bc
4(a b + bc + ca)
+1 ≥
+ 2,
2
2
b +c
a2 + b2 + c 2
X (b + c)2
b2 + c 2
≥
2(a + b + c)2
.
a2 + b2 + c 2
By the Cauchy-Schwarz inequality, we have
X (b + c)2
b2 + c 2
P
2
(b + c)
2(a + b + c)2
≥ P
= 2
.
a + b2 + c 2
(b2 + c 2 )
P 1.80. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
2a2 + 3bc
2b2 + 3ca
2c 2 + 3a b
+
+
≥ 5.
b2 + bc + c 2 c 2 + ca + a2 a2 + a b + b2
(Vasile Cîrtoaje, 2005)
118
Vasile Cîrtoaje
Solution. We apply the SOS method. Write the inequality as
X 3(2a2 + 3bc)
− 5 ≥ 0,
b2 + bc + c 2
or
X 6a2 + 4bc − 5b2 − 5c 2
b2 + bc + c 2
≥ 0.
Since
2(a2 − bc) = (a − b)(a + c) + (a − c)(a + b),
we have
6a2 + 4bc − 5b2 − 5c 2 = 5(2a2 − b2 − c 2 ) − 4(a2 − bc)
= 5(a2 − b2 ) − 2(a − b)(a + c) + 5(a2 − c 2 ) − 2(a − c)(a + b)
= (a − b)(3a + 5b − 2c) + (a − c)(3a + 5c − 2b).
Thus, we can write the desired inequality as follows
X (a − b)(3a + 5b − 2c)
b2 + bc + c 2
X (a − b)(3a + 5b − 2c)
+
+
X (a − c)(3a + 5c − 2b)
b2 + bc + c 2
X (b − a)(3b + 5a − 2c)
b2 + bc + c 2
c 2 + ca + a2
X (a − b)2 (3a2 + 3b2 − 4c 2 + 8a b + bc + ca)
(b2 + bc + c 2 )(c 2 + ca + a2 )
≥ 0,
≥ 0,
,
(b − c)2 Sa + (c − a)2 S b + (a − b)2 Sc ≥ 0,
where
Sa = (b2 + bc + c 2 )(3b2 + 3c 2 − 4a2 + a b + 8bc + ca).
Assume that a ≥ b ≥ c. Since
S b = (c 2 + ca + a2 )[(a − b)(3a + 4b) + c(8a + b + c)] ≥ 0
and
Sc = (a2 + a b + b2 )[(b − c)(3b + 4c) + a(8b + c + a)] > 0,
it suffices to show that Sa + S b ≥ 0. We have
Sa + S b ≥ (b2 + bc + c 2 )(b − a)(3b + 4a) + (c 2 + ca + a2 )(a − b)(3a + 4b)
= (a − b)2 [3(a + b)(a + b + c) + a b − c 2 ] ≥ 0.
The equality holds for a = b = c, and for a = 0 and b = c (or any cyclic permutation).
Symmetric Rational Inequalities
119
P 1.81. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
21
2a2 + 5bc 2b2 + 5ca 2c 2 + 5a b
+
+
≥
.
(b + c)2
(c + a)2
(a + b)2
4
(Vasile Cîrtoaje, 2005)
Solution. Use the SOS method.Write the inequality as follows
X 2a2 + 5bc 7 ≥ 0,
−
(b + c)2
4
X 4(a2 − b2 ) + 4(a2 − c 2 ) − 3(b − c)2
(b + c)2
X c 2 − b2
≥ 0,
X (b − c)2
X b2 − c 2
+
4
−
3
≥ 0,
(c + a)2
(a + b)2
(b + c)2
X (b − c)2 (b + c)(2a + b + c)
X (b − c)2
4
−
3
≥ 0.
(c + a)2 (a + b)2
(b + c)2
Substituting b + c = x, c + a = y and a + b = z, we can rewrite the inequality in the
form
( y − z)2 S x + (z − x)2 S y + (x − y)2 Sz ≥ 0,
4
where
S x = 4x 3 ( y + z) − 3 y 2 z 2 , S y = 4 y 3 (z + x) − 3z 2 x 2 , Sz = 4z 3 (x + y) − 3x 2 y 2 .
Without loss of generality, assume that 0 < x ≤ y ≤ z, z ≤ x + y. We have Sz > 0 and
S y ≥ 4x 2 y(z + x) − 3x 2 z(x + y) = x 2 [4x y + z( y − 3x)] ≥ 0,
since for the nontrivial case y − 3x < 0, we get
4x y + z( y − 3x) ≥ 4x y + (x + y)( y − 3x) = x 2 (3x + y)( y − x) ≥ 0.
Thus, it suffices to show that S x + S y ≥ 0. Since
S x + S y = 4x y(x 2 + y 2 ) + 4(x 3 + y 3 )z − 3(x 2 + y 2 )z 2
≥ 4x y(x 2 + y 2 ) + 4(x 3 + y 3 )z − 3(x 2 + y 2 )(x + y)z
= 4x y(x 2 + y 2 ) + (x 2 − 4x y + y 2 )(x + y)z,
for the nontrivial case x 2 − 4x y + y 2 < 0, we get
S x + S y ≥ 4x y(x 2 + y 2 ) + (x 2 − 4x y + y 2 )(x + y)2
≥ 2x y(x + y)2 + (x 2 − 4x y + y 2 )(x + y)2
= (x − y)2 (x + y)2 .
The equality holds for a = b = c, and for a = 0 and b = c (or any cyclic permutation).
120
Vasile Cîrtoaje
P 1.82. Let a, b, c be nonnegative real numbers, no two of which are zero. If k > −2, then
X 2a2 + (2k + 1)bc
b2 + k bc + c 2
≥
3(2k + 3)
.
k+2
(Vasile Cîrtoaje, 2005)
First Solution. There are two cases to consider.
Case 1: −2 < k ≤ −1/2. Write the inequality as
X 2a2 + (2k + 1)bc 2k + 1 6
−
≥
,
2
2
b + k bc + c
k+2
k+2
X 2(k + 2)a2 − (2k + 1)(b − c)2
b2 + k bc + c 2
≥ 6.
Since 2(k + 2)a2 − (2k + 1)(b − c)2 ≥ 0 for −2 < k ≤ −1/2, we can apply the CauchySchwarz inequality. Thus, it suffices to show that
P
P
[2(k + 2) a2 − (2k + 1) (b − c)2 ]2
P
≥ 6,
[2(k + 2)a2 − (2k + 1)(b − c)2 ](b2 + k bc + c 2 )
which is equivalent to each of the following inequalities
P
P
2[(1 − k) a2 + (2k + 1) a b]2
P
≥ 3,
[2(k + 2)a2 − (2k + 1)(b − c)2 ](b2 + k bc + c 2 )
X
X
X
X
2(k + 2)
a4 + 2(k + 2)a bc
a − (2k + 1)
a b(a2 + b2 ) ≥ 6
a2 b2 ,
X
X
X
X
2(k + 2)[
a4 + a bc
a−
a b(a2 + b2 )] + 3
a b(a − b)2 ≥ 0.
The last inequality is true since, by Schur’s inequality of degree four, we have
X
X
X
a4 + a bc
a−
a b(a2 + b2 ) ≥ 0.
Case 2: k ≥ −9/5. Use the SOS method. Without loss of generality, assume that a ≥
b ≥ c. Write the inequality as
X 2a2 + (2k + 1)bc 2k + 3 −
≥ 0,
b2 + k bc + c 2
k+2
X 2(k + 2)a2 − (2k + 3)(b2 + c 2 ) + 2(k + 1)bc
b2 + k bc + c 2
X (2k + 3)(2a2 − b2 − c 2 ) − 2(k + 1)(a2 − bc)
b2 + k bc + c 2
≥ 0,
≥ 0.
Symmetric Rational Inequalities
121
Since
2(a2 − bc) = (a − b)(a + c) + (a − c)(a + b),
we have
(2k + 3)(2a2 − b2 − c 2 ) − 2(k + 1)(a2 − bc) =
= (2k + 3)(a2 − b2 ) − (k + 1)(a − b)(a + c) + (2k + 3)(a2 − c 2 ) − (k + 1)(a − c)(a + b)
= (a − b)[(k + 2)a + (2k + 3)b − (k + 1)c] + (a − c)[(k + 2)a + (2k + 3)c − (k + 1)b).
Thus, we can write the desired inequality as
X (a − b)[(k + 2)a + (2k + 3)b − (k + 1)c]
b2 + k bc + c 2
+
or
X (a − c)[(k + 2)a + (2k + 3)c − (k + 1)b]
b2 + k bc + c 2
+
≥ 0,
X (a − b)[(k + 2)a + (2k + 3)b − (k + 1)c]
+
b2 + k bc + c 2
X (b − a)[(k + 2)b + (2k + 3)a − (k + 1)c]
≥ 0,
c 2 + kca + a2
or
(b − c)2 R a Sa + (c − a)2 R b S b + (a − b)2 R c Sc ≥ 0,
where
R a = b2 + k bc + c 2 , R b = c 2 + kca + a2 , R c = a2 + ka b + b2 ,
Sa = (k + 2)(b2 + c 2 ) − (k + 1)2 a2 + (3k + 5)bc + (k2 + k − 1)a(b + c)
= −(a − b)[(k + 1)2 a + (k + 2)b] + c[(k2 + k − 1)a + (3k + 5)b + (k + 2)c],
S b = (k + 2)(c 2 + a2 ) − (k + 1)2 b2 + (3k + 5)ca + (k2 + k − 1)b(c + a)
= (a − b)[(k + 2)a + (k + 1)2 b] + c[(3k + 5)a + (k2 + k − 1)b + (k + 2)c],
Sc = (k + 2)(a2 + b2 ) − (k + 1)2 c 2 + (3k + 5)a b + (k2 + k − 1)c(a + b)
= (k + 2)(a2 + b2 ) + (3k + 5)a b + c[(k2 + k − 1)(a + b) − (k + 1)2 c]
≥ (5k + 9)a b + c[(k2 + k − 1)(a + b) − (k + 1)2 c].
We have S b ≥ 0, since for the nontrivial case
(3k + 5)a + (k2 + k − 1)b + (k + 2)c < 0,
we get
S b ≥ (a − b)[(k + 2)a + (k + 1)2 b] + b[(3k + 5)a + (k2 + k − 1)b + (k + 2)c]
= (k + 2)(a2 − b2 ) + (k + 2)2 a b + (k + 2)bc > 0.
122
Vasile Cîrtoaje
Also, we have Sc ≥ 0 for k ≥ −9/5, since
(5k + 9)a b + c[(k2 + k − 1)(a + b) − (k + 1)2 c] ≥
≥ (5k + 9)ac + c[(k2 + k − 1)(a + b) − (k + 1)2 c]
= (k + 2)(k + 4)ac + (k2 + k − 1)bc − (k + 1)2 c 2
≥ (2k2 + 7k + 7)bc − (k + 1)2 c 2
≥ (k + 2)(k + 3)c 2 ≥ 0.
Therefore, it suffices to show that R a Sa + R b S b ≥ 0. Since
bR b − aR a = (a − b)(a b − c 2 ) ≥ 0
and
R a Sa + R b S b ≥ R a (Sa +
it suffices to show that
Sa +
a
S b ),
b
a
S b ≥ 0.
b
We have
bSa + aS b = (k + 2)(a + b)(a − b)2 + c f (a, b, c)
≥ 2(k + 2)b(a − b)2 + c f (a, b, c),
Sa +
c
a
S b ≥ 2(k + 2)(a − b)2 + f (a, b, c),
b
b
where
f (a, b, c) = b[(k2 + k − 1)a + (3k + 5)b] + a[(3k + 5)a + (k2 + k − 1)b]
+(k + 2)c(a + b) = (3k + 5)(a2 + b2 ) + 2(k2 + k − 1)a b + (k + 2)c(a + b).
For the nontrivial case f (a, b, c) < 0, we have
Sa +
a
S b ≥ 2(k + 2)(a − b)2 + f (a, b, c)
b
≥ 2(k + 2)(a − b)2 + (3k + 5)(a2 + b2 ) + 2(k2 + k − 1)a b
= (5k + 9)(a2 + b2 ) + 2(k2 − k − 5)a b ≥ 2(k + 2)2 a b ≥ 0.
The proof is completed. The equality holds for a = b = c, and for a = 0 and b = c (or
any cyclic permutation).
Second Solution. Let
p = a + b + c,
q = a b + bc + ca.
Symmetric Rational Inequalities
123
Write the inequality as f6 (a, b, c) ≥ 0, where
X
f6 (a, b, c) = (k + 2)
[2a2 + (2k + 1)bc](a2 + ka b + b2 )(a2 + kac + c 2 )
−3(2k + 3)
Y
(b2 + k bc + c 2 ).
Since
(a2 + ka b + b2 )(a2 + kac + c 2 ) = (p2 − 2q + ka b − c 2 )(p2 − 2q + kac − b2 ),
b2 + k bc + c 2 = p2 − 2q + k bc − a2 ,
f6 (a, b, c has the same highest coefficient A as
(k + 2)P2 (a, b, c) − 3(2k + 3)P3 (a, b, c),
where
P2 (a, b, c) =
X
Y
[2a2 +(2k +1)bc](ka b − c 2 )(kac − b2 ), P3 (a, b, c) =
(b2 + k bc + c 2 ).
According to Remark 2 from the proof of P 2.75 in Volume 1, we have
A = (k + 2)P2 (1, 1, 1) − 3(2k + 3)P3 (1, 1, 1) = 9(2k + 3)(k − 1)2 .
On the other hand,
f6 (a, 1, 1) = 2(k + 2)a(a2 + ka + 1)(a − 1)2 (a + k + 2) ≥ 0,
f6 (0, b, c) = (b − c)2 2(k + 2)(b2 + c 2 )2 + 2(k + 2)2 bc(b2 + c 2 ) + (4k2 + 6k − 1)b2 c 2 .
For −2 < k ≤ −3/2, we have A ≤ 0. According to P 3.76-(a) in Volume 1, it suffices
to show that f6 (a, 1, 1) ≥ 0 and f6 (0, b, c) ≥ 0 for all a, b, c ≥ 0. The first condition is
clearly satisfied. The second condition is also satisfied since
2(k + 2)(b2 + c 2 )2 + (4k2 + 6k − 1)b2 c 2 ≥ [8(k + 2) + 4k2 + 6k − 1]b2 c 2
= (4k2 + 14k + 15)b2 c 2 ≥ 0.
For k > −3/2, when A ≥ 0, we will apply the highest coefficient cancellation method.
Consider two cases: p2 ≤ 4q and p2 > 4q.
Case 1: p2 ≤ 4q. Since
f6 (1, 1, 1) = f6 (0, 1, 1) = 0,
define the homogeneous function
P(a, b, c) = a bc + B(a + b + c)3 + C(a + b + c)(a b + bc + ca)
124
Vasile Cîrtoaje
such that P(1, 1, 1) = P(0, 1, 1) = 0; that is,
4
1
P(a, b, c) = a bc + (a + b + c)3 − (a + b + c)(a b + bc + ca).
9
9
We will prove the sharper inequality g6 (a, b, c) ≥ 0, where
g6 (a, b, c) = f6 (a, b, c) − 9(2k + 3)(k − 1)2 P 2 (a, b, c).
Clearly, g6 (a, b, c) has the highest coefficient A = 0. Then, according to Remark 1 from
the proof of P 3.76 in Volume 1, it suffices to prove that g6 (a, 1, 1) ≥ 0 for 0 ≤ a ≤ 4.
We have
a(a − 1)2
P(a, 1, 1) =
,
9
hence
g6 (a, 1, 1) = f6 (a, 1, 1) − 9(2k + 3)(k − 1)2 P 2 (a, 1, 1) =
a(a − 1)2 g(a)
,
9
where
g(a) = 18(k + 2)(a2 + ka + 1)(a + k + 2) − (2k + 3)(k − 1)2 a(a − 1)2 .
Since a2 + ka + 1 ≥ (k + 2)a, it suffices to show that
18(k + 2)2 (a + k + 2) ≥ (2k + 3)(k − 1)2 (a − 1)2 .
Also, since (a − 1)2 ≤ 2a + 1, it is enough to prove that h(a) ≥ 0, where
h(a) = 18(k + 2)2 (a + k + 2) − (2k + 3)(k − 1)2 (2a + 1).
Since h(a) is a linear function, the inequality h(a) ≥ 0 is true if h(0) ≥ 0 and h(4) ≥ 0.
Setting x = 2k + 3, x > 0, we get
h(0) = 18(k + 2)3 − (2k + 3)(k − 1)2 =
1
(8x 3 + 37x 2 + 2x + 9) > 0.
4
Also,
1
h(4) = 2(k + 2)2 (k + 6) − (2k + 3)(k − 1)2 = 3(7k2 + 20k + 15) > 0.
9
Case 2: p2 > 4q. We will prove the sharper inequality g6 (a, b, c) ≥ 0, where
g6 (a, b, c) = f6 (a, b, c) − 9(2k + 3)(k − 1)2 a2 b2 c 2 .
Symmetric Rational Inequalities
125
We see that g6 (a, b, c) has the highest coefficient A = 0. According to Remark 1 from
the proof of P 3.76 in Volume 1, it suffices to prove that g6 (a, 1, 1) ≥ 0 for a > 4 and
g6 (0, b, c) ≥ 0 for all b, c ≥ 0. We have
g6 (a, 1, 1) = f6 (a, 1, 1) − 9(2k + 3)(k − 1)2 a2
= a[2(k + 2)(a2 + ka + 1)(a − 1)2 (a + k + 2) − 9(2k + 3)(k − 1)2 a].
Since
a2 + ka + 1 > (k + 2)a,
(a − 1)2 > 9,
it suffices to show that
2(k + 2)2 (a + k + 2) ≥ (2k + 3)(k − 1)2 .
Indeed,
2(k + 2)2 (a + k + 2) − (2k + 3)(k − 1)2 > 2(k + 2)2 (k + 6) − (2k + 3)(k − 1)2
= 3(7k2 + 20k + 15) > 0.
Also,
g6 (0, b, c) = f6 (0, b, c) ≥ 0.
P 1.83. Let a, b, c be nonnegative real numbers, no two of which are zero. If k > −2, then
X
3bc − 2a2
3
≤
.
b2 + k bc + c 2
k+2
(Vasile Cîrtoaje, 2011)
First Solution. Write the inequality as
X 2a2 − 3bc
3
6
+
≥
,
b2 + k bc + c 2 k + 2
k+2
X 2(k + 2)a2 + 3(b − c)2
b2 + k bc + c 2
≥ 6.
Applying the Cauchy-Schwarz inequality, it suffices to show that
P
P
[2(k + 2) a2 + 3 (b − c)2 ]2
P
≥ 6,
[2(k + 2)a2 + 3(b − c)2 ](b2 + k bc + c 2 )
126
Vasile Cîrtoaje
which is equivalent to each of the following inequalities
P
P
2[(k + 5) a2 − 3 a b]2
P
≥ 3,
[2(k + 2)a2 + 3(b − c)2 ](b2 + k bc + c 2 )
X
X
X
X
2(k + 8)
a4 + 2(2k + 19)
a2 b2 ≥ 6(k + 2)a bc
a + 21
a b(a2 + b2 ),
2(k + 2) f (a, b, c) + 3g(a, b, c) ≥ 0,
where
X
X
X
f (a, b, c) =
a4 + 2
a2 b2 − 3a bc
a,
X
X
X
a4 + 10
a2 b2 − 7
a b(a2 + b2 ).
g(a, b, c) = 4
We need to show that f (a, b, c) ≥ 0 and g(a, b, c) ≥ 0. Indeed,
X
X
X
X
f (a, b, c) = (
a2 )2 − 3a bc
a≥(
a b)2 − 3a bc
a≥0
and
X
[2(a4 + b4 ) + 10a2 b2 − 7a b(a2 + b2 )]
X
=
(a − b)2 (2a2 − 3a b + 2b2 ) ≥ 0.
g(a, b, c) =
The equality occurs for a = b = c.
Second Solution. Write the inequality in P 1.82 as
X
2a2 + (2k + 1)bc
3
2−
≤
,
b2 + k bc + c 2
k+2
X 2(b2 + c 2 ) − bc − 2a2
b2
+ k bc
+ c2
≤
3
.
k+2
Since b + c ≥ 2bc, we get
2
2
X
3bc − 2a2
3
≤
,
b2 + k bc + c 2
k+2
which is just the desired inequality.
P 1.84. If a, b, c are nonnegative real numbers, no two of which are zero, then
a2 + 16bc b2 + 16ca c 2 + 16a b
+ 2
+ 2
≥ 10.
b2 + c 2
c + a2
a + b2
(Vasile Cîrtoaje, 2005)
Symmetric Rational Inequalities
127
Solution. Let a ≤ b ≤ c, and
E(a, b, c) =
a2 + 16bc b2 + 16ca c 2 + 16a b
+ 2
+ 2
.
b2 + c 2
c + a2
a + b2
Consider two cases.
Case 1: 16b3 ≥ ac 2 . We will show that
E(a, b, c) ≥ E(0, b, c) ≥ 10.
We have
E(a, b, c) − E(0, b, c) =
a2
a(16c 3 − a b2 ) a(16b3 − ac 2 )
+
+ 2 2
≥ 0,
b2 + c 2
c 2 (c 2 + a2 )
b (a + b2 )
since c 3 − a b2 ≥ 0 and 16b3 − ac 2 ≥ 0. Also,
16bc
b2 c 2
+
+ 2 − 10
b2 + c 2 c 2
b
(b − c)4 (b2 + c 2 + 4bc)
≥ 0.
=
b2 c 2 (b2 + c 2 )
E(0, b, c) − 10 =
Case 2: 16b3 ≤ ac 2 . It suffices to show that
c 2 + 16a b
≥ 10.
a2 + b2
Indeed,
16b3
+ 16a b
c + 16a b
a
−
10
≥
− 10
a2 + b2
a2 + b2
16b
=
− 10 ≥ 16 − 10 > 0.
a
2
This completes the proof. The equality holds for a = 0 and b = c (or any cyclic permutation).
P 1.85. If a, b, c are nonnegative real numbers, no two of which are zero, then
a2 + 128bc b2 + 128ca c 2 + 128a b
+
+
≥ 46.
b2 + c 2
c 2 + a2
a2 + b2
(Vasile Cîrtoaje, 2005)
128
Vasile Cîrtoaje
Solution. Let a ≤ b ≤ c, and
E(a, b, c) =
a2 + 128bc b2 + 128ca c 2 + 128a b
+
+
.
b2 + c 2
c 2 + a2
a2 + b2
Consider two cases.
Case 1: 128b3 ≥ ac 2 . We will show that
E(a, b, c) ≥ E(0, b, c) ≥ 46.
We have
E(a, b, c) − E(0, b, c) =
a(128c 3 − a b2 ) a(128b3 − ac 2 )
a2
+
+
≥ 0,
b2 + c 2
c 2 (c 2 + a2 )
b2 (a2 + b2 )
since c 3 − a b2 ≥ 0 and 128b3 − ac 2 ≥ 0. Also,
128bc
b2 c 2
+
+ 2 − 46
b2 + c 2 c 2
b
2
2
2 2
(b + c − 4bc) (b + c 2 + 8bc)
≥ 0.
=
b2 c 2 (b2 + c 2 )
E(0, b, c) − 46 =
Case 2: 128b3 ≤ ac 2 . It suffices to show that
c 2 + 128a b
≥ 46.
a2 + b2
Indeed,
128b3
+ 128a b
c + 128a b
a
−
46
≥
− 46
a2 + b2
a2 + b2
128b
− 46 ≥ 128 − 46 > 0.
=
a
2
This completes the proof. The equality holds for a = 0 and
permutation).
c
b
+ = 4 (or any cyclic
c
b
P 1.86. If a, b, c are nonnegative real numbers, no two of which are zero, then
a2 + 64bc b2 + 64ca c 2 + 64a b
+
+
≥ 18.
(b + c)2
(c + a)2
(a + b)2
(Vasile Cîrtoaje, 2005)
Symmetric Rational Inequalities
129
Solution. Let a ≤ b ≤ c, and
E(a, b, c) =
a2 + 64bc b2 + 64ca c 2 + 64a b
+
+
.
(b + c)2
(c + a)2
(a + b)2
Consider two cases.
Case 1: 64b3 ≥ c 2 (a + 2b). We will show that
E(a, b, c) ≥ E(0, b, c) ≥ 18.
We have
E(a, b, c) − E(0, b, c) =
≥
a2
a[64c 3 − b2 (a + 2c)] a[64b3 − c 2 (a + 2b)]
+
+
(b + c)2
c 2 (c + a)2
b2 (a + b)2
a[64c 3 − b2 (a + 2c])
a[64b2 c − b2 (c + 2c)]
61a b2 c
≥
=
≥ 0.
c 2 (c + a)2
c 2 (c + a)2
c 2 (c + a)2
Also,
64bc
b2 c 2
+
+ 2 − 18
(b + c)2 c 2
b
4 2
2
(b − c) (b + c + 6bc)
=
≥ 0.
b2 c 2 (b + c)2
E(0, b, c) − 18 =
Case 2: 64b3 ≤ c 2 (a + 2b). It suffices to show that
c 2 + 64a b
≥ 18.
(a + b)2
Indeed,
64b3
+ 64a b
c + 64a b
a + 2b
−
18
≥
− 18
(a + b)2
(a + b)2
64b
64
=
− 18 ≥
− 18 > 0.
a + 2b
3
2
This completes the proof. The equality holds for a = 0 and b = c (or any cyclic permutation).
P 1.87. Let a, b, c be nonnegative real numbers, no two of which are zero. If k ≥ −1, then
X a2 (b + c) + ka bc
b2 + k bc + c 2
≥ a + b + c.
130
Vasile Cîrtoaje
Solution. We apply the SOS method. Write the inequality as follows
X a2 (b + c) + ka bc
b2 + k bc + c 2
− a ≥ 0,
X a(a b + ac − b2 − c 2 )
≥ 0,
b2 + k bc + c 2
X a b(a − b)
X ac(a − c)
+
≥ 0,
b2 + k bc + c 2
b2 + k bc + c 2
X a b(a − b)
X ba(b − a)
+
≥ 0,
b2 + k bc + c 2
c 2 + kca + a2
X
a b(a2 + ka b + b2 )(a + b + kc)(a − b)2 ≥ 0.
Without loss of generality, assume that a ≥ b ≥ c. Since a + b + kc ≥ a + b − c > 0, it
suffices to show that
b(b2 + k bc + c 2 )(b + c + ka)(b − c)2 + a(c 2 + kca + a2 )(c + a + k b)(c − a)2 ≥ 0.
Since c + a + k b ≥ c + a − b ≥ 0 and c 2 + kca + a2 ≥ b2 + k bc + c 2 , it is enough to prove
that
b(b + c + ka)(b − c)2 + a(c + a + k b)(c − a)2 ≥ 0.
We have
b(b + c + ka)(b − c)2 + a(c + a + k b)(c − a)2 ≥
≥ [b(b + c + ka) + a(c + a + k b)](b − c)2
= [a2 + b2 + 2ka b + c(a + b)](b − c)2
≥ [(a − b)2 + c(a + b)](b − c)2 ≥ 0.
The equality holds for a = b = c, and for a = 0 and b = c (or any cyclic permutation).
P 1.88. Let a, b, c be nonnegative real numbers, no two of which are zero. If k ≥
X a3 + (k + 1)a bc
b2 + k bc + c 2
−3
, then
2
≥ a + b + c.
(Vasile Cîrtoaje, 2009)
Symmetric Rational Inequalities
131
Solution. Use the SOS method. Write the inequality as follows
X a3 + (k + 1)a bc
X a3 − a(b2 − bc + c 2 )
−
a
≥
0,
≥ 0,
b2 + k bc + c 2
b2 + k bc + c 2
X (b + c)a3 − a(b3 + c 3 )
(b + c)(b2 + k bc + c 2 )
X
≥ 0,
X a b(a2 − b2 ) + ac(a2 − c 2 )
(b + c)(b2 + k bc + c 2 )
≥ 0,
X
ba(b2 − a2 )
a b(a2 − b2 )
+
≥ 0,
(b + c)(b2 + k bc + c 2 )
(c + a)(c 2 + kca + a2 )
X
(a2 − b2 )2 a b(a2 + ka b + b2 )[a2 + b2 + a b + (k + 1)c(a + b + c)] ≥ 0,
X
(b2 − c 2 )2 bc(b2 + k bc + c 2 )Sa ≥ 0,
where
Sa = b2 + c 2 + bc + (k + 1)a(a + b + c).
Since Sc > 0, it suffices to show that
(b2 − c 2 )2 b(b2 + k bc + c 2 )Sa + (c 2 − a2 )2 a(c 2 + kca + a2 )S b ≥ 0.
Since (c 2 − a2 )2 ≥ (b2 − c 2 )2 , a ≥ b,
c 2 + kca + a2 − (b2 + k bc + c 2 ) = (a − b)(a + b + kc) ≥ 0,
and
S b = a2 + c 2 + ac + (k + 1)b(a + b + c) ≥ a2 + c 2 + ac −
=
(a − b)(2a + b) + c(2a + 2c − b)
≥ 0,
2
1
b(a + b + c)
2
it is enough to show that Sa + S b ≥ 0. Indeed,
Sa + S b = a2 + b2 + 2c 2 + c(a + b) + (k + 1)(a + b)(a + b + c)
1
≥ a2 + b2 + 2c 2 + c(a + b) − (a + b)(a + b + c)
2
(a − b)2 + c(a + b + 4c)
=
≥ 0.
2
This completes the proof. The equality holds for a = b = c, and for a = 0 and b = c (or
any cyclic permutation).
132
Vasile Cîrtoaje
P 1.89. Let a, b, c be nonnegative real numbers, no two of which are zero. If k > 0, then
2a k − b k − c k 2b k − c k − a k 2c k − a k − b k
+ 2
+ 2
≥ 0.
b2 − bc + c 2
c − ca + a2
a − a b + b2
(Vasile Cîrtoaje, 2004)
Solution. Let
X = bk − c k ,
Y = c k − ak ,
Z = ak − bk ,
A = b2 − bc + c 2 , B = c 2 − ca + a2 , C = a2 − a b + b2 .
Without loss of generality, assume that a ≥ b ≥ c. This involves A ≤ B, A ≤ C, X ≥ 0,
and Z ≥ 0. Since
X 2a k − b k − c k
b2
− bc
+ c2
X + 2Z X − Z 2X + Z
+
−
B‹
A
C
‹
1 1 2
2 1 1
=X
+ −
+Z
− −
,
A B C
A B C
=
it suffices to prove that
1 1 2
+ − ≥ 0.
A B C
Write this inequality as
1 1
1 1
− ≥ − ,
A C
C B
that is,
(a − c)(a + c − b)(a2 − ac + c 2 ) ≥ (b − c)(a − b − c)(b2 − bc + c 2 ).
For the nontrivial case a > b + c, we can obtain this inequality from
a + c − b ≥ a − b − c,
a − c ≥ b − c,
a2 − ac + c 2 > b2 − bc + c 2 .
This completes the proof. The equality holds for a = b = c, and for a = 0 and b = c (or
any cyclic permutation).
Symmetric Rational Inequalities
133
P 1.90. If a, b, c are the lengths of the sides of a triangle, then
c+a−b
a+b−c
2(a + b + c)
b+c−a
+ 2
+ 2
≥ 2
;
2
2
2
− bc + c
c − ca + a
a − ab + b
a + b2 + c 2
(a)
b2
a2 − 2bc
b2 − 2ca
c 2 − 2a b
+
+
≤ 0.
b2 − bc + c 2 c 2 − ca + a2 a2 − a b + b2
(b)
(Vasile Cîrtoaje, 2009)
Solution. (a) By the Cauchy-Schwarz inequality, we get
P
X b+c−a
[ (b + c − a)]2
≥P
b2 − bc + c 2
(b + c − a)(b2 − bc + c 2 )
P
( a)2
P
= P
.
2 a3 − a2 (b + c) + 3a bc
On the other hand, from
(b + c − a)(c + a − b)(a + b − c) ≥ 0,
we get
2a bc ≤
X
a2 (b + c) −
X
a3 ,
and hence
2
X
3
a −
X
P
2
a (b + c) + 3a bc ≤
Therefore,
X
a3 +
P
a2 (b + c)
2
=
P
P
( a)( a2 )
2
.
P
2 a
b+c−a
≥ P .
b2 − bc + c 2
a2
The equality holds for a degenerate triangle with a = b + c (or any cyclic permutation).
(b) Since
a2 − 2bc
(b − c)2 + (b + c)2 − a2
=
2
−
,
b2 − bc + c 2
b2 − bc + c 2
we can write the inequality as
X
X
X b+c−a
(b − c)2
+
(a
+
b
+
c)
≥ 6.
b2 − bc + c 2
b2 − bc + c 2
Using the inequality in (a), it suffices to prove that
X
(b − c)2
2(a + b + c)2
+
≥ 6.
b2 − bc + c 2
a2 + b2 + c 2
134
Vasile Cîrtoaje
Write this inequality as
X
X 2(b − c)2
(b − c)2
≥
,
b2 − bc + c 2
a2 + b2 + c 2
or, equivalently,
X (b − c)2 (a − b + c)(a + b − c)
≥ 0,
b2 − bc + c 2
which is true. The equality holds for degenerate triangles with either a/2 = b = c (or
any cyclic permutation), or a = 0 and b = c (or any cyclic permutation).
Remark. The following generalization of the inequality in (b) holds (Vasile Cîrtoaje,
2009):
• Let a, b, c be the lengths of the sides of a triangle. If k ≥ −1, then
X a2 − 2(k + 2)bc
b2 + k bc + c 2
≤ 0.
with equality for a = 0 and b = c (or any cyclic permutation).
P 1.91. If a, b, c are nonnegative real numbers, then
a2
b2
c2
1
+
+
≤ .
2
2
2
2
2
2
5a + (b + c)
5b + (c + a)
5c + (a + b)
3
(Vo Quoc Ba Can, 2009)
Solution. Apply the Cauchy-Schwarz inequality in the following manner
(1 + 2)2
1
2
9
=
≤ 2
+ 2
.
2
2
2
2
2
2
2
2
5a + (b + c)
(a + b + c ) + 2(2a + bc)
a +b +c
2a + bc
Then,
X
X
X 2a2
X
9a2
a2
bc
≤
+
=
4
−
,
5a2 + (b + c)2
a2 + b2 + c 2
2a2 + bc
2a2 + bc
and it remains to show that
bc
≥ 1.
+ bc
This is a known inequality, which can be proved by the Cauchy-Schwarz inequality, as
follows
P
X
( bc)2
bc
≥P
= 1.
2a2 + bc
bc(2a2 + bc)
X
2a2
The equality holds for a = b = c, and for a = 0 and b = c (or any cyclic permutation).
Symmetric Rational Inequalities
135
P 1.92. If a, b, c are nonnegative real numbers, then
c 2 + a2 − b2
a2 + b2 − c 2
1
b2 + c 2 − a2
+
+
≥ .
2
2
2
2
2
2
2a + (b + c)
2b + (c + a)
2c + (a + b)
2
(Vasile Cîrtoaje, 2011)
Solution. We apply the SOS method. Write the inequality as follows
X b2 + c 2 − a2
1
−
≥ 0,
2a2 + (b + c)2 6
X 5(b2 + c 2 − 2a2 ) + 2(a2 − bc)
2a2 + (b + c)2
≥ 0;
X 5(b2 − a2 ) + 5(c 2 − a2 ) + (a − b)(a + c) + (a − c)(a + b)
≥ 0;
2a2 + (b + c)2
X (b − a)[5(b + a) − (a + c)] X (c − a)[5(c + a) − (a + b)]
+
≥ 0;
2a2 + (b + c)2
2a2 + (b + c)2
X (b − a)[5(b + a) − (a + c)] X (a − b)[5(a + b) − (b + c)]
+
≥ 0;
2a2 + (b + c)2
2b2 + (c + a)2
X
(a − b)2 [2c 2 + (a + b)2 ][2(a2 + b2 ) + c 2 + 3a b − 3c(a + b)] ≥ 0,
X
(b − c)2 R a Sa ≥ 0,
where
R a = 2a2 + (b + c)2 ,
Sa = a2 + 2(b2 + c 2 ) + 3bc − 3a(b + c).
Without loss of generality, assume that a ≥ b ≥ c. We have
S b = b2 + 2(c 2 + a2 ) + 3ca − 3b(c + a) = (a − b)(2a − b) + 2c 2 + 3c(a − b) ≥ 0,
Sc = c 2 + 2(a2 + b2 ) + 3a b − 3c(a + b) ≥ 7a b − 3c(a + b) ≥ 3a(b − c) + 3b(a − c) ≥ 0,
Sa + S b = 3(a − b)2 + 4c 2 ≥ 0.
Since
X
(b − c)2 R a Sa ≥ (b − c)2 R a Sa + (c − a)2 R b S b
= (b − c)2 R a (Sa + S b ) + [(c − a)2 R b − (b − c)2 R a ]S b ,
it suffices to prove that
(a − c)2 R b ≥ (b − c)2 R a .
We can get this by multiplying the inequalities
b2 (a − c)2 ≥ a2 (b − c)2
136
Vasile Cîrtoaje
and
a2 R b ≥ b2 R a .
The equality holds for a = b = c, and for a = 0 and b = c (or any cyclic permutation).
P 1.93. Let a, b, c be positive real numbers. If k > 0, then
3a2 − 2bc
3b2 − 2ca
3c 2 − 2a b
3
+
+
≤ .
2
2
2
2
2
2
ka + (b − c)
k b + (c − a)
kc + (a − b)
k
(Vasile Cîrtoaje, 2011)
Solution. Use the SOS method. Write the inequality as follows
X1
3a2 − 2bc
−
≥ 0,
k ka2 + (b − c)2
X b2 + c 2 − 2a2 + 2(k − 1)(bc − a2 )
≥ 0;
ka2 + (b − c)2
X (b2 − a2 ) + (c 2 − a2 ) + (k − 1)[(a + b)(c − a) + (a + c)(b − a)]
≥ 0;
ka2 + (b − c)2
X (b − a)[b + a + (k − 1)(a + c)] X (c − a)[c + a + (k − 1)(a + b)]
+
≥ 0;
ka2 + (b − c)2
ka2 + (b − c)2
X (b − a)[b + a + (k − 1)(a + c)] X (a − b)[a + b + (k − 1)(b + c)]
+
≥ 0;
ka2 + (b − c)2
k b2 + (c − a)2
X
(a − b)2 [kc 2 + (a − b)2 ][(k − 1)c 2 + 2c(a + b) + (k2 − 1)(a b + bc + ca)] ≥ 0.
For k ≥ 1, the inequality is clearly true. Consider further that 0 < k < 1. Since
(k − 1)c 2 + 2c(a + b) + (k2 − 1)(a b + bc + ca) >
> −c 2 + 2c(a + b) − (a b + bc + ca) = (b − c)(c − a),
it suffices to prove that
(a − b)(b − c)(c − a)
X
(a − b)[kc 2 + (a − b)2 ] ≥ 0.
Since
X
X
X
(a − b)[kc 2 + (a − b)2 ] = k
(a − b)c 2 +
(a − b)3
= (3 − k)(a − b)(b − c)(c − a),
Symmetric Rational Inequalities
137
we have
(a − b)(b − c)(c − a)
X
(a − b)[kc 2 + (a − b)2 ] =
= (3 − k)(a − b)2 (b − c)2 (c − a)2 ≥ 0.
This completes the proof. The equality holds for a = b = c.
P 1.94. Let a, b, c be nonnegative real numbers, no two of which are zero. If k ≥ 3 +
then
(a)
(b)
a2
ka2
p
7,
b
c
9
a
+ 2
+ 2
≥
;
+ k bc b + kca c + ka b
(1 + k)(a + b + c)
1
1
1
9
+ 2
+ 2
≥
.
+ bc k b + ca kc + a b
(k + 1)(a b + bc + ca)
(Vasile Cîrtoaje, 2005)
Solution. (a) Assume that a = max{a, b, c}. Setting t = (b + c)/2, t ≤ a, by the
Cauchy-Schwarz inequality, we get
b
c
(b + c)2
4t 2
+
≥
=
b2 + kca c 2 + ka b
b(b2 + kca) + c(c 2 + ka b) 8t 3 − 6bc t + 2ka bc
2t 2
2
2t 2
≥ 3
=
.
= 3
2
4t + (ka − 3t)bc
4t + (ka − 3t)t
t + ka
On the other hand,
a2
a
a
≥ 2
.
+ k bc
a + kt 2
Therefore, it suffices to prove that
2
9
a
+
≥
,
a2 + kt 2 t + ka
(k + 1)(a + 2t)
which is equivalent to
(a − t)2 [(k2 − 6k + 2)a + k(4k − 5)t] ≥ 0.
This inequality is true, since k2 − 6k + 2 ≥ 0 and 4k − 5 > 0. The equality holds for
a = b = c.
(b) For a = 0, the inequality becomes
1
1
k(8 − k)
+ 2≥
.
2
b
c
(k + 1)bc
138
We have
Vasile Cîrtoaje
1
k(8 − k)
2
k(8 − k)
k2 − 6k + 2
1
+
−
≥
−
=
≥ 0.
b2 c 2 (k + 1)bc
bc (k + 1)bc
(k + 1)bc
For a, b, c > 0, the desired inequality follows from the inequality in (a) by substituting
a, b, c with
p 1/a, 1/b, 1/c, respectively. The equality holds for a = b = c. In the case
k = 3 + 7, the equality also holds for a = 0 and b = c (or any cyclic permutation).
P 1.95. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
1
1
6
1
+
+
≥ 2
.
2a2 + bc 2b2 + ca 2c 2 + a b
a + b2 + c 2 + a b + bc + ca
(Vasile Cîrtoaje, 2005)
Solution. Applying the Cauchy-Schwarz inequality, we have
X
X
1
(b + c)2
4(a + b + c)2
P
=
≥
.
2a2 + bc
(b + c)2 (2a2 + bc)
(b + c)2 (2a2 + bc)
Thus, it suffices to show that
2(a + b + c)2 (a2 + b2 + c 2 + a b + bc + ca) ≥ 3
X
(b + c)2 (2a2 + bc),
which is equivalent to
X
X
X
X
2
a4 + 3
a b(a2 + b2 ) + 2a bc
a ≥ 10
a2 b2 .
This follows by adding Schur’s inequality
X
X
X
2
a4 + 2a bc
a≥2
a b(a2 + b2 )
to the inequality
5
X
a b(a2 + b2 ) ≥ 10
X
a2 b2 .
The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).
P 1.96. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
22a2
1
1
1
1
+
+
≥
.
2
2
+ 5bc 22b + 5ca 22c + 5a b
(a + b + c)2
(Vasile Cîrtoaje, 2005)
Symmetric Rational Inequalities
139
Solution. Applying the Cauchy-Schwarz inequality, we have
X
X
(b + c)2
4(a + b + c)2
1
P
=
≥
.
22a2 + 5bc
(b + c)2 (22a2 + 5bc)
(b + c)2 (22a2 + 5bc)
Thus, it suffices to show that
4(a + b + c)4 ≥
X
(b + c)2 (22a2 + 5bc),
which is equivalent to
X
X
X
X
4
a4 + 11
a b(a2 + b2 ) + 4a bc
a ≥ 30
a2 b2 .
This follows by adding Schur’s inequality
X
X
X
4
a4 + 4a bc
a≥4
a b(a2 + b2 )
to the inequality
15
X
a b(a2 + b2 ) ≥ 30
X
a2 b2 .
The equality holds for a = b = c.
P 1.97. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
1
1
8
1
+
+
≥
.
2a2 + bc 2b2 + ca 2c 2 + a b
(a + b + c)2
(Vasile Cîrtoaje, 2005)
First Solution. Applying the Cauchy-Schwarz inequality, we have
X
X
1
(b + c)2
4(a + b + c)2
P
=
≥
.
2a2 + bc
(b + c)2 (2a2 + bc)
(b + c)2 (2a2 + bc)
Thus, it suffices to show that
(a + b + c)4 ≥ 2
X
(b + c)2 (2a2 + bc),
which is equivalent to
X
X
X
X
a4 + 2
ab(a2 + b2 ) + 4a bc
a≥6
a2 b2 .
We will prove the sharper inequality
X
X
X
X
a4 + 2
a b(a2 + b2 ) + a bc
a≥6
a2 b2 .
140
Vasile Cîrtoaje
This follows by adding Schur’s inequality
X
X
X
a4 + a bc
a≥
a b(a2 + b2 )
to the inequality
3
X
a b(a2 + b2 ) ≥ 6
X
a2 b2 .
The equality holds for a = 0 and b = c (or any cyclic permutation).
Second Solution. Without loss of generality, we may assume that a ≥ b ≥ c. Since the
equality holds for c = 0 and a = b, write the inequality as
2a2
1
1
1
8
1
+ 2
+ 2
+ 2
≥
+ bc 2b + ca 4c + 2a b 4c + 2a b
(a + b + c)2
and then apply the Cauchy-Schwarz inequality. It suffices to prove that
(2a2
+
bc) + (2b2
8
16
≥
,
2
2
+ ca) + (4c + 2a b) + (4c + 2a b) (a + b + c)2
which is equivalent to the obvious inequality
c(a + b − 2c) ≥ 0.
P 1.98. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
1
1
1
12
+
+
≥
.
a2 + bc b2 + ca c 2 + a b
(a + b + c)2
(Vasile Cîrtoaje, 2005)
Solution. Due to homogeneity, we may assume that a + b + c = 1. On this assumption,
write the inequality as
‹
X 1
− 1 ≥ 9,
a2 + bc
X 1 − a2 − bc
≥ 9.
a2 + bc
Since
1 − a2 − bc = (a + b + c)2 − a2 − bc > 0,
by the Cauchy-Schwarz inequality, we have
X 1 − a2 − bc
a2 + bc
[
P
(1 − a2 − bc)]2
≥P
.
(1 − a2 − bc)(a2 + bc)
Symmetric Rational Inequalities
141
Then, it suffices to prove that
X
X
X
X
(3 −
a2 −
bc)2 ≥ 9
(a2 + bc) − 9
(a2 + bc)2 ,
which is equivalent to
(1 − 4q)(4 − 7q) + 36a bc ≥ 0,
where q = a b + bc + ca. For q ≤ 1/4, this inequality is clearly true. Consider further
that q > 1/4. By Schur’s inequality of degree three
(a + b + c)3 + 9a bc ≥ 4(a + b + c)(a b + bc + ca),
we get 1 + 9a bc ≥ 4q, and hence 36a bc ≥ 16q − 4. Thus,
(1 − 4q)(4 − 7q) + 36a bc ≥ (1 − 4q)(4 − 7q) + 16q − 4 = 7q(4q − 1) > 0.
The equality holds for a = 0 and b = c (or any cyclic permutation).
P 1.99. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
(a)
(b)
a2
1
1
1
1
2
+ 2
+ 2
≥ 2
+
;
2
2
+ 2bc b + 2ca c + 2a b
a +b +c
a b + bc + ca
a(b + c)
b(c + a)
c(a + b)
a b + bc + ca
+ 2
+ 2
≥1+ 2
.
2
a + 2bc b + 2ca c + 2a b
a + b2 + c 2
(Darij Grinberg and Vasile Cîrtoaje, 2005)
Solution. (a) Write the inequality as
P 2
(b + 2ca)(c 2 + 2a b)
(a2 + 2bc)(b2 + 2ca)(c 2 + 2a b)
≥
a b + bc + ca + 2a2 + 2b2 + 2c 2
.
(a2 + b2 + c 2 )(a b + bc + ca)
Since
X
(b2 + 2ca)(c 2 + 2a b) = (a b + bc + ca)(a b + bc + ca + 2a2 + 2b2 + 2c 2 ),
it suffices to show that
(a2 + b2 + c 2 )(a b + bc + ca)2 ≥ (a2 + 2bc)(b2 + 2ca)(c 2 + 2a b),
which is just the inequality (a) in P 2.16 in Volume 1. The equality holds for a = b, or
b = c, or c = a.
142
Vasile Cîrtoaje
(b) Write the inequality in (a) as
X a b + bc + ca
a2
+ 2bc
≥2+
a b + bc + ca
,
a2 + b2 + c 2
or
X a(b + c) X
bc
a b + bc + ca
+
≥2+ 2
.
2
2
a + 2bc
a + 2bc
a + b2 + c 2
The desired inequality follows by adding this inequality to
bc
.
+ 2bc
1≥
X
X
a2
≥ 1,
a2 + 2bc
a2
The last inequality is equivalent to
which follows by applying the AM-GM inequality as follows
X
X
a2
a2
≥
= 1.
a2 + 2bc
a2 + b2 + c 2
The equality holds for a = b = c.
P 1.100. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
(a)
(b)
a2
b
c
a+b+c
a
+ 2
+ 2
≤
;
+ 2bc b + 2ca c + 2a b
a b + bc + ca
a(b + c)
b(c + a)
c(a + b)
a2 + b2 + c 2
+
+
≤
1
+
.
a2 + 2bc b2 + 2ca c 2 + 2a b
a b + bc + ca
(Vasile Cîrtoaje, 2008)
Solution. (a) Use the SOS method. Write the inequality as
‹
X 
a b + bc + ca
a 1−
≥ 0,
a2 + 2bc
X a(a − b)(a − c)
≥ 0.
a2 + 2bc
Assume that a ≥ b ≥ c. Since (c − a)(c − b) ≥ 0, it suffices to show that
a(a − b)(a − c) b(b − a)(b − c)
+
≥ 0.
a2 + 2bc
b2 + 2ca
Symmetric Rational Inequalities
143
This inequality is equivalent to
c(a − b)2 [2a(a − c) + 2b(b − c) + 3a b] ≥ 0,
which is clearly true. The equality holds for a = b = c, and for a = 0 and b = c (or any
cyclic permutation).
(b) Since
a(b + c)
a(a + b + c)
a2
=
−
,
a2 + 2bc
a2 + 2bc
a2 + 2bc
we can write the inequality as
(a + b + c)
X
a2 + b2 + c 2 X
a2
a
≤
1
+
+
.
a2 + 2bc
a b + bc + ca
a2 + 2bc
According to the inequality in (a), it suffices to show that
a2
(a + b + c)2
a2 + b2 + c 2 X
≤1+
+
,
a b + bc + ca
a b + bc + ca
a2 + 2bc
which is equivalent to
X
a2
≥ 1.
a2 + 2bc
Indeed,
X
a2
a2
≥
= 1.
a2 + 2bc
a2 + b2 + c 2
The equality holds for a = b = c.
X
P 1.101. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
(a)
(b)
2a2
b
c
a+b+c
a
;
+ 2
+ 2
≥ 2
+ bc 2b + ca 2c + a b
a + b2 + c 2
b+c
c+a
a+b
6
+ 2
+ 2
≥
.
2
2a + bc 2b + ca 2c + a b
a+b+c
(Vasile Cîrtoaje, 2008)
Solution. Assume that a ≥ b ≥ c.
(a) Multiplying by a + b + c, we can write the inequality as follows
X a(a + b + c)
2a2 + bc
≥
(a + b + c)2
,
a2 + b2 + c 2
144
Vasile Cîrtoaje
•
˜
(a + b + c)2 X
a(a + b + c)
3− 2
1−
≥
,
a + b2 + c 2
2a2 + bc
X
X (a − b)(a − c)
,
2
(a − b)(a − c) ≥ (a2 + b2 + c 2 )
2a2 + bc
X 3a2 − (b − c)2
(a − b)(a − c) ≥ 0,
2a2 + bc
3 f (a, b, c) + (a − b)(b − c)(c − a)g(a, b, c) ≥ 0,
where
X a2 (a − b)(a − c)
b−c
.
+ bc
2a2 + bc
It suffices to show that f (a, b, c) ≥ 0 and g(a, b, c) ≤ 0. We have
f (a, b, c) =
2a2
,
g(a, b, c) =
X
a2 (a − b)(a − c) b2 (b − a)(b − c)
+
2a2 + bc
2b2 + ca
2
2
a (a − b)(b − c) b (b − a)(b − c)
≥
+
2a2 + bc
2b2 + ca
2
2
2
a c(a − b) (b − c)(a + a b + b2 )
=
≥ 0.
(2a2 + bc)(2b2 + ca)
f (a, b, c) ≥
Also,
(a − b) + (b − c)
a−b
b−c
−
+ 2
2
2
2a + bc
2b + ca

‹ 2c + a b 
‹
1
1
1
1
= (a − b)
−
+ (b − c)
−
2c 2 + a b 2b2 + ca
2a2 + bc 2b2 + ca
•
˜
(a − b)(b − c) 2b + 2c − a 2b + 2a − c
=
−
=
2b2 + ca
2c 2 + a b
2a2 + bc
2(a − b)(b − c)(c − a)(a2 + b2 + c 2 − a b − bc − ca)
≤ 0.
=
(2a2 + bc)(2b2 + ca)(2c 2 + a b)
g(a, b, c) =
The equality holds for a = b = c, and for a = 0 and b = c (or any cyclic permutation).
(b) We apply the SOS method. Write the inequality as follows
˜
X • (b + c)(a + b + c)
−
2
≥ 0,
2a2 + bc
X (b2 + a b − 2a2 ) + (c 2 + ca − 2a2 )
≥ 0,
2a2 + bc
X (b − a)(b + 2a) + (c − a)(c + 2a)
≥ 0,
2a2 + bc
X (b − a)(b + 2a) X (a − b)(a + 2b)
+
≥ 0,
2a2 + bc
2b2 + ca
Symmetric Rational Inequalities
145

‹
X
b + 2a
a + 2b
(a − b)
−
≥ 0,
2b2 + ca 2a2 + bc
X
(a − b)2 (2c 2 + a b)(a2 + b2 + 3a b − ac − bc) ≥ 0.
Since
a2 + b2 + 3a b − ac − bc ≥ a2 + b2 + 2a b − ac − bc = (a + b)(a + b − c),
it suffices to show that
X
(a − b)2 (2c 2 + a b)(a + b)(a + b − c) ≥ 0.
This inequality is true if
(b − c)2 (2a2 + bc)(b + c)(b + c − a) + (c − a)2 (2b2 + ca)(c + a)(c + a − b) ≥ 0;
that is,
(a − c)2 (2b2 + ca)(a + c)(a + c − b) ≥ (b − c)2 (2a2 + bc)(b + c)(a − b − c).
Since a + c ≥ b + c and a + c − b ≥ a − b − c, it is enough to prove that
(a − c)2 (2b2 + ca) ≥ (b − c)2 (2a2 + bc).
We can obtain this inequality by multiplying the inequalities
b2 (a − c)2 ≥ a2 (b − c)2
and
a2 (2b2 + ca) ≥ b2 (2a2 + bc).
The equality holds for a = b = c, and for a = 0 and b = c (or any cyclic permutation).
P 1.102. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
a(b + c) b(c + a) c(a + b)
(a + b + c)2
+
+
≥
.
a2 + bc
b2 + ca
c2 + a b
a2 + b2 + c 2
(Pham Huu Duc, 2006)
146
Vasile Cîrtoaje
Solution. Assume that a ≥ b ≥ c and write the inequality as follows
3−

‹
(a + b + c)2 X
a b + ac
≥
1
−
,
a2 + b2 + c 2
a2 + bc
X
X (a − b)(a − c)
(a − b)(a − c) ≥ (a2 + b2 + c 2 )
,
a2 + bc
X (a − b)(a − c)(a + b − c)(a − b + c)
≥ 0.
a2 + bc
It suffices to show that
2
(b − c)(b − a)(b + c − a)(b − c + a) (c − a)(c − b)(c + a − b)(c − a + b)
+
≥ 0,
b2 + ca
c2 + a b
which is equivalent to the obvious inequality
(b − c)2 (c − a + b)2 (a2 + bc)
≥ 0.
(b2 + ca)(c 2 + a b)
The equality holds for a = b = c, and for a = 0 and b = c (or any cyclic permutation).
P 1.103. Let a, b, c be nonnegative real numbers, no two of which are zero. If k > 0, then
p
p
p
p
3(2 + 3)
b2 + c 2 + 3bc c 2 + a2 + 3ca a2 + b2 + 3a b
+
+
≥
.
a2 + k bc
b2 + kca
c 2 + ka b
1+k
(Vasile Cîrtoaje, 2013)
Solution. Write the inequality in the form f6 (a, b, c) ≥ 0, where
X
p
f6 (a, b, c) = (1 + k)
(b2 + c 2 + 3bc)(b2 + kca)(c 2 + ka b)
−3(2 +
p
3 )(a2 + k bc)(b2 + kca)(c 2 + ka b).
Clearly, f6 (a, b, c) has the same highest coefficient as f (a, b, c), where
Xp
f (a, b, c) = (1 + k)
( 3bc − a2 )(b2 + kca)(c 2 + ka b)
−3(2 +
p
3 )(a2 + k bc)(b2 + kca)(c 2 + ka b));
therefore,
p
p
A = 3(1 + k)3 ( 3 − 1) − 3(2 + 3 )(1 + k)3
= −9(1 + k)3 .
Symmetric Rational Inequalities
147
Since A ≤ 0, according to P 3.76-(a) in Volume 1, it suffices to prove the original inequality for b = c = 1 and for a = 0.
In the first case, this inequality is equivalent to
p p 2 ™
1
+
3
3
(k + 1)a2 − 1 +
(k − 2)a + k −
≥ 0.
2
2
–
(a − 1)2
For the nontrivial case k > 2, we have
p 2
p p
1+ 3
1+ 3
(k + 1)a + k −
a
≥2 k+1 k−
2
2
2
p p p
1+ 3
3
≥2 3 k−
a ≥ 1+
(k − 2)a.
2
2
In the second case (a = 0), the original inequality can be written as
p
‹ 2

1 b c p
b
c2
3(2 + 3)
+ + 3 + 2 + 2 ≥
,
k c
b
c
b
1+k
and is true if
p
p
2+ 3
3(2 + 3)
+2≥
,
k
1+k
which is equivalent to
p 2
1+ 3
≥ 0.
k−
2
p
1+ 3
The equality holds for a = b = c. If k =
, then the equality holds also for a = 0
2
and b = c (or any cyclic permutation).
P 1.104. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
a2
1
1
8
6
1
+ 2
+ 2
+ 2
≥
.
2
2
2
2
2
+b
b +c
c +a
a +b +c
a b + bc + ca
(Vasile Cîrtoaje, 2013)
Solution. Multiplying by a2 + b2 + c 2 , the inequality becomes
a2
b2
c2
6(a2 + b2 + c 2 )
+
+
+
11
≥
.
b2 + c 2 c 2 + a2 a2 + b2
a b + bc + ca
148
Vasile Cîrtoaje
Since
b2
c2
a2
(a2 b2 + b2 c 2 + c 2 a2 ) =
+
+
b2 + c 2 c 2 + a2 a2 + b2

‹
1
1
1
4
4
4
2 2 2
=a +b +c +a b c
+
+
≥ a4 + b4 + c 4 ,
a2 + b2 b2 + c 2 c 2 + a2
it suffices to show that
6(a2 + b2 + c 2 )
a4 + b4 + c 4
+
11
≥
,
a2 b2 + b2 c 2 + c 2 a2
a b + bc + ca
which is equivalent to
(a2 + b2 + c 2 )2
6(a2 + b2 + c 2 )
+
9
≥
.
a2 b2 + b2 c 2 + c 2 a2
a b + bc + ca
Clearly, it is enough to prove that
a2 + b2 + c 2
a b + bc + ca
2
+9≥
6(a2 + b2 + c 2 )
,
a b + bc + ca
which is
a2 + b2 + c 2
−3
a b + bc + ca
The equality holds for a = 0 and
2
≥ 0.
b c
+ = 3 (or any cyclic permutation).
c
b
P 1.105. If a, b, c are the lengths of the sides of a triangle, then
a(b + c)
b(c + a)
c(a + b)
+ 2
+ 2
≤ 2.
2
a + 2bc b + 2ca c + 2a b
(Vo Quoc Ba Can and Vasile Cîrtoaje, 2010)
Solution. Write the inequality as
‹
X
a b + ac)
≥ 1,
1− 2
a + 2bc
X a2 + 2bc − a b − ac
a2 + 2bc
≥ 1.
Since
a2 + 2bc − a b − ac = bc − (a − c)(b − a) ≥ |a − c||b − a| − (a − c)(b − a) ≥ 0,
Symmetric Rational Inequalities
149
by the Cauchy-Schwarz inequality, we have
X a2 + 2bc − a b − ac
a2 + 2bc
[
P 2
(a + 2bc − a b − ac)]2
≥P
.
(a2 + 2bc)(a2 + 2bc − a b − ac)
Thus, it suffices to prove that
(a2 + b2 + c 2 )2 ≥
X
(a2 + 2bc)(a2 + 2bc − a b − ac),
which reduces to the obvious inequality
a b(a − b)2 + bc(b − c)2 + ca(c − a)2 ≥ 0.
The equality holds for an equilateral triangle, and for a degenerate triangle with a = 0
and b = c (or any cyclic permutation).
P 1.106. If a, b, c are real numbers, then
b2 − ca
c2 − a b
a2 − bc
+
+
≥ 0.
2a2 + b2 + c 2 2b2 + c 2 + a2 2c 2 + a2 + b2
(Nguyen Anh Tuan, 2005)
First Solution. Rewrite the inequality as
X1
a2 − bc
3
−
≤ ,
2 2a2 + b2 + c 2
2
(b + c)2
≤ 3.
2a2 + b2 + c 2
If two of a, b, c are zero, then the inequality is trivial. Otherwise, applying the CauchySchwarz inequality, we get
X
X
(b + c)2
(b + c)2
=
2a2 + b2 + c 2
(a2 + b2 ) + (a2 + c 2 )
X
X b2
X a2
c2
b2
≤
+
=
+
= 3.
a2 + b2 a2 + c 2
a2 + b2
b2 + a2
X
The equality holds for a = b = c.
Second Solution. Use the SOS method. We have
2
X
X (a − b)(a + c) + (a − c)(a + b)
a2 − bc
=
2a2 + b2 + c 2
2a2 + b2 + c 2
150
Vasile Cîrtoaje
=
X (a − b)(a + c)
+
X (b − a)(b + c)
2a2 + b2 + c 2
2b2 + c 2 + a2
‹

X
a+c
b+c
=
(a − b)
−
2a2 + b2 + c 2 2b2 + c 2 + a2
X
(a − b)2
= (a2 + b2 + c 2 − a b − bc − ca)
≥ 0.
(2a2 + b2 + c 2 )(2b2 + c 2 + a2 )
P 1.107. If a, b, c are nonnegative real numbers, then
3a2 − bc
3b2 − ca
3c 2 − a b
3
+
+
≤ .
2
2
2
2
2
2
2
2
2
2a + b + c
2b + c + a
2c + a + b
2
(Vasile Cîrtoaje, 2008)
First Solution. Write the inequality as
X3
3a2 − bc
−
≥ 3,
2 2a2 + b2 + c 2
X 8bc + 3(b − c)2
2a2 + b2 + c 2
By the Cauchy-Schwarz inequality, we have
8bc + 3(b − c)2 ≥
≥ 6.
2(b + c)4
3[4bc + (b − c)2 ]2
=
.
6bc + (b − c)2
b2 + c 2 + 4bc
Therefore, it suffices to prove that
X
(b + c)4
≥ 2.
(2a2 + b2 + c 2 )(b2 + c 2 + 4bc)
Using again the Cauchy-Schwarz inequality, we get
X
P
[ (b + c)2 ]2
(b + c)4
≥P
= 2.
(2a2 + b2 + c 2 )(b2 + c 2 + 4bc)
(2a2 + b2 + c 2 )(b2 + c 2 + 4bc)
The equality holds for a = b = c, for a = 0 and b = c, and for b = c = 0 (or any cyclic
permutation).
Second Solution. Write the inequality as
X1
3a2 − bc
−
≥ 0,
2 2a2 + b2 + c 2
Symmetric Rational Inequalities
151
X (b + c + 2a)(b + c − 2a)
≥ 0,
2a2 + b2 + c 2
X (b + c + 2a)(b − a) + (b + c + 2a)(c − a)
≥ 0,
2a2 + b2 + c 2
X (b + c + 2a)(b − a) X (c + a + 2b)(a − b)
+
≥ 0,
2a2 + b2 + c 2
2b2 + c 2 + a2
‹

X
b + c + 2a
c + a + 2b
−
≥ 0,
(a − b)
2b2 + c 2 + a2 2a2 + b2 + c 2
X
(3a b + bc + ca − c 2 )(2c 2 + a2 + b2 )(a − b)2 ≥ 0.
Since 3a b + bc + ca − c 2 ≥ c(a + b − c), it suffices to show that
X
c(a + b − c)(2c 2 + a2 + b2 )(a − b)2 ≥ 0.
Assume that a ≥ b ≥ c. It is enough to prove that
a(b + c − a)(2a2 + b2 + c 2 )(b − c)2 + b(c + a − b)(2b2 + c 2 + a2 )(c − a)2 ≥ 0;
that is,
b(c + a − b)(2b2 + c 2 + a2 )(a − c)2 ≥ a(a − b − c)(2a2 + b2 + c 2 )(b − c)2 .
Since c + a − b ≥ a − b − c, it suffices to prove that
b(2b2 + c 2 + a2 )(a − c)2 ≥ a(2a2 + b2 + c 2 )(b − c)2 .
We can obtain this inequality by multiplying the inequalities
b2 (a − c)2 ≥ a2 (b − c)2
and
a(2b2 + c 2 + a2 ) ≥ b(2a2 + b2 + c 2 ).
The last inequality is equivalent to
(a − b)[(a − b)2 + a b + c 2 ] ≥ 0.
P 1.108. If a, b, c are nonnegative real numbers, then
(b + c)2
(c + a)2
(a + b)2
+
+
≥ 2.
4a2 + b2 + c 2 4b2 + c 2 + a2 4c 2 + a2 + b2
(Vasile Cîrtoaje, 2005)
152
Vasile Cîrtoaje
Solution. By the Cauchy-Schwarz inequality, we have
P
X (b + c)2
[ (b + c)2 ]2
≥P
4a2 + b2 + c 2
(b + c)2 (4a2 + b2 + c 2 )
P
P
P
P
2[ a4 + 3 a2 b2 + 4a bc a + 2 a b(a2 + b2 )]
P
P
P
= P
≥ 2,
a4 + 5 a2 b2 + 4a bc a + a b(a2 + b2 )
since
X
a b(a2 + b2 ) ≥ 2
X
a2 b2 .
The equality holds for a = b = c, and for b = c = 0 (or any cyclic permutation).
P 1.109. If a, b, c are positive real numbers, then
P
(a)
(b)
11a2
P
3
1
≤
;
2
2
+ 2b + 2c
5(a b + bc + ca)
1
1
1
≤
+
.
4a2 + b2 + c 2
2(a2 + b2 + c 2 ) a b + bc + ca
(Vasile Cîrtoaje, 2008)
Solution. We will prove that
X
ka2
k+2
11 − 2k
2(k − 1)
≤ 2
+
2
2
2
2
+b +c
a +b +c
a b + bc + ca
for any k > 1. Due to homogeneity, we may assume that a2 + b2 + c 2 = 3. On this
hypothesis, we need to show that
X
k+2
11 − 2k
2(k − 1)
≤
+
.
2
(k − 1)a + 3
3
a b + bc + ca
Using the substitution m = 3/(k − 1), m > 0, the inequality can be written as
m(m + 1)
X
a2
1
6
≤ 3m − 2 +
.
+m
a b + bc + ca
By the Cauchy-Schwarz inequality, we have
p
p
(a2 + m)[m + (m + 1 − a)2 ] ≥ [a m + m(m + 1 − a)]2 = m(m + 1)2 ,
and hence
m(m + 1)
a2 − 1
≤
+ m + 2 − 2a,
a2 + m
m+1
Symmetric Rational Inequalities
m(m + 1)
153
X
X
1
≤
3(m
+
2)
−
2
a.
a2 + m
Thus, it suffices to show that
3(m + 2) − 2
X
a ≤ 3m − 2 +
6
;
a b + bc + ca
that is,
(4 − a − b − c)(a b + bc + ca) ≤ 3.
Let p = a + b + c. Since
2(a b + bc + ca) = (a + b + c)2 − (a2 + b2 + c 2 ) = p2 − 3,
we get
6 − 2(4 − a − b − c)(a b + bc + ca) = 6 − (4 − p)(p2 − 3)
= (p − 3)2 (p + 2) ≥ 0.
This completes the proof. The equality holds for a = b = c.
P 1.110. If a, b, c are nonnegative real numbers such that a b + bc + ca = 3, then
p
p
p
a
b
c
3
+
+
≥ .
b+c c+a a+b
2
(Vasile Cîrtoaje, 2006)
Solution. By the Cauchy-Schwarz inequality, we have
P 3/4 2
X pa
a
1 €X 3/4 Š2
≥P
=
a
.
b+c
a(b + c) 6
Thus, it suffices to show that
a3/4 + b3/4 + c 3/4 ≥ 3,
which follows immediately by Remark 1 from the proof of the inequality in P 3.33 in
Volume 1. The equality occurs for a = b = c = 1.
Remark. Analogously, according to Remark 2 from the proof of P 3.33 in Volume 1, we
can prove that
ak
bk
ck
3
+
+
≥
b+c c+a a+b
2
4 ln 2
4 ln 2
for all k ≥ 3 −
≈ 0.476. For k = 3 −
, the equality occurs for a = b = c = 1,
ln 3
ln 3
p
and also for a = 0 and b = c = 3 (or any cyclic permutation).
154
Vasile Cîrtoaje
P 1.111. If a, b, c are nonnegative real numbers such that a b + bc + ca ≥ 3, then
1
1
1
1
1
1
+
+
≥
+
+
.
2+a 2+ b 2+c
1+ b+c 1+c+a 1+a+ b
(Vasile Cîrtoaje, 2014)
Solution. Denote
E(a, b, c) =
1
1
1
1
1
1
+
+
−
−
−
.
2+a 2+ b 2+c 1+ b+c 1+c+a 1+a+ b
Consider first the case a b + bc + ca = 3. We will show that
1
1
1
1
1
1
+
+
≥1≥
+
+
.
2+a 2+ b 2+c
1+ b+c 1+c+a 1+a+ b
By direct calculation, we can show that the left inequality is equivalent to a bc ≤ 1.
Indeed, applying the AM-GM inequality, we get
p
3 = a b + bc + ca ≥ 3 a b · bc · ca.
Also, the right inequality is equivalent to
a + b + c ≥ 2 + a bc.
Since a bc ≤ 1, it suffices to show that
a + b + c ≥ 3.
Indeed,
(a + b + c)2 ≥ 3(a b + bc + ca) = 9.
Consider further that a b+ bc+ca > 3. Without loss of generality, assume that a ≥ b ≥ c,
a > 1. For c ≥ 1, that is, a ≥ b ≥ c ≥ 1, the desired inequality follows by summing the
obvious inequalities
1
1
≥
,
2+a
1+c+a
1
1
≥
,
2+ b
1+a+ b
1
1
≥
.
2+c
1+ b+c
Therefore, assume now that c < 1. Consider the cases b + c ≥ 2 and b + c < 2.
Case 1: b + c ≥ 2, a > 1, c < 1. We will show that
E(a, b, c) ≥ E(1, b, c) ≥ 0.
Symmetric Rational Inequalities
155
We have
‹ 
‹ 
‹
1
1
1
1
1
1
E(a, b, c) − E(1, b, c) =
−
+
−
+
−
2+a 3
2+ b 1+a+ b
2+c 1+c+a
˜
•
1
1
−1
+
+
= (a − 1)
3(2 + a) (2 + b)(1 + a + b) (2 + c)(1 + c + a)
˜
•
1
−1
+
> (a − 1)
3(2 + a) (2 + c)(1 + c + a)
(a − 1)(1 − c)(4 + c + a)
=
>0
3(2 + a)(2 + c)(1 + c + a)

and
E(1, b, c) =
b+c−2
≥ 0.
3(1 + b + c)
Case 2: b + c < 2, a > 1, c < 1. From b + c < 2, it follows that
bc ≤

b+c
2
‹2
< 1.
For fixed b and c, define the function
f (x) = E(x, b, c).
Since
1
1
−1
1
−1
+
+
>
+
2
2
2
2
(2 + x)
(1 + c + x)
(1 + x + b)
(2 + x)
(1 + c + x)2
(1 − c)(3 + 2x + c)
=
> 0,
(2 + x)2 (1 + c + x)2
f 0 (x) =
f (x) is strictly increasing for x ≥ 0. Since
a>
3 − bc
,
b+c
‹

‹
3 − bc
3 − bc
. Therefore, it suffices to prove that f
≥ 0, which
b+c
b+c
3 − bc
, that is, for a b + bc + ca = 3. But this was
is equivalent to E(a, b, c) ≥ 0 for a =
b+c
proved in the first part of the proof. So, the proof is completed. The equality occurs for
a = b = c = 1.
we have f (a) > f

156
Vasile Cîrtoaje
P 1.112. If a, b, c are the lengths of the sides of a triangle, then
(a)
b2 − ca
c2 − a b
a2 − bc
+
+
≤ 0;
3a2 + b2 + c 2 3b2 + c 2 + a2 3c 2 + a2 + b2
(b)
a4 − b2 c 2
b4 − c 2 a2
c 4 − a2 b2
+
+
≤ 0.
3a4 + b4 + c 4 3b4 + c 4 + a4 3c 4 + a4 + b4
(Nguyen Anh Tuan and Vasile Cîrtoaje, 2006)
Solution. (a) Apply the SOS method. We have
2
X
X (a − b)(a + c) + (a − c)(a + b)
a2 − bc
=
3a2 + b2 + c 2
3a2 + b2 + c 2
=
X (a − b)(a + c)
+
X (b − a)(b + c)
3a2 + b2 + c 2
3b2 + c 2 + a2
‹

X
b+c
a+c
−
=
(a − b)
3a2 + b2 + c 2 3b2 + c 2 + a2
= (a2 + b2 + c 2 − 2a b − 2bc − 2ca)
X
(a − b)2
.
(3a2 + b2 + c 2 )(3b2 + c 2 + a2 )
Since
a2 + b2 + c 2 − 2a b − 2bc − 2ca = a(a − b − c) + b(b − c − a) + c(c − a − b) ≤ 0,
the conclusion follows. The equality holds for an equilateral triangle, and for a degenerate triangle with a = 0 and b = c (or any cyclic permutation).
(b) Using the same way as above, we get
2
X
X
a4 − b2 c 2
(a2 − b2 )2
=
A
,
3a4 + b4 + c 4
(3a4 + b4 + c 4 )(3b4 + c 4 + a4 )
where
A = a4 + b4 + c 4 − 2a2 b2 − 2b2 c 2 − 2c 2 a2
= −(a + b + c)(a + b − c)(b + c − a)(c + a − b) ≤ 0.
The equality holds for an equilateral triangle, and for a degenerate triangle with a = b+c
(or any cyclic permutation).
Symmetric Rational Inequalities
157
P 1.113. If a, b, c are the lengths of the sides of a triangle, then
4a2
ca
ab
1
bc
+ 2
+ 2
≥ .
2
2
2
2
2
2
+b +c
4b + c + a
4c + a + b
2
(Vasile Cîrtoaje and Vo Quoc Ba Can, 2010)
Solution. We apply the SOS method. Write the inequality as
X
X
b2 c 2
2bc
≥ 0,
−
4a2 + b2 + c 2
a2 b2 + b2 c 2 + c 2 a2
X bc(2a2 − bc)(b − c)2
≥ 0.
4a2 + b2 + c 2
Without loss of generality, assume that a ≥ b ≥ c. Then, it suffices to prove that
c(2b2 − ca)(c − a)2 b(2c 2 − a b)(a − b)2
+
≥ 0.
4b2 + c 2 + a2
4c 2 + a2 + b2
Since
2b2 − ca ≥ c(b + c) − ca = c(b + c − a) ≥ 0
and
(2b2 − ca) + (2c 2 − a b) = 2(b2 + c 2 ) − a(b + c) ≥ (b + c)2 − a(b + c)
= (b + c)(b + c − a) ≥ 0,
it is enough to show that
b(a − b)2
c(a − c)2
≥
.
4b2 + c 2 + a2
4c 2 + a2 + b2
This follows by multiplying the inequalities
c 2 (a − c)2 ≥ b2 (a − b)2
and
b
c
≥ 2
.
4b2 + c 2 + a2
4c + a2 + b2
These inequalities are true, since
c(a − c) − b(a − b) = (b − c)(b + c − a) ≥ 0,
b(4c 2 + a2 + b2 ) − c(4b2 + c 2 + a2 ) = (b − c)[(b − c)2 + a2 − bc] ≥ 0.
The equality occurs for an equilateral triangle, and for a degenerate triangle with a = 0
and b = c (or any cyclic permutation).
158
Vasile Cîrtoaje
P 1.114. If a, b, c are the lengths of the sides of a triangle, then
b2
1
1
1
9
+ 2
+ 2
≤
.
2
2
2
+c
c +a
a +b
2(a b + bc + ca)
(Vo Quoc Ba Can, 2008)
Solution. Apply the SOS method. Write the inequality as
X • 3 a b + bc + ca ˜
−
≥ 0,
2
b2 + c 2
X 3(b2 + c 2 ) − 2(a b + bc + ca)
b2 + c 2
≥ 0,
X 3b(b − a) + 3c(c − a) + c(a − b) + b(a − c)
b2 + c 2
≥ 0,
X (a − b)(c − 3b) + (a − c)(b − 3c)
≥ 0,
b2 + c 2
X (a − b)(c − 3b) X (b − a)(c − 3a)
+
≥ 0,
b2 + c 2
c 2 + a2
X
(a2 + b2 )(a − b)2 (ca + c b + 3c 2 − 3a b) ≥ 0.
Without loss of generality, assume that a ≥ b ≥ c. Since a b + ac + 3a2 − 3bc > 0, it
suffices to prove that
(a2 + b2 )(a − b)2 (ca + c b + 3c 2 − 3a b) + (a2 + c 2 )(a − c)2 (a b + bc + 3b2 − 3ac) ≥ 0,
or, equivalently,
(a2 + c 2 )(a − c)2 (a b + bc + 3b2 − 3ac) ≥ (a2 + b2 )(a − b)2 (3a b − 3c 2 − ca − c b).
Since
2
a b + bc + 3b − 3ac = a
=
bc + 3b2
bc + 3b2
+ b − 3c ≥ a
+ b − 3c
a
b+c
a(b − c)(4b + 3c)
≥0
b+c
and
(a b + bc + 3b2 − 3ac) − (3a b − 3c 2 − ca − c b) = 3(b2 + c 2 ) + 2bc − 2a(b + c)
≥ 3(b2 + c 2 ) + 2bc − 2(b + c)2 = (b − c)2 ≥ 0,
Symmetric Rational Inequalities
159
it suffices to show that
(a2 + c 2 )(a − c)2 ≥ (a2 + b2 )(a − b)2 ).
This is true, since is equivalent to (b − c)A ≥ 0, where
A = 2a3 − 2a2 (b + c) + 2a(b2 + bc + c 2 ) − (b + c)(b2 + c 2 )

‹
b + c 2 a(3b2 + 2bc + 3c 2 )
= 2a a −
− (b + c)(b2 + c 2 )
+
2
2
≥
b(3b2 + 2bc + 3c 2 )
− (b + c)(b2 + c 2 )
2
=
(b − c)(b2 + bc + 2c 2 )
≥ 0.
2
The equality occurs for an equilateral triangle, and for a degenerate triangle with a = 0
and b = c (or any cyclic permutation).
P 1.115. If a, b, c are the lengths of the sides of a triangle, then
(a)
a + b b + c c + a
a − b + b − c + c − a > 5;
(b)
2
a + b2 b2 + c 2 c 2 + a2 a2 − b2 + b2 − c 2 + c 2 − a2 ≥ 3.
(Vasile Cîrtoaje, 2003)
Solution. Since the inequalities are symmetric, we consider that a > b > c.
(a) Let x = a − c and y = b − c. From a > b > c and a ≤ b + c, it follows that
x > y > 0 and c ≥ x − y. We have
2c + x + y 2c + y 2c + x
a+b b+c c+a
+
+
=
+
−
a−b b−c c−a
x−y
y
x

‹
x+y
1
1 1
2c x + y
= 2c
+ −
+
>
+
x−y
y x
x−y
y
x−y

‹
2(x − y) x + y
x−y
y
≥
+
=2
+
+ 1 ≥ 5.
y
x−y
y
x−y
160
Vasile Cîrtoaje
(b) We will show that
a2 + b2 b2 + c 2 c 2 + a2
+
+
≥ 3;
a2 − b2 b2 − c 2 c 2 − a2
that is,
b2
c2
a2
+
≥
.
a2 − b2 b2 − c 2
a2 − c 2
Since
(b + c)2
a2
≤
,
a2 − c 2
a2 − c 2
it suffices to prove that
b2
c2
(b + c)2
+
≥
.
a2 − b2 b2 − c 2
a2 − c 2
This is equivalent to each of the following inequalities:


‹
‹
1
1
1
1
2bc
2
2
b
− 2
+c
− 2
≥ 2
,
2
2
2
2
2
2
a −b
a −c
b −c
a −c
a − c2
b2 (b2 − c 2 ) c 2 (a2 − b2 )
+
≥ 2bc,
a2 − b2
b2 − c 2
[b(b2 − c 2 ) − c(a2 − b2 )]2 ≥ 0.
This completes the proof. If a > b > c, then the equality holds for a degenerate triangle
with a = b + c and b/c = x 1 , where x 1 ≈ 1.5321 is the positive root of the equation
x 3 − 3x − 1 = 0.
P 1.116. If a, b, c are the lengths of the sides of a triangle, then

‹
b+c c+a a+b
a
b
c
+
+
+3≥6
+
+
.
a
b
c
b+c c+a a+b
Solution. We apply the SOS method. Write the inequality as
X
‹
X b+c
2a
−6≥3
−3 .
a
b+c
Since
and
X b+c
X b c ‹
X (b − c)2
−6=
+
−6=
a
c
b
bc
X 2a
X 2a − b − c X a − b X a − c
−3=
=
+
b+c
b+c
b+c
b+c
Symmetric Rational Inequalities
=
X a−b
b+c
+
161
X b − a X (a − b)2
X
(b − c)2
=
=
,
c+a
(b + c)(c + a)
(c + a)(a + b)
we can rewrite the inequality as
X
a(b + c)Sa (b − c)2 ≥ 0,
where
Sa = a(a + b + c) − 2bc.
Without loss of generality, assume that a ≥ b ≥ c. Since Sa > 0,
S b = b(a + b + c) − 2ca = (b − c)(a + b + c) + c(b + c − a) ≥ 0
and
X
a(b + c)Sa (b − c)2 ≥ b(c + a)S b (c − a)2 + c(a + b)Sc (a − b)2
≥ (a − b)2 [b(c + a)S b + c(a + b)Sc ],
it suffices to prove that
b(c + a)S b + c(a + b)Sc ≥ 0.
This is equivalent to each of the following inequalities
(a + b + c)[a(b2 + c 2 ) + bc(b + c)] ≥ 2a bc(2a + b + c),
a(a + b + c)(b − c)2 + (a + b + c)[2a bc + bc(b + c)] ≥ 2a bc(2a + b + c),
a(a + b + c)(b − c)2 + bc(2a + b + c)(b + c − a) ≥ 0.
Since the last inequality is true, the proof is completed. The equality occurs for an
equilateral triangle, and for a degenerate triangle with a/2 = b = c (or any cyclic
permutation).
P 1.117. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
X 3a(b + c) − 2bc
3
≥ .
(b + c)(2a + b + c) 2
(Vasile Cîrtoaje, 2009)
Solution. Use the SOS method. Write the inequality as follows
˜
X • 3a(b + c) − 2bc
1
−
≥ 0,
(b + c)(2a + b + c) 2
162
Vasile Cîrtoaje
X 4a(b + c) − 6bc − b2 − c 2
(b + c)(2a + b + c)
≥ 0,
X b(a − b) + c(a − c) + 3b(a − c) + 3c(a − b)
≥ 0,
(b + c)(2a + b + c)
X (a − b)(b + 3c) + (a − c)(c + 3b)
≥ 0,
(b + c)(2a + b + c)
X (a − b)(b + 3c)
X (b − a)(a + 3c)
+
≥ 0,
(b + c)(2a + b + c)
(c + a)(2b + c + a)
˜
•
X
a + 3c
b + 3c
−
≥ 0,
(a − b)
(b + c)(2a + b + c) (c + a)(2b + c + a)
X
(a − b)(b − c)(c − a)
(a2 − b2 )(a + b + 2c) ≥ 0.
Since
X
(a2 − b2 )(a + b + 2c) = (a − b)(b − c)(c − a),
the conclusion follows. The equality holds for a = b, or b = c, or c = a.
P 1.118. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
X
a(b + c) − 2bc
≥ 0.
(b + c)(3a + b + c)
(Vasile Cîrtoaje, 2009)
Solution. We apply the SOS method. Since
X
=
X b(a − c) + c(a − b)
a(b + c) − 2bc
=
(b + c)(3a + b + c)
(b + c)(3a + b + c)
X
c(b − a)
c(a − b)
+
(c + a)(3b + c + a)
(b + c)(3a + b + c)
X
c(a + b − c)(a − b)2
=
,
(b + c)(c + a)(3a + b + c)(3b + c + a)
X
the inequality is equivalent to
X
c(a + b)(3c + a + b)(a + b − c)(a − b)2 ≥ 0.
Without loss of generality, assume that a ≥ b ≥ c. Since a + b − c ≥ 0, it suffices to show
that
b(c + a)(3b + c + a)(c + a − b)(a − c)2 ≥ a(b + c)(3a + b + c)(a − b − c)(b − c)2 .
Symmetric Rational Inequalities
163
This is true since
c + a − b ≥ a − b − c,
b2 (a − c)2 ≥ a2 (b − c)2 ,
c + a ≥ b + c,
a(3b + c + a) ≥ b(3a + b + c).
The equality holds for a = b = c, and for a = 0 and b = c (or any cyclic permutation).
P 1.119. Let a, b, c be positive real numbers such that a2 + b2 + c 2 ≥ 3. Prove that
a5 − a2
b5 − b2
c5 − c2
+
+
≥ 0.
a5 + b2 + c 2 b5 + c 2 + a2 c 5 + a2 + b2
(Vasile Cîrtoaje, 2005)
Solution. The inequality is equivalent to
a5
1
1
1
3
+ 5
+ 5
≤ 2
.
2
2
2
2
2
2
+b +c
b +c +a
c +a +b
a + b2 + c 2
Setting a = t x, b = t y and c = tz, where t > 0 and x, y, z > 0 such that x 2 + y 2 +z 2 = 3,
the condition a2 + b2 + c 2 ≥ 3 implies t ≥ 1, and the inequality becomes
1
1
1
+
+
≤ 1.
t 3 x 5 + y 2 + z2 t 3 y 5 + z2 + x 2 t 3z5 + x 2 + y 2
We see that it suffices to prove this inequality for t = 1, when it becomes
1
1
1
+
+
≤ 1.
x 5 − x 2 + 3 y 5 − y 2 + 3 z5 − z2 + 3
Without loss of generality, assume that x ≥ y ≥ z. There are two cases to consider.
p
Case 1: z ≤ y ≤ x ≤ 2. The desired inequality follows by adding the inequalities
3 − y2
1
3 − x2
1
1
3 − z2
≤
,
≤
,
≤
.
x5 − x2 + 3
6
y5 − y2 + 3
6
z5 − z2 + 3
6
We have
1
3 − x2
(x − 1)2 (x 5 + 2x 4 − 3x 2 − 6x − 3)
−
=
≤ 0,
x5 − x2 + 3
6
6(x 5 − x 2 + 3)
since
x 5 + 2x 4 − 3x 2 − 6x − 3 = x 2 (x 3 + 2x 2 − 3 −
6
3
− 2)
x x
164
Vasile Cîrtoaje
p
p
p
3
1
≤ x 2 (2 2 + 4 − 3 − 3 2 − ) = −x 2 ( 2 + ) < 0.
2
2
Case 2: x >
p
2. From x 2 + y 2 + z 2 = 3, it follows that y 2 + z 2 < 1. Since
x5
1
1
1
1
< p
<
<
p
2
2
− x + 3 (2 2 − 1)x + 3 2(2 2 − 1) + 3 6
and
y5
1
1
1
1
+ 5
<
+
,
2
2
2
− y +3 z −z +3 3− y
3 − z2
it suffices to prove that
1
5
1
+
≤ .
2
2
3− y
3−z
6
Indeed, we have
1
1
5 9( y 2 + z 2 − 1) − 5 y 2 z 2
+
−
=
< 0,
3 − y 2 3 − z2 6
6(3 − y 2 )(3 − z 2 )
which completes the proof. The equality occurs for a = b = c = 1.
p
3
Remark. Since a bc ≥ 1 involves a2 + b2 + c 2 ≥ 3 a2 b2 c 2 ≥ 3, the original inequality
is also true for a bc ≥ 1, which is a problem from IMO-2005 (by Hojoo Lee). A proof of
this inequality is the following:
X
=
X
a5 − a2
a3 − 1
≥
a5 + b2 + c 2
a(a2 + b2 + c 2 )
X
X
1
1
1
2
(a
−
)
≥
(a2 − bc)
a2 + b2 + c 2
a
a2 + b2 + c 2
X
1
=
(a − b)2 ≥ 0.
2(a2 + b2 + c 2 )
P 1.120. Let a, b, c be positive real numbers such that a2 + b2 + c 2 = a3 + b3 + c 3 . Prove
that
a2
b2
c2
3
+
+
≥ .
b+c c+a a+b
2
(Pham Huu Duc, 2008)
Symmetric Rational Inequalities
165
First Solution. By the Cauchy-Schwarz inequality, we have
P
P
P
X a2
( a3 )( a2 )
( a3 )2
P
P .
= P
≥P
b+c
a4 (b + c) ( a3 )( a b) − a bc a2
Therefore, it is enough to show that
X
X
X
X
X
2(
a3 )(
a2 ) + 3a bc
a2 ≥ 3(
a3 )(
a b).
Write this inequality as follows
X
X
X
X
X
3(
a3 )(
a2 −
a b) − (
a3 − 3a bc)
a2 ≥ 0,
X
X
X
X
X
X
X
a3 )(
a2 −
a b) − (
a)(
a2 −
a b)
a2 ≥ 0,
3(
X
X
X
X
X
(
a2 −
a b)[3
a3 − (
a)(
a2 )] ≥ 0.
The last inequality is true, since
X
X
X
2(
a2 −
a b) =
(a − b)2 ≥ 0
and
3
X
X
X
X
X
X
a3 − (
a)(
a2 ) =
(a3 + b3 ) −
a b(a + b) =
(a + b)(a − b)2 ≥ 0.
The equality occurs for a = b = c = 1.
Second Solution. Write the inequality in the homogeneous form A ≥ B, where
A=2
X
X a2
−
a,
b+c
B=
3(a3 + b3 + c 3 ) X
−
a.
a2 + b2 + c 2
Since
A=
X a(a − b) + a(a − c)
and
B=
b+c
X b(b − a)
b+c
c+a
X (a − b)2
= (a + b + c)
(b + c)(c + a)
=
X a(a − b)
P 3
P
(a + b3 ) − a b(a + b)
+
P
(a + b)(a − b)2
=
,
a2 + b2 + c 2
a2 + b2 + c 2
we can write the inequality as
˜
X• a + b + c
a+b
−
(a − b)2 ≥ 0,
(b + c)(c + a) a2 + b2 + c 2
X (a − b)2
(a3 + b3 + c 3 − 2a bc)
≥ 0.
(b + c)(c + a)
Since a3 + b3 + c 3 ≥ 3a bc, the conclusion follows.
166
Vasile Cîrtoaje
P 1.121. If a, b, c ∈ [0, 1], then
(a)
b
c
a
+
+
≤ 1;
bc + 2 ca + 2 a b + 2
(b)
ab
bc
ca
+
+
≤ 1.
2bc + 1 2ca + 1 2a b + 1
(Vasile Cîrtoaje, 2010)
Solution. (a) It suffices to show that
b
c
a
+
+
≤ 1,
a bc + 2 a bc + 2 a bc + 2
which is equivalent to
a bc + 2 ≥ a + b + c.
We have
a bc + 2 − a − b − c = (1 − b)(1 − c) + (1 − a)(1 − bc) ≥ 0.
The equality holds for a = b = c = 1, and for a = 0 and b = c = 1 (or any cyclic
permutation).
(b) It suffices to prove that
ab
bc
ca
+
+
≤ 1;
2a bc + 1 2a bc + 1 2a bc + 1
that is,
2a bc + 1 ≥ a b + bc + ca.
Since
a + b + c − (a b + bc + ca) = a(1 − b) + b(1 − c) + c(1 − a) ≥ 0,
we have
2a bc + 1 − a b − bc − ca ≥ 2a bc + 1 − a − b − c
= (1 − b)(1 − c) + (1 − a)(1 − bc) ≥ 0.
The equality holds for a = b = c = 1, and for a = 0 and b = c = 1 (or any cyclic
permutation).
P 1.122. Let a, b, c be positive real numbers such that a + b + c = 2. Prove that

‹
1
1
1
5(1 − a b − bc − ca)
+
+
+ 9 ≥ 0.
1 − a b 1 − bc 1 − ca
(Vasile Cîrtoaje, 2011)
Symmetric Rational Inequalities
167
Solution. Write the inequality as
24 −
5a(b + c) 5b(c + a) 5c(a + b)
−
−
≥ 0.
1 − bc
1 − ca
1 − ab
Since
4(1 − bc) ≥ 4 − (b + c)2 = (a + b + c)2 − (b + c)2 = a(a + 2b + 2c),
it suffices to show that

6−5
which is equivalent to
‹
c+a
a+b
b+c
−
−
≥ 0,
a + 2b + 2c b + 2c + 2a c + 2a + 2b
‹
b+c
≥ 9,
a + 2b + 2c
X
1
5(a + b + c)
≥ 9,
a + 2b + 2c
˜
”X
— •X
1
≥ 9.
(a + 2b + 2c)
a + 2b + 2c
X 
5 1−
The last inequality follows immediately from the AM-HM inequality. The equality holds
for a = b = c = 2/3.
P 1.123. Let a, b, c be nonnegative real numbers such that a + b + c = 2. Prove that
2 − a2 2 − b2
2 − c2
+
+
≤ 3.
2 − bc 2 − ca 2 − a b
(Vasile Cîrtoaje, 2011)
First Solution. Write the inequality as follows
X
2 − a2
1−
≥ 0,
2 − bc
X a2 − bc
≥ 0,
2 − bc
X
(a2 − bc)(2 − ca)(2 − a b) ≥ 0,
X
(a2 − bc)[4 − 2a(b + c) + a2 bc] ≥ 0,
X
X
X
4
(a2 − bc) − 2
a(b + c)(a2 − bc) + a bc
a(a2 − bc) ≥ 0.
168
Vasile Cîrtoaje
By virtue of the AM-GM inequality,
X
a(a2 − bc) = a3 + b3 + c 3 − 3a bc ≥ 0.
Then, it suffices to prove that
X
X
2
(a2 − bc) ≥
a(b + c)(a2 − bc).
Indeed, we have
X
X
X
a(b + c)(a2 − bc) =
a3 (b + c) − a bc
(b + c)
X
X
X
=
a(b3 + c 3 ) − a bc
(b + c) =
a(b + c)(b − c)2
X
X
X • a + (b + c) ˜2
(b − c)2 =
(b − c)2 = 2
(a2 − bc).
≤
2
The equality holds for a = b = c = 2/3, and for a = 0 and b = c = 1 (or any cyclic
permutation).
Second Solution. We apply the SOS method. Write the inequality as follows
X a2 − bc
2 − bc
≥ 0,
X (a − b)(a + c) + (a − c)(a + b)
≥ 0,
2 − bc
X (a − b)(a + c) X (b − a)(b + c)
+
≥ 0,
2 − bc
2 − ca
X (a − b)2 [2 − c(a + b) − c 2 ]
≥ 0,
(2 − bc)(2 − ca)
X
(a − b)2 (2 − a b)(1 − c) ≥ 0.
Assuming that a ≥ b ≥ c, it suffices to prove that
(b − c)2 (2 − bc)(1 − a) + (c − a)2 (2 − ca)(1 − b) ≥ 0.
Since 2(1 − b) = a − b + c ≥ 0 and (c − a)2 ≥ (b − c)2 , it suffices to show that
(2 − bc)(1 − a) + (2 − ca)(1 − b) ≥ 0.
We have
(2 − bc)(1 − a) + (2 − ca)(1 − b) = 4 − 2(a + b) − c(a + b) + 2a bc
˜
•
(a + b) + (2 + c) 2
≥ 4 − (a + b)(2 + c) ≥ 4 −
= 0.
2
Symmetric Rational Inequalities
169
P 1.124. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that
3 + 5a2 3 + 5b2 3 + 5c 2
+
+
≥ 12.
3 − bc
3 − ca
3 − ab
(Vasile Cîrtoaje, 2010)
Solution. Use the SOS method. Write the inequality as follows
X 3 + 5a2
− 4 ≥ 0,
3 − bc
X 5a2 + 4bc − 9
3 − bc
≥ 0,
X 4a2 − b2 − c 2 − 2a b + 2bc − 2ca
3 − bc
≥ 0,
X (2a2 − b2 − c 2 ) + 2(a − b)(a − c)
≥ 0,
3 − bc
X [(a − b)(a + b) + (a − b)(a − c)] + [(a − c)(a + c) + (a − c)(a − b)]
≥ 0,
3 − bc
X (a − b)(2a + b − c) + (a − c)(2a + c − b)
≥ 0,
3 − bc
X (a − b)(2a + b − c) X (b − a)(2b + a − c)
+
≥ 0,
3 − bc
3 − ca
X (a − b)2 [3 − 2c(a + b) + c 2 ]
≥ 0,
(3 − bc)(3 − ca)
X (a − b)2 (c − 1)2
≥ 0.
(3 − bc)(3 − ca)
The equality holds for a = b = c = 1, and for a = 0 and b = c = 3/2 (or any cyclic
permutation).
P 1.125. Let a, b, c be nonnegative real numbers such that a + b + c = 2. If
−1
7
≤m≤ ,
7
8
then
a2 + m
b2 + m
c2 + m
3(4 + 9m)
+
+
≥
.
3 − 2bc 3 − 2ca 3 − 2a b
19
(Vasile Cîrtoaje, 2010)
170
Vasile Cîrtoaje
Solution. We apply the SOS method. Write the inequality as
X a2 + m 4 + 9m −
≥ 0,
3 − 2bc
19
X 19a2 + 2(4 + 9m)bc − 12 − 8m
3 − 2bc
≥ 0.
Since
19a2 + 2(4 + 9m)bc − 12 − 8m = 19a2 + 2(4 + 9m)bc − (3 + 2m)(a + b + c)2
= (16 − 2m)a2 − (3 + 2m)(b2 + c 2 + 2a b + 2ac) + 2(1 + 7m)bc
= (3 + 2m)(2a2 − b2 − c 2 ) + 2(5 − 3m)(a2 + bc − a b − ac) + (4 − 10m)(a b + ac − 2bc)
= (3 + 2m)(a2 − b2 ) + (5 − 3m)(a − b)(a − c) + (4 − 10m)c(a − b)
+(3 + 2m)(a2 − c 2 ) + (5 − 3m)(a − c)(a − b) + (4 − 10m)b(a − c)
= (a − b)B + (a − c)C,
where
B = (8 − m)a + (3 + 2m)b − (1 + 7m)c,
C = (8 − m)a + (3 + 2m)c − (1 + 7m)b,
the inequality can be written as
B1 + C1 ≥ 0,
where
B1 =
X (a − b)[(8 − m)a + (3 + 2m)b − (1 + 7m)c]
,
3 − 2bc
X (b − a)[(8 − m)b + (3 + 2m)a − (1 + 7m)c]
C1 =
.
3 − 2ca
We have
B1 + C1 =
X
(a − b)2 Ec
,
(3 − 2bc)(3 − 2ca)
where
Ec =3(5 − 3m) − 2(8 − m)c(a + b) + 2(1 + 7m)c 2
=6(2m + 3)c 2 − 4(8 − m)c + 3(5 − 3m)
•
˜2
8−m
(1 + 7m)(7 − 8m)
=6(2m + 3) c −
+
.
3(2m + 3)
3(2m + 3)
Since Ec ≥ 0 for −1/7 ≤ m ≤ 7/8, we get B1 + C1 ≥ 0. Thus, the proof is completed.
The equality holds for a = b = c = 2/3. When m = −1/7, the equality holds for
Symmetric Rational Inequalities
171
a = b = c = 2/3, and for a = 0 and b = c = 1 (or any cyclic permutation). When
m = 7/8, the equality holds for a = b = c = 2/3, and for a = 1 and b = c = 1/2 (or any
cyclic permutation).
Remark. The inequalities in P 1.124 and P 1.125 are particular cases (k = 3 and k =
8/3, respectively) of the following more general result:
• Let a, b, c be nonnegative real numbers such that a + b + c = 3. For 0 < k ≤ 3 and
m1 ≤ m ≤ m2 , where

3

0<k≤

 −∞,
2
,
m1 =
(3
−
k)(4
−
k)
3


,
<
k
≤
3

2(3 − 2k)
2
p
36 − 4k − k2 + 4(9 − k) 3(3 − k)
m2 =
,
72 + k
then
a2 + mbc b2 + mca c 2 + ma b
3(1 + m)
+
+
≥
,
9 − k bc
9 − kca
9 − ka b
9−k
with equality for a = b = c = 1. When m = m1 and 3/2 < k ≤ 3, the equality holds also
for
3
a = 0, b = c = .
2
When m = m2 , the equality holds also for
p
p
3k − 6 + 2 3(3 − k)
3 − 3(3 − k)
a=
, b=c=
.
k
k
P 1.126. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that
47 − 7a2 47 − 7b2 47 − 7c 2
+
+
≥ 60.
1 + bc
1 + ca
1 + ab
(Vasile Cîrtoaje, 2011)
Solution. We apply the SOS method. Write the inequality as follows
X 47 − 7a2
− 20 ≥ 0,
1 + bc
172
Vasile Cîrtoaje
X 27 − 7a2 − 20bc
1 + bc
≥ 0,
X 3(a + b + c)2 − 7a2 − 20bc
1 + bc
≥ 0,
X −3(2a2 − b2 − c 2 ) + 2(a − b)(a − c) + 8(a b − 2bc + ca)
≥ 0,
1 + bc
X −3(a − b)(a + b) + (a − b)(a − c) + 8c(a − b)
+
1 + bc
X −3(a − c)(a + c) + (a − c)(a − b) + 8b(a − c)
+
≥ 0,
1 + bc
X (a − b)(−2a − 3b + 7c) X (a − c)(−2a − 3c + 7b)
+
≥ 0,
1 + bc
1 + bc
X (a − b)(−2a − 3b + 7c) X (b − a)(−2b − 3a + 7c)
+
≥ 0,
1 + bc
1 + ca
X (a − b)2 [1 − 2c(a + b) + 7c 2 ]
≥ 0,
(1 + bc)(1 + ca)
X (a − b)2 (3c − 1)2
≥ 0,
(1 + bc)(1 + ca)
The equality holds for a = b = c = 1, and for a = 7/3 and b = c = 1/3 (or any cyclic
permutation).
Remark. The inequality in P 1.126 is a particular cases (k = 9) of the following more
general result:
• Let a, b, c be nonnegative real numbers such that a + b + c = 3. For k > 0 and m ≥ m1 ,
where
p

36 + 4k − k2 + 4(9 + k) 3(3 + k)


, k 6= 72

72 − k
m1 =
,
238


,
k
=
72

5
then
a2 + mbc b2 + mca c 2 + ma b
3(1 + m)
+
+
≤
,
9 + k bc
9 + kca
9 + ka b
9+k
with equality for a = b = c = 1. When m = m1 , the equality holds also for
p
p
3k + 6 − 2 3(3 + k)
3(3 + k) − 3
a=
, b=c=
.
k
k
Symmetric Rational Inequalities
173
P 1.127. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that
57
26 − 7a2 26 − 7b2 26 − 7c 2
+
+
≤
.
1 + bc
1 + ca
1 + ab
2
(Vasile Cîrtoaje, 2011)
Solution. Use the SOS method. Write the inequality as follows
X 19 26 − 7a2 −
≥ 0,
2
1 + bc
X 14a2 + 19bc − 33
1 + bc
≥ 0,
X 42a2 + 57bc − 11(a + b + c)2
1 + bc
≥ 0,
X 11(2a2 − b2 − c 2 ) + 9(a − b)(a − c) − 13(a b − 2bc + ca)
≥ 0,
1 + bc
X 22(a − b)(a + b) + 9(a − b)(a − c) − 26c(a − b)
+
1 + bc
X 22(a − c)(a + c) + 9(a − c)(a − b) − 26b(a − c)
≥ 0,
+
1 + bc
X (a − b)(31a + 22b − 35c) X (a − c)(31a + 22c − 35b)
+
≥ 0,
1 + bc
1 + bc
X (a − b)(31a + 22b − 35c) X (b − a)(31b + 22a − 35c)
+
≥ 0,
1 + bc
1 + ca
X (a − b)2 [9 + 31c(a + b) − 35c 2 ]
≥ 0,
(1 + bc)(1 + ca)
X
(a − b)2 (1 + a b)(1 + 11c)(3 − 2c) ≥ 0.
Assume that a ≥ b ≥ c. Since 3 − 2c > 0, it suffices to show that
(b − c)2 (1 + bc)(1 + 11a)(3 − 2a) + (c − a)2 (1 + a b)(1 + 11b)(3 − 2b) ≥ 0;
that is,
(a − c)2 (1 + a b)(1 + 11b)(3 − 2b) ≥ (b − c)2 (1 + bc)(1 + 11a)(2a − 3).
Since 3 − 2b = a − b + c ≥ 0, we get this inequality by multiplying the inequalities
3 − 2b ≥ 2a − 3,
a(1 + a b) ≥ b(1 + bc),
174
Vasile Cîrtoaje
a(1 + 11b) ≥ b(1 + 11a),
b2 (a − c)2 ≥ a2 (b − c)2 .
The equality holds for a = b = c = 1, and for a = 0 and b = c = 3/2 (or any cyclic
permutation).
Remark. The inequalities in P 1.127 is a particular cases (k = 9) of the following more
general result:
• Let a, b, c be nonnegative real numbers such that a + b + c = 3. For k > 0 and m ≤ m2 ,
where
(3 + k)(4 + k)
m2 =
,
2(3 + 2k)
then
a2 + mbc b2 + mca c 2 + ma b
3(1 + m)
+
+
≥
,
9 + k bc
9 + kca
9 + ka b
9+k
with equality for a = b = c = 1. When m = m2 , the equality holds also for a = 0 and
b = c = 3/2 (or any cyclic permutation).
P 1.128. Let a, b, c be nonnegative real numbers, no all are zero. Prove that
X 5a(b + c) − 6bc
≤ 3.
a2 + b2 + c 2 + bc
(Vasile Cîrtoaje, 2010)
First Solution. Apply the SOS method. If two of a, b, c are zero, then the inequality is
trivial. Consider further that a2 + b2 + c 2 = 1, a ≥ b ≥ c, b > 0, and write the inequality
as follows
˜
X•
5a(b + c) − 6bc
1− 2
≥ 0,
a + b2 + c 2 + bc
X a2 + b2 + c 2 − 5a(b + c) + 7bc
≥ 0,
a2 + b2 + c 2 + bc
X (7b + 2c − a)(c − a) − (7c + 2b − a)(a − b)
≥ 0,
1 + bc
X (7c + 2a − b)(a − b) X (7c + 2b − a)(a − b)
−
≥ 0,
1 + ca
1 + bc
X
(a − b)2 (1 + a b)(3 + ac + bc − 7c 2 ] ≥ 0.
Symmetric Rational Inequalities
175
Since
3 + ac + bc − 7c 2 = 3a2 + 3b2 + ac + bc − 4c 2 > 0,
it suffices to prove that
(1 + bc)(3 + a b + ac − 7a2 )(b − c)2 + (1 + ac)(3 + a b + bc − 7b2 )(a − c)2 ≥ 0.
Since
3 + a b + ac − 7b2 = 3(a2 − b2 ) + 3c 2 + b(a − b) + bc ≥ 0
and 1 + ac ≥ 1 + bc, it is enough to show that
(3 + a b + ac − 7a2 )(b − c)2 + (3 + a b + bc − 7b2 )(a − c)2 ≥ 0.
From b(a − c) ≥ a(b − c) ≥ 0, we get b2 (a − c)2 ≥ a2 (b − c)2 , and hence b(a − c)2 ≥
a(b − c)2 . Thus, it suffices to show that
b(3 + a b + ac − 7a2 ) + a(3 + a b + bc − 7b2 ) ≥ 0.
This is true if
b(3 + a b − 7a2 ) + a(3 + a b − 7b2 ) ≥ 0.
Indeed,
b(3 + a b − 7a2 ) + a(3 + a b − 7b2 ) = 3(a + b)(1 − 2a b) ≥ 0,
since
1 − 2a b = (a − b)2 + c 2 ≥ 0.
The equality holds for a = b = c, and for a = 0 and b = c (or any cyclic permutation).
Second Solution. Without loss of generality, assume that a2 + b2 +c 2 = 1 and a ≤ b ≤ c.
Setting p = a + b + c, q = a b + bc + ca and r = a bc, the inequality becomes
X 5q − 11bc
≤ 3,
1 + bc
Y
X
3
(1 + bc) +
(11bc − 5q)(1 + ca)(1 + a b) ≥ 0,
3(1 + q + pr + r 2 ) + 11(q + 2pr + 3r 2 ) − 5q(3 + 2q + pr) ≥ 0,
36r 2 + 5(5 − q)pr + 3 − q − 10q2 ≥ 0.
Since p2 − 2q = 1, the inequality has the homogeneous form
36r 2 + 5(5p2 − 11q)pr + 3(p2 − 2q)3 − q(p2 − 2q)2 − 10q2 (p2 − 2q) ≥ 0.
According to P 3.57-(a) in Volume 1, for fixed p and q, the product r = a bc is minimal
when b = c or a = 0. Therefore, since 5p2 − 11q > 0, it suffices to prove the inequality
for a = 0, and for b = c = 1. For a = 0, the original inequality becomes
b2
−6bc
10bc
+ 2
≤ 3,
2
+ c + bc b + c 2
176
Vasile Cîrtoaje
which reduces to
(b − c)2 (3b2 + 5bc + 3b2 ) ≥ 0,
while for b = c = 1, we get
5−a
10a − 6
+2 2
≤ 3,
a2 + 3
a +a+2
which is equivalent to
a(3a + 1)(a − 1)2 ≥ 0.
Remark. Similarly, we can prove the following generalization:
• Let a, b, c be nonnegative real numbers, no all are zero. If k > 0, then
X (2k + 3)a(b + c) + (k + 2)(k − 3)bc
a2 + b2 + c 2 + k bc
≤ 3k,
with equality for a = b = c, and for a = 0 and b = c (or any cyclic permutation).
P 1.129. Let a, b, c be nonnegative real numbers, no two of which are zero, and let
x=
a2 + b2 + c 2
.
a b + bc + ca
Prove that
(a)
(b)
(c)
a
b
c
1
1
+
+
+ ≥x+ ;
b+c c+a a+b 2
x

‹
a
b
c
4
6
+
+
≥ 5x + ;
b+c c+a a+b
x

‹
a
b
c
3 1
1
+
+
− ≥
x−
.
b+c c+a a+b 2 3
x
(Vasile Cîrtoaje, 2011)
Solution. We will prove the more general inequality
2a
2b
2c
2(1 − k)
+
+
+ 1 − 3k ≥ (2 − k)x +
,
b+c c+a a+b
x
p
where 0 ≤ k ≤ (21+6 6)/25. For k = 0, k = 1/3 and k = 4/3, we get the inequalities in
(a), (b) and (c), respectively. Let p = a+b+c and q = a b+bc+ca. Since x = (p2 −2q)/q,
we can write the inequality as follows
a
b
c
+
+
≥ f (p, q),
b+c c+a a+b
Symmetric Rational Inequalities
177
X a
+ 1 ≥ 3 + f (p, q),
b+c
p(p2 + q)
≥ 3 + f (p, q).
pq − a bc
According to P 3.57-(a) in Volume 1, for fixed p and q, the product a bc is minimal when
b = c or a = 0. Therefore, it suffices to prove the inequality for a = 0, and for b = c = 1.
For a = 0, using the substitution y = b/c + c/b, the desired inequality becomes
2 y + 1 − 3k ≥ (2 − k) y +
2(1 − k)
,
y
( y − 2)[k( y − 1) + 1]
≥ 0.
y
Since y ≥ 2, this inequality is clearly true. For b = c = 1, the desired inequality becomes
a+
4
(2 − k)(a2 + 2) 2(1 − k)(2a + 1)
+ 1 − 3k ≥
+
,
a+1
2a + 1
a2 + 2
which is equivalent to
a(a − 1)2 [ka2 + 3(1 − k)a + 6 − 4k] ≥ 0.
p
For 0 ≤ k ≤ 1, this is obvious, and for 1 < k ≤ (21 + 6 6)/25, we have
Æ
ka2 + 3(1 − k)a + 6 − 4k ≥ [2 k(6 − 4k) + 3(1 − k)]a ≥ 0.
The equality holds for a = b = c, and for a = 0 and b = c (or any cyclic permutation).
P 1.130. If a, b, c are real numbers, then
a2
1
1
1
9
+ 2
+ 2
≤
.
2
2
2
2
2
2
+ 7(b + c ) b + 7(c + a ) c + 7(a + b ) 5(a + b + c)2
(Vasile Cîrtoaje, 2008)
Solution. Let p = a + b + c and q = a b + bc + ca. Write the inequality as f6 (a, b, c) ≥ 0,
where
Y
X
f6 (a, b, c) = 9
(a2 + 7b2 + 7c 2 ) − 5p2
(b2 + 7c 2 + 7a2 )(c 2 + 7a2 + 7b2 ).
Since
Y
Y
(a2 + 7b2 + 7c 2 ) =
[7(p2 − 2q) − 6a2 ],
178
Vasile Cîrtoaje
f6 (a, b, c) has the highest coefficient
A = 9(−6)3 < 0.
According to P 2.75 in Volume 1, it suffices to prove the original inequality for b = c = 1,
when the inequality reduces to
(a − 1)2 (a − 4)2 ≥ 0.
Thus, the proof is completed. The equality holds for a = b = c, and for a/4 = b = c (or
any cyclic permutation).
P 1.131. If a, b, c are real numbers, then
3a2
ca
ab
3
bc
+ 2
+ 2
≤ .
2
2
2
2
2
2
+b +c
3b + c + a
3c + a + b
5
(Vasile Cîrtoaje and Pham Kim Hung, 2005)
Solution. Write the inequality as f6 (a, b, c) ≥ 0, where
Y
X
f6 (a, b, c) = 3
(3a2 + b2 + c 2 ) − 5
bc(3b2 + c 2 + a2 )(3c 2 + a2 + b2 ).
Let p = a + b + c and q = a b + bc + ca. From
Y
X
f6 (a, b, c) = 3
(2a2 + p2 − 2q) − 5
bc(2b2 + p2 − 2q)(2c 2 + p2 − 2q),
it follows that f6 (a, b, c) has the same highest coefficient A as
X
24a2 b2 c 2 − 20
b3 c 3 ;
that is,
A = 24 − 60 < 0.
According to P 2.75 in Volume 1, it suffices to prove the original inequality for b = c = 1,
when the inequality is equivalent to
(a − 1)2 (3a − 2)2 ≥ 0.
Thus, the proof is completed. The equality holds for a = b = c, and for 3a/2 = b = c
(or any cyclic permutation).
Remark. The inequality in P 1.131 is a particular case (k = 3) of the following more
general result (Vasile Cîrtoaje, 2008):
• Let a, b, c be real numbers. If k > 1, then
X k(k − 3)a2 + 2(k − 1)bc
3(k + 1)(k − 2)
,
+
k+2
with equality for a = b = c, and for ka/2 = b = c (or any cyclic permutation).
ka2
b2
+ c2
≤
Symmetric Rational Inequalities
179
P 1.132. If a, b, c are real numbers such that a + b + c = 3, then
(a)
1
1
3
1
+
+
≤ ;
2
2
2
2
2
2
2+ b +c
2+c +a
2+a + b
4
(b)
1
1
1
1
+
+
≤ .
2
2
2
2
2
2
8 + 5(b + c ) 8 + 5(c + a ) 8 + 5(a + b ) 6
(Vasile Cîrtoaje, 2006, 2009)
Solution. (a) Rewrite the inequality as follows
‹
X
1
1
3 3
−
≤ − ,
2
2
2+ b +c
2
4 2
3
b2 + c 2
≥ .
2
2
2+ b +c
2
By the Cauchy-Schwarz inequality, we have
X
X
2
Pp
P 2 Pp
b2 + c 2
a +
(a2 + b2 )(a2 + c 2 )
b2 + c 2
P
P
≥
=
.
2 + b2 + c 2
(2 + b2 + c 2 )
3 + a2
Thus, it suffices to show that
X
XÆ
(a2 + b2 )(a2 + c 2 ) ≥
a2 + 9.
2
Indeed, applying again the Cauchy-Schwarz inequality, we get
2
XÆ
(a2 + b2 )(a2 + c 2 ) ≥ 2
X
X
€X Š2 X
(a2 + bc) =
a2 +
a =
a2 + 9.
The equality holds for a = b = c = 1.
(b) Denote p = a + b + c and q = a b + bc + ca, we write the inequality in the
homogeneous form
8p2
1
1
1
1
+ 2
+ 2
≤ 2,
2
2
2
2
2
2
+ 45(b + c ) 8p + 45(c + a ) 8p + 45(a + b ) 6p
which is equivalent to f6 (a, b, c) ≥ 0, where
Y
f6 (a, b, c) =
(53p2 − 90q − 45a2 )
−6p2
X
(53p2 − 90q − 45b2 )(53p2 − 90q − 45c 2 ).
Clearly, f6 (a, b, c) has the highest coefficient
A = (−45)3 < 0.
180
Vasile Cîrtoaje
By P 2.75 in Volume 1, it suffices to prove the original inequality for b = c. In this case,
the inequality is equivalent to
(a − 1)2 (a − 13)2 ≥ 0.
The equality holds for a = b = c = 1, and for a = 13/5 and b = c = 1/5 (or any cyclic
permutation).
P 1.133. If a, b, c are real numbers, then
(a + b)(a + c)
(b + c)(b + a)
(c + a)(c + b)
4
+
+
≤ .
a2 + 4(b2 + c 2 ) b2 + 4(c 2 + a2 ) c 2 + 4(a2 + b2 ) 3
(Vasile Cîrtoaje, 2008)
Solution. Let p = a + b + c and q = a b + bc + ca. Write the inequality as f6 (a, b, c) ≥ 0,
where
Y
f6 (a, b, c) = 4
(a2 + 4b2 + 4c 2 )
X
−3
(a + b)(a + c)(b2 + 4c 2 + 4a2 )(c 2 + 4a2 + 4b2 )
Y
X
=4
(4p2 − 8q − 3a2 ) − 3
(a2 + q)(4p2 − 8q − 3b2 )(4p2 − 8q − 3c 2 ).
Thus, f6 (a, b, c) has the highest coefficient
A = 4(−3)3 − 34 < 0.
By P 2.75 in Volume 1, it suffices to prove the original inequality for b = c = 1, when
the inequality is equivalent to
(a − 1)2 (2a − 7)2 ≥ 0.
The equality holds for a = b = c, and for 2a/7 = b = c (or any cyclic permutation).
P 1.134. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
X
1
1
≤
.
(b + c)(7a + b + c) 2(a b + bc + ca)
(Vasile Cîrtoaje, 2009)
Symmetric Rational Inequalities
181
First Solution. Write the inequality as
˜
X•
4(a b + bc + ca)
1−
≥ 1,
(b + c)(7a + b + c)
X (b − c)2 + 3a(b + c)
≥ 1.
(b + c)(7a + b + c)
By the Cauchy-Schwarz inequality, we have
X (b − c)2 + 3a(b + c)
4(a + b + c)4
≥P
.
(b + c)(7a + b + c)
[(b − c)2 + 3a(b + c)](b + c)(7a + b + c)
Therefore, it suffices to show that
X
4(a + b + c)4 ≥
(b2 + c 2 − 2bc + 3ca + 3a b)(b + c)(7a + b + c).
Write this inequality as
X
X
X
X
a4 + a bc
a+3
a b(a2 + b2 ) − 8
a2 b2 ≥ 0,
X
X
X
X
a4 + a bc
a−
a b(a2 + b2 ) + 4
a b(a − b)2 ≥ 0.
P 4
P
P
Since
a + a bc a − a b(a2 + b2 ) ≥ 0 (Schur’s inequality of degree four), the
conclusion follows. The equality holds for a = b = c, and also for a = 0 and b = c (or
any cyclic permutation).
Second Solution. Let p = a + b + c and q = a b + bc + ca. We need to prove that
f6 (a, b, c) ≥ 0, where
Y
f6 (a, b, c) =
(b + c)(7a + b + c)
X
−2(a b + bc + ca)
(a + b)(a + c)(7b + c + a)(7c + a + b)
Y
X
=
(p − a)(p + 6a) − 2q
(p − b)(p − c)(p + 6b)(p + 6c).
Clearly, f6 (a, b, c) has the highest coefficient A = (−6)3 < 0. Thus, by P 3.76-(a) in
Volume 1, it suffices to prove the original inequality for b = c = 1, and for a = 0. For
b = c = 1, the inequality reduces to a(a − 1)2 ≥ 0, which is obviously true. For a = 0,
the inequality can be written as
1
1
1
1
+
+
≤
,
2
(b + c)
c(7b + c) b(7c + b) 2bc
1
b2 + c 2 + 14bc
1
+
≤
,
2
2
2
(b + c)
bc[7(b + c ) + 50bc] 2bc
1
x + 14
1
+
≤ ,
x + 2 7x + 50 2
where x = b/c +c/b, x ≥ 2. This reduces to the obvious inequality (x −2)(5x +28) ≥ 0.
182
Vasile Cîrtoaje
P 1.135. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
X
b2
+ c2
1
9
≤
.
+ 4a(b + c) 10(a b + bc + ca)
(Vasile Cîrtoaje, 2009)
Solution. Let p = a + b + c and q = a b + bc + ca. We need to prove that f6 (a, b, c) ≥ 0,
where
Y
f6 (a, b, c) = 9
[b2 + c 2 + 4a(b + c)]
X
−10(a b + bc + ca)
[a2 + b2 + 4c(a + b)][a2 + c 2 + 4b(a + c)]
Y
X
=9
(p2 + 2q − a2 − 4bc) − 10q
(p2 + 2q − c 2 − 4a b)(p2 + 2q − b2 − 4ca).
Clearly, f6 (a, b, c) has the same highest coefficient A as f (a, b, c), where
Y
X
X
f (a, b, c) = −9
(a2 + 4bc) = −9(65a2 b2 c 2 + 16a bc
a3 + 4
a3 b3 );
that is,
A = −9(65 + 48 + 12) < 0.
Thus, by P 3.76-(a) in Volume 1, it suffices to prove the original inequality for b = c = 1,
and for a = 0. For b = c = 1, the inequality reduces to a(a − 1)2 ≥ 0, which is obviously
true. For a = 0, the inequality becomes
b2
1
1
1
9
+ 2
+ 2
≤
,
2
+c
b + 4bc c + 4bc
10bc
1
b2 + c 2 + 8bc
9
+
≤
.
2
2
2
2
2
2
b +c
4bc(b + c ) + 17b c
10bc
1
x +8
9
+
≤
,
x 4x + 17 10
where x = b/c +c/b, x ≥ 2. The inequality is true, since it is equivalent to (x −2)(26x +
85) ≥ 0. The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic
permutation).
P 1.136. If a, b, c are nonnegative real numbers such that a + b + c = 3, then
1
1
1
9
+
+
≤
.
3 − a b 3 − bc 3 − ca
2(a b + bc + ca)
(Vasile Cîrtoaje, 2011)
Symmetric Rational Inequalities
183
First Solution. We apply the SOS method. Write the inequality as
X  3 a b + bc + ca ‹
−
≥ 0.
2
3 − bc
X 9 − 2a(b + c) − 5bc
≥ 0,
3 − bc
X a2 + b2 + c 2 − 3bc
≥ 0.
3 − bc
Since
2(a2 + b2 + c 2 − 3bc) = 2(a2 − bc) + 2(b2 + c 2 − a b − ac) + 2(a b + ac − 2bc)
= (a − b)(a + c) + (a − c)(a + b) − 2b(a − b) − 2c(a − c) + 2c(a − b) + 2b(a − c)
= (a − b)(a − 2b + 3c) + (a − c)(a − 2c + 3b),
the required inequality is equivalent to
X (a − b)(a − 2b + 3c) + (a − c)(a − 2c + 3b)
≥ 0,
3 − bc
X (a − b)(a − 2b + 3c) X (b − a)(b − 2a + 3c)
+
≥ 0,
3 − bc
3 − ca
X (a − b)2 [9 − c(a + b + 3c)]
≥ 0,
(3 − bc)(3 − ca)
X
(a − b)2 (3 − a b)(3 + c)(3 − 2c) ≥ 0.
Without loss of generality, assume that a ≥ b ≥ c. Then it suffices to prove that
(b − c)2 (3 − bc)(3 + a)(3 − 2a) + (c − a)2 (3 − ca)(3 + b)(3 − 2b) ≥ 0,
which is equivalent to
(a − c)2 (3 − ac)(3 + b)(3 − 2b) ≥ (b − c)2 (3 − bc)(a + 3)(2a − 3).
Since 3−2b = a− b+c ≥ 0, we can obtain this inequality by multiplying the inequalities
b2 (a − c)2 ≥ a2 (b − c)2 ,
a(3 − ac) ≥ b(3 − bc) ≥ 0,
a(3 + b)(3 − 2b) ≥ b(a + 3)(2a − 3) ≥ 0.
We have
a(3 − ac) − b(3 − bc) = (a − b)[3 − c(a + b)] = (a − b)(3 − 3c + c 2 )
≥ 3(a − b)(1 − c) ≥ 0.
184
Vasile Cîrtoaje
Also, since a + b ≤ a + b + c = 3, we have
a(3 + b)(3 − 2b) − b(a + 3)(2a − 3) = 9(a + b) − 6a b − 2a b(a + b)
≥ 9(a + b) − 12a b ≥ 3(a + b)2 − 12a b = 3(a − b)2 ≥ 0.
The equality holds for a = b = c = 1, and for a = 0 and b = c = 3/2 (or any cyclic
permutation).
Second Solution. Let p = a + b + c and q = a b + bc + ca. We need to prove that
f6 (a, b, c) ≥ 0, where
Y
X
f6 (a, b, c) = 3
(p2 − 3bc) − 2q
(p2 − 2ca)(p2 − 2a b).
Clearly, f6 (a, b, c) has the highest coefficient
A = 3(−3)3 < 0.
Thus, by P 3.76-(a) in Volume 1, it suffices to prove the original inequality for b = c,
and for a = 0. For b = c = 3 − a, the inequality reduces to
a(9 − a)(a − 1)2 ≥ 0,
which is obviously true. For a = 0, which yields b + c = 3, the inequality can be written
as
(9 − 4bc)(9 − bc) ≥ 0.
Indeed,
(9 − 4bc)(9 − bc) = (b − c)2 (b2 + c 2 + bc) ≥ 0.
P 1.137. If a, b, c are nonnegative real numbers such that a + b + c = 3, then
a2
bc
ca
ab
3
+ 2
+ 2
≤ .
+a+6 b + b+6 c +c+6 8
(Vasile Cîrtoaje, 2009)
Solution. Write the inequality as
X
3a2
bc
1
≤ ,
2
+ ap + 2p
8
where p = a + b + c. We need to prove that f6 (a, b, c) ≥ 0, where
Y
X
f6 (a, b, c) =
(3a2 + ap + 2p2 ) − 8
bc(3b2 + bp + 2p2 )(3c 2 + c p + 2p2 ).
Symmetric Rational Inequalities
185
Clearly, f6 (a, b, c) has the same highest coefficient as
X
27a2 b2 c 2 − 72
b3 c 3 ,
that is,
A = 27 − 216 < 0.
Thus, by P 3.76-(a) in Volume 1, it suffices to prove that f6 (a, 1, 1) ≥ 0 and f6 (0, b, c) ≥ 0
for all a, b, c ≥ 0. Indeed, we have
f6 (a, 1, 1) = 2a(a2 + 9a + 3)(a − 1)2 (6a + 1) ≥ 0
and
f6 (0, b, c) = 2(b − c)2 (5b2 + 5bc + 2c 2 )(2b2 + 5bc + 5c 2 ) ≥ 0.
The equality holds for a = b = c = 1, and for a = 0 and b = c = 3/2 (or any cyclic
permutation).
P 1.138. If a, b, c are nonnegative real numbers such that a b + bc + ca = 3, then
8a2
1
1
1
1
+ 2
+ 2
≥ .
− 2bc + 21 8b − 2ca + 21 8c − 2a b + 21 9
(Michael Rozenberg, 2013)
Solution. Let
q = a b + bc + ca.
Write the inequality in the homogeneous form f6 (a, b, c) ≥ 0, where
X
Y
f6 (a, b, c) = 3q
(8b2 − 2ca + 7q)(8c 2 − 2a b + 7q) −
(8a2 − 2bc + 7q).
Clearly, f6 (a, b, c) has the same highest coefficient as f (a, b, c), where
Y
X
X
f (a, b, c) = −8
(4a2 − bc) = −8(63a2 b2 c 2 − 16
a3 b3 + 4a bc
a3 );
that is,
A = −8(63 − 48 + 12) < 0.
By P 3.76-(a) in Volume 1, it suffices to prove that f6 (a, 1, 1) ≥ 0 and f6 (0, b, c) ≥ 0 for
all a, b, c ≥ 0. Indeed, we have
f6 (a, 1, 1) = 0
and
f6 (0, b, c) = 8b2 c 2 (b − c)2 ≥ 0.
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Vasile Cîrtoaje
The equality holds when two of a, b, c are equal.
Remark. The following identity holds:
X
Q
8 (a − b)2
9
−1= Q
.
8a2 − 2bc + 21
(a2 − 2bc + 21)
P 1.139. Let a, b, c be real numbers, no two of which are zero. Prove that
(a + b + c)2
a2 + bc b2 + ca c 2 + a b
+
+
≥
;
b2 + c 2
c 2 + a2 a2 + b2
a2 + b2 + c 2
(a)
(b)
6(a b + bc + ca)
a2 + 3bc b2 + 3ca c 2 + 3a b
+ 2
+ 2
≥
.
b2 + c 2
c + a2
a + b2
a2 + b2 + c 2
(Vasile Cîrtoaje, 2014)
Solution. (a) Using the known inequality
X
a2
3
≥
2
2
b +c
2
and the Cauchy-Schwarz inequality yields
X a2 + bc
‹
X bc
X1
a2
bc
+
≥
+
b2 + c 2
b2 + c 2
b2 + c 2
2 b2 + c 2
P
2
X (b + c)2
(b + c)
(a + b + c)2
P
=
=
≥
.
2(b2 + c 2 )
a2 + b2 + c 2
2(b2 + c 2 )
=
X
The equality holds for a = b = c.
(b) We have
X a2 + 3bc
b2 + c 2
X 3bc
a2
3 X 3bc
+
≥
+
b2 + c 2
b2 + c 2
2
b2 + c 2
‹
X1
X
bc
(b + c)2
= −3 + 3
+ 2
=
−3
+
3
2 b + c2
2(b2 + c 2 )
P
2
P 2
3
(b + c)
3
a
6(a b + bc + ca)
≥ −3 + P
= −3 + P
=
.
2
2
2
a2 + b2 + c 2
2(b + c )
a
=
X
The equality holds for a = b = c.
Symmetric Rational Inequalities
187
P 1.140. Let a, b, c be real numbers such that a b + bc + ca ≥ 0 and no two of which are
zero. Prove that
a(b + c) b(c + a) c(a + b)
3
+ 2
+ 2
≥
.
b2 + c 2
c + a2
a + b2
10
(Vasile Cîrtoaje, 2014)
Solution. Since the problem remains unchanged by replacing a, b, c by −a, −b, −c, it
suffices to consider the cases a, b, c ≥ 0 and a < 0, b ≥ 0, c ≥ 0.
Case 1: a, b, c ≥ 0. We have
X a(b + c)
b2 + c 2
≥
=
X a(b + c)
X
(b + c)2
3
3
a
≥ >
.
b+c
2 10
Case 2: a < 0, b ≥ 0, c ≥ 0. Replacing a by −a, we need to show that
b(c − a) c(b − a) a(b + c)
3
+ 2
− 2
≥
2
2
2
2
a +c
a +b
b +c
10
for any nonnegative numbers a, b, c such that
a≤
We show first that
where x =
bc
.
b+c
b(c − a)
b(c − x)
≥ 2
,
a2 + c 2
x + c2
bc
, x ≥ a. This is equivalent to
b+c
b(x − a)[(c − a)x + ac + c 2 ] ≥ 0,
which is true because
(c − a)x + ac + c 2 =
c 2 (a + 2b + c)
≥ 0.
b+c
Similarly, we can show that
c(b − a)
c(b − x)
≥ 2
.
2
2
a +b
x + b2
In addition,
a(b + c)
x(b + c)
≤ 2
.
2
2
b +c
b + c2
188
Vasile Cîrtoaje
Therefore, it suffices to prove that
3
b(c − x) c(b − x) x(b + c)
.
+ 2
− 2
≥
2
2
2
2
x +c
x +b
b +c
10
Denote
p=
Since
and
we need to show that
Since
b
,
b+c
q=
p
b(c − x)
=
,
2
2
x +c
1 + p2
c
,
b+c
p + q = 1.
q
c(b − x)
=
2
2
x +b
1 + q2
pq
x(b + c)
bc
= 2
=
,
2
2
2
b +c
b +c
1 − 2pq
p
q
pq
3
+
−
≥
.
2
2
1+p
1+q
1 − 2pq
10
q
1 + pq
p
+
=
,
1 + p2 1 + q2
2 − 2pq + p2 q2
the inequality can be written as
(pq + 2)2 (1 − 4pq) ≥ 0,
which is true since
1 − 4pq = (p + q)2 − 4pq = (p − q)2 ≥ 0.
The equality holds for −2a = b = c (or any cyclic permutation).
P 1.141. If a, b, c are positive real numbers such that a bc > 1, then
1
1
4
+
≥
.
a + b + c − 3 a bc − 1
a b + bc + ca − 3
(Vasile Cîrtoaje, 2011)
Solution (by Vo Quoc Ba Can). By the AM-GM inequality, we have
p
3
a + b + c ≥ 3 a bc > 3,
p
3
a b + bc + ca ≥ a2 b2 c 2 > 3.
Symmetric Rational Inequalities
189
Without loss of generality, assume that a = min{a, b, c}. By the Cauchy-Schwarz inequality, we have

1
1
+
a + b + c − 3 a bc − 1
‹•
˜ 
‹
p
1 2
a bc − 1
a+ p
a(a + b + c − 3) +
≥
.
a
a
Therefore, it suffices to prove that
(a + 1)
≥
4a
2
Since
a(a + b + c − 3) +
a bc − 1
a
.
a b + bc + ca − 3
a(a + b + c − 3) +
a bc − 1
(a − 1)3
= a b + bc + ca − 3 +
,
a
a
this inequality can be written as follows
(a + 1)2
(a − 1)3
−1≥
,
4a
a(a b + bc + ca − 3)
(a − 1)2
(a − 1)3
≥
,
4a
a(a b + bc + ca − 3)
(a − 1)2 (a b + bc + ca + 1 − 4a) ≥ 0.
This is true since
bc ≥
Æ
3
(a bc)2 > 1,
and hence
a b + bc + ca + 1 − 4a > a2 + 1 + a2 + 1 − 4a = 2(a − 1)2 ≥ 0.
The equality holds for a > 1 and b = c = 1 (or any cyclic permutation).
Remark. Using this inequality, we can prove P 3.84 in Volume 1, which states that
‹
1 1 1
1
(a + b + c − 3)
+ + − 3 + a bc +
≥2
a b c
a bc

for any positive real numbers a, b, c. This inequality is clearly true for a bc = 1. In
addition, it remains unchanged by substituting a, b, c with 1/a, 1/b, 1/c,
respectively.
p
3
Therefore, it suffices to consider the case a bc > 1. Since a + b + c ≥ 3 a bc > 3, we
can write the required inequality as E ≥ 0, where
E = a b + bc + ca − 3a bc +
(a bc − 1)2
.
a+ b+c−3
190
Vasile Cîrtoaje
According to the inequality in P 1.141, we have

2
E ≥ a b + bc + ca − 3a bc + (a bc − 1)
= (a b + bc + ca − 3) +
≥2
v
t
(a b + bc + ca − 3) ·
1
4
−
a b + bc + ca − 3 a bc − 1
‹
4(a bc − 1)2
− 4(a bc − 1)
a b + bc + ca − 3
4(a bc − 1)2
− 4(a bc − 1) = 0.
a b + bc + ca − 3
P 1.142. Let a, b, c be positive real numbers, no two of which are zero. Prove that
X (4b2 − ac)(4c 2 − a b)
b+c
≤
27
a bc.
2
(Vasile Cîrtoaje, 2011)
Solution. Since
X (4b2 − ac)(4c 2 − a b)
b+c
X bc(16bc + a2 )
X a(b3 + c 3 )
−4
b+c
b+c
X bc(16bc + a2 )
X
=
−4
a(b2 + c 2 ) + 12a bc
b
+
c
X a2
16bc
=
bc
+
− 4(b + c) + 12a bc
b+c b+c
X a2
(b − c)2
=
bc
−4
+ 12a bc
b+c
b+c
=
we can write the inequality as follows
X a
a2
4(b − c)2
bc
−
+
≥ 0,
2 b+c
b+c
8
X bc(b − c)2
X 2a − b − c
≥ a bc
.
b+c
b+c
In addition, since
X 2a − b − c
b+c
=
=
X (a − b) + (a − c)
X
=
X a−b
+
X b−a
c+a
b+c
b+c
X
(a − b)2
(b − c)2
=
,
(b + c)(c + a)
(c + a)(a + b)
Symmetric Rational Inequalities
191
the inequality can be restated as
8
X bc(b − c)2
X
(b − c)2
≥ a bc
,
b+c
(c + a)(a + b)
X bc(b − c)2 (8a2 + 8bc + 7a b + 7ac)
≥ 0.
(a + b)(b + c)(c + a)
Since the last form is obvious, the proof is completed. The equality holds for a = b = c,
and also for a = 0 and b = c (or any cyclic permutation).
P 1.143. Let a, b, c be nonnegative real numbers, no two of which are zero, such that
a + b + c = 3.
Prove that
a
b
c
2
+
+
≥ .
3a + bc 3b + ca 3c + a b
3
Solution. Since
3a + bc = a(a + b + c) + bc = (a + b)(a + c),
we can write the inequality as follows
a(b + c) + b(c + a) + c(a + b) ≥
2
(a + b)(b + c)(c + a),
3
6(a b + bc + ca) ≥ 2[(a + b + c)(a b + bc + ca) − a bc],
2a bc ≥ 0.
The equality holds for a = 0, or b = 0, or c = 0.
P 1.144. Let a, b, c be positive real numbers such that

‹
1 1 1
(a + b + c)
+ +
= 10.
a b c
Prove that
19
a
b
c
5
≤
+
+
≤ .
12
b+c c+a a+b
3
(Vasile Cîrtoaje, 2012)
192
Vasile Cîrtoaje
First Solution. Write the hypothesis
‹
1 1 1
+ +
= 10
(a + b + c)
a b c

as
b+c c+a a+b
+
+
=7
a
b
c
and
(a + b)(b + c)(c + a) = 9a bc.
b+c
c+a
a+b
, y =
and z =
, we need to show that
a
b
c
x + y + z = 7 and x yz = 9 involve
Using the substitutions x =
1 1 1 5
19
≤ + + ≤ ,
12
x
y z
3
or, equivalently,
19
1 x(7 − x) 5
≤ +
≤ .
12
x
9
3
Clearly, x, y, z ∈ (0, 7). The left inequality is equivalent to
(x − 4)(2x − 3)2 ≤ 0,
while the right inequality is equivalent to
(x − 1)(x − 3)2 ≥ 0.
These inequalities are true if 1 ≤ x ≤ 4. To show that 1 ≤ x ≤ 4, from ( y + z)2 ≥ 4 yz,
we get
36
(7 − x)2 ≥
,
x
(x − 1)(x − 4)(x − 9) ≥ 0,
1 ≤ x ≤ 4.
Thus, the proof is completed. The left inequality is an equality for 2a = b = c (or any
cyclic permutation), and the right inequality is an equality for a/2 = b = c (or any cyclic
permutation).
Second Solution. Due to homogeneity, assume that b + c = 2; this involves bc ≤ 1.
From the hypothesis

‹
1 1 1
(a + b + c)
+ +
= 10,
a b c
we get
2a(a + 2)
bc =
.
9a − 2
Symmetric Rational Inequalities
193
Since
bc − 1 =
(a − 2)(2a − 1)
,
9a − 2
from the condition bc ≤ 1, we get
1
≤ a ≤ 2.
2
We have
b
c
a(b + c) + b2 + c 2
2a + 4 − 2bc
+
= 2
= 2
c+a a+b
a + (b + c)a + bc
a + 2a + bc
2
2(7a + 12a − 4) 2(7a − 2)
=
=
,
9a2 (a + 2)
9a2
and hence
b
c
a 2(7a − 2) 9a3 + 28a − 8
a
+
+
= +
=
.
b+c c+a a+b
2
9a2
18a2
Thus, we need to show that
19 9a3 + 28a − 8 5
≤
≤ .
12
18a2
3
These inequalities are true, since the left inequality is equivalent to
(2a − 1)(3a − 4)2 ≥ 0,
and the right inequality is equivalent to
(a − 2)(3a − 2)2 ≤ 0.
Remark. Similarly, we can prove the following generalization.
• Let a, b, c be positive real numbers such that
‹
1 1 1
8k2
(a + b + c)
+ +
=9+
,
a b c
1 − k2

where k ∈ (0, 1). Then,
k2
a
b
c
3
k2
≤
+
+
− ≤
.
1+k
b+c c+a a+ b 2 1−k
194
Vasile Cîrtoaje
P 1.145. Let a, b, c be nonnegative real numbers, no two of which are zero, such that
a + b + c = 3. Prove that
a
b
c
9
<
+
+
≤ 1.
10 2a + bc 2b + ca 2c + a b
(Vasile Cîrtoaje, 2012)
Solution. (a) Since
a
1
−bc
− =
,
2a + bc 2 2(2a + bc)
we can write the right inequality as
X
bc
≥ 1.
2a + bc
According to the Cauchy-Schwarz inequality, we have
P
P 2 2
P
X bc
( bc)2
b c + 2a bc a
P
≥P
=
= 1.
2a + bc
bc(2a + bc)
6a bc + b2 c 2
The equality holds for a = b = c = 1, and also for a = 0, or b = 0, or c = 0.
(b) First Solution. For the nontrivial case a, b, c > 0, we can write the left inequality
as
X 1
9
.
>
bc
10
2+
a
Using the substitutions
x=
v
t bc
a
, y=
we need to show that
X
s
v
t ab
ca
, z=
,
b
c
1
9
>
2 + x2
10
for all positive real numbers x, y, z satisfying x y + yz + z x = 3. By expanding, the
inequality becomes
X
X
4
x 2 + 48 ≥ 9x 2 y 2 z 2 + 8
x 2 y 2.
Since
X
X
X
X
x2 y2 = (
x y)2 − 2x yz
x = 9 − 2x yz
x,
we can write the desired inequality as
X
X
4
x 2 + 16x yz
x ≥ 9x 2 y 2 z 2 + 24,
Symmetric Rational Inequalities
195
which is equivalent to
16x yz(x + y + z) ≥ 9x 2 y 2 z 2 + 4(2x y + 2 yz + 2z x − x 2 − y 2 − z 2 ).
Using Schur’s inequality
9x yz
≥ 2x y + 2 yz + 2z x − x 2 − y 2 − z 2 ,
x + y +z
it suffices to prove that
16x yz(x + y + z) ≥ 9x 2 y 2 z 2 +
This is true if
16(x + y + z) ≥ 9x yz +
36x yz
.
x + y +z
36
.
x + y +z
Since
x + y +z ≥
and
1=
Æ
3(x y + yz + z x) = 3
x y + yz + z x Æ
3
≥ x 2 y 2z2,
3
we have
16(x + y + z) − 9x yz −
36
≥ 48 − 9x yz − 12 = 9(4 − x yz) > 0.
x + y +z
Second Solution. As it is shown at the first solution, it suffices to show that
X
1
9
>
2
2+ x
10
for all positive real numbers x, y, z satisfying x y + yz + z x = 3. Rewrite this inequality
as
X x2
6
< .
2
2+ x
5
Let p and q be two positive real numbers such that
p
p + q = 3.
By the Cauchy-Schwarz inequality, we have
(p x + q x)2
x2
3x 2
=
=
2 + x2
2(x y + yz + z x) + 3x 2
2x(x + y + z) + (x 2 + 2 yz)
≤
p2 x
q2 x 2
+ 2
.
2(x + y + z) x + 2 yz
196
Vasile Cîrtoaje
Therefore,
X
X
X q2 x 2
X
p2 x
p2
x2
x2
2
≤
+
=
+
q
.
2 + x2
2(x + y + z)
x 2 + 2 yz
2
x 2 + 2 yz
Thus, it suffices to prove that
X
p2
x2
6
+ q2
< .
2
2
x + 2 yz
5
We claim that
X
x2
< 2.
x 2 + 2 yz
Under this assumption, we only need to show that
p2
6
+ 2q2 ≤ .
2
5
p
p
p
p2
4 3
3
6
Indeed, choosing p =
and q =
, we have p + q = 3 and
+ 2q2 = . To
5
5
2
5
P
x2
complete the proof, we need to prove the homogeneous inequality
< 2,
x 2 + 2 yz
which is equivalent to
X
yz
1
> .
x 2 + 2 yz
2
By the Cauchy-Schwarz inequality, we get
P
P 2 2
P
X
( yz)2
y z + 2x yz x
yz
1
P
P
≥P
> .
=
2
2
2
2
x + 2 yz
2
x yz x + 2 y z
yz(x + 2 yz)
P 1.146. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
b3
c3
a3 + b3 + c 3
a3
+
+
≤
.
2a2 + bc 2b2 + ca 2c 2 + a b
a2 + b2 + c 2
(Vasile Cîrtoaje, 2011)
Solution. Write the inequality as follows
X
a3
a3
−
≥ 0,
a2 + b2 + c 2 2a2 + bc
Symmetric Rational Inequalities
197
X a3 (a2 + bc − b2 − c 2 )
2a2 + bc
≥ 0,
X a3 [a2 (b + c) − b3 − c 3 ]
(b + c)(2a2 + bc)
≥ 0,
X a3 b(a2 − b2 ) + a3 c(a2 − c 2 )
≥ 0,
(b + c)(2a2 + bc)
X a3 c(a2 − c 2 )
X a3 b(a2 − b2 )
+
≥ 0,
(b + c)(2a2 + bc)
(b + c)(2a2 + bc)
X a3 b(a2 − b2 )
X b3 a(b2 − a2 )
+
≥ 0,
(b + c)(2a2 + bc)
(c + a)(2b2 + ca)
X a b(a + b)(a − b)2 [2a2 b2 + c(a3 + a2 b + a b2 + b3 ) + c 2 (a2 + a b + b2 )]
(b + c)(c + a)(2a2 + bc)(2b2 + ca)
≥ 0.
The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).
P 1.147. Let a, b, c be positive real numbers, no two of which are zero. Prove that
b3
c3
a+b+c
a3
+
+
≥
.
2
2
2
4a + bc 4b + ca 4c + a b
5
(Vasile Cîrtoaje, 2011)
Solution. Assume that a ≥ b ≥ c, and write the inequality as follows
X
a3
a
−
≥ 0,
4a2 + bc 5
X a(a2 − bc)
4a2 + bc
≥ 0,
X a[(a − b)(a + c) + (a − c)(a + b)]
≥ 0,
4a2 + bc
X a(a − b)(a + c) X a(a − c)(a + b)
+
≥ 0,
4a2 + bc
4a2 + bc
X a(a − b)(a + c) X b(b − a)(b + c)
+
≥ 0,
4a2 + bc
4b2 + ca
X c(a − b)2 [(a − b)2 + bc + ca − a b]
≥ 0.
(4a2 + bc)(4b2 + ca)
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Vasile Cîrtoaje
Clearly, it suffices to show that
X c(a − b)2 (bc + ca − a b)
(4a2 + bc)(4b2 + ca)
≥ 0,
we can be written as
X
(a − b)2 (bc + ca − a b)(4c 3 + a bc) ≥ 0.
Since ca + a b − bc > 0, it is enough to prove that
(c − a)2 (a b + bc − ca)(4b3 + a bc) + (a − b)2 (bc + ca − a b)(4c 3 + a bc) ≥ 0.
In addition, since (c − a)2 ≥ (a − b)2 , 4b3 + a bc ≥ 4c 3 + a bc and a b + bc − ca > 0, we
only need to show that
(a − b)2 (a b + bc − ca)(4c 3 + a bc) + (a − b)2 (bc + ca − a b)(4c 3 + a bc) ≥ 0.
This is equivalent to the obvious inequality
a bc(a − b)2 (4c 2 + bc) ≥ 0.
The equality holds for a = b = c.
P 1.148. If a, b, c are positive real numbers, then
1
1
3
1
+
+
≥
.
(2 + a)2 (2 + b)2 (2 + c)2
6 + a b + bc + ca
(Vasile Cîrtoaje, 2013)
Solution. By the Cauchy-Schwarz inequality, we have
X
1
4(a + b + c)2
P
≥
.
(2 + a)2
(2 + a)2 (b + c)2
Thus, it suffices to show that
4(a + b + c)2 (6 + a b + bc + ca) ≥
X
(2 + a)2 (b + c)2 .
This inequality is equivalent to
2p2 q − 3q2 + 3pr + 12q ≥ 6(pq + 3r),
Symmetric Rational Inequalities
199
where
p = a + b + c, q = a b + bc + ca,
r = a bc.
According to AM-GM inequality,
2p2 q − 3q2 + 3pr + 12q ≥ 2
Æ
12q(2p2 q − 3q2 + 3pr).
Therefore, it is enough to prove the homogeneous inequality
4q(2p2 q − 3q2 + 3pr) ≥ 3(pq + 3r)2 ,
which can be written as
5p2 q2 ≥ 12q3 + 6pqr + 27r 2 .
Since pq ≥ 9r, we have
3(5p2 q2 − 12q3 − 6pqr − 27r 2 ) ≥ 15p2 q2 − 36q3 − 2p2 q2 − p2 q2
= 12q2 (p2 − 3q) ≥ 0.
The equality holds for a = b = c = 1.
P 1.149. If a, b, c are positive real numbers, then
1
1
1
3
+
+
≥
.
1 + 3a 1 + 3b 1 + 3c
3 + a bc
(Vasile Cîrtoaje, 2013)
Solution. Set
p = a + b + c, q = a b + bc + ca,
r=
p
3
a bc,
and write the inequality as follows
X
(3 + r 3 )
(1 + 3b)(1 + 3c) ≥ 3(1 + 3a)(1 + 3b)(1 + 3c),
(3 + r 3 )(3 + 6p + 9q) ≥ 3(1 + 3p + 9q + 27r 3 ),
r 3 (2p + 3q) + 2 + 3p ≥ 26r 3 .
By virtue of the AM-GM inequality, we have
p ≥ 3r, q ≥ 3r 2 .
Therefore, it suffices to show that
r 3 (6r + 9r 2 ) + 2 + 9r ≥ 26r 3 ,
200
Vasile Cîrtoaje
which is equivalent to the obvious inequality
(r − 1)2 (9r 3 + 24r 2 + 13r + 2) ≥ 0.
The equality holds for a = b = c = 1.
P 1.150. Let a, b, c be real numbers, no two of which are zero. If 1 ≤ k ≤ 3, then

‹
‹
‹
2a b
2bc
2ca
k+ 2
k
+
k
+
≥ (k − 1)(k2 − 1).
a + b2
b2 + c 2
c 2 + a2
(Vasile Cîrtoaje and Vo Quoc Ba Can, 2011)
Solution. If a, b, c are the same sign, then the inequality is obvious since

‹
‹
‹
2a b
2bc
2ca
k+ 2
k+ 2
k+ 2
≥ k3 > (k − 1)(k2 − 1).
a + b2
b + c2
c + a2
Since the inequality remains unchanged by replacing a, b, c with −a, −b, −c, it suffices
to consider further that a ≤ 0, b ≥ 0, c ≥ 0. Setting −a for a, we need to show that

‹
‹
‹
2a b
2bc
2ca
k− 2
k+ 2
k− 2
≥ (k − 1)(k2 − 1)
a + b2
b + c2
c + a2
for a, b, c ≥ 0. Since
•

‹
‹ ˜
2a b
(a − c)
2ca
(a − b)2
k− 2
k
−
1
+
k
−
=
k
−
1
+
a + b2
c 2 + a2
a2 + b2
c 2 + a2
(a − b)2 (a − c)2
2
≥ (k − 1) + (k − 1)
+ 2
,
a2 + b2
c + a2
it suffices to prove that
(a − b)2 (a − c)2
k−1+ 2
+ 2
a + b2
c + a2

k+
2bc
b2 + c 2
‹
≥ k2 − 1.
According to the inequality (a) from P 2.19 in Volume 3, we have
(a − b)2 (a − c)2
(b − c)2
+
≥
.
a2 + b2
c 2 + a2
(b + c)2
Thus, it suffices to show that
(b − c)2
k−1+
(b + c)2

2bc
k+ 2
b + c2
‹
≥ k2 − 1,
Symmetric Rational Inequalities
201
which is equivalent to the obvious inequality
(b − c)4 + 2(3 − k)bc(b − c)2 ≥ 0.
The equality holds for a = b = c.
P 1.151. If a, b, c are non-zero and distinct real numbers, then
•
˜

‹
1
1
1
1
1
1
1
1
1
+
+
+3
+
+
≥4
+
+
.
a2 b2 c 2
(a − b)2 (b − c)2 (c − a)2
a b bc ca
Solution. Write the inequality as
X
‹
X
X 1
1 X 1
1
−
+
3
≥
3
.
a2
bc
(b − c)2
bc
In virtue of the AM-GM inequality, it suffices to prove that
v 
˜
X 1
t X 1 X 1 ‹ •X
1
2 3
−
≥
3
,
a2
bc
(b − c)2
bc
which is true if
X
‹ •X
˜
X ‹2
1 X 1
1
1
4
−
≥3
.
2
2
a
bc
(b − c)
bc
Rewrite this inequality as
X
X
X
X
4(
a2 b2 − a bc
a)(
a2 −
a b)2 ≥ 3(a + b + c)2 (a − b)2 (b − c)2 (c − a)2 .
Using the notations
p = a + b + c, q = a b + bc + ca, r = a bc,
and the identity
(a − b)2 (b − c)2 (c − a)2 = −27r 2 − 2(2p2 − 9q)pr + p2 q2 − 4q3 ,
we can write the inequality as
4(q2 − 3pr)(p2 − 3q)2 ≥ 3p2 [−27r 2 − 2(2p2 − 9q)pr + p2 q2 − 4q3 ],
which is equivalent to
(9pr + p2 q − 6q2 )2 ≥ 0.
202
Vasile Cîrtoaje
P 1.152. Let a, b, c be positive real numbers, and let
A=
a b
+ + k,
b a
B=
b c
+ + k,
c
b
C=
c a
+ + k,
a b
where −2 < k ≤ 4. Prove that
1
4
1 1 1
+ + ≤
+
.
A B C
k + 2 A + B + C − (k + 2)
(Vasile Cîrtoaje, 2009)
Solution. Let us denote
x=
We need to show that
X
a
,
b
y=
b
c
, z= .
c
a
x
1
4
P
≤
+P
x2 + kx + 1
k+2
x + x y + 2k − 2
for all positive real numbers x, y, z satisfying x yz = 1. Write this inequality as follows:
‹
X 1
x
2
4
P
,
− 2
≥
−P
k + 2 x + kx + 1
k+2
x + x y + 2k − 2
P
X (x − 1)2
2 yz(x − 1)2
P
≥P
,
x2 + kx + 1
x + x y + 2k − 2
X (x − 1)2 [−x + y + z + x( y + z) − yz − 2]
≥ 0.
x2 + kx + 1
Since
−x + y + z + x( y + z) − yz − 2 = (x + 1)( y + z) − (x + yz + 2)
= (x + 1)( y + z) − (x + 1)( yz + 1) = −(x + 1)( y − 1)(z − 1),
the inequality is equivalent to
−(x − 1)( y − 1)(z − 1)
X
x2 − 1
≥ 0,
x2 + kx + 1
or E ≥ 0, where
E = −(x − 1)( y − 1)(z − 1)
We have
X
(x 2 − 1)( y 2 + k y + 1)(z 2 + kz + 1).
X
(x 2 − 1)( y 2 + k y + 1)(z 2 + kz + 1) =
X Š €X
X Š
€X
= k(2 − k)
xy−
x +
x2 y2 −
x2
Symmetric Rational Inequalities
203
= −k(2 − k)(x − 1)( y − 1)(z − 1) − (x 2 − 1)( y 2 − 1)(z 2 − 1)
= −(x − 1)( y − 1)(z − 1)[(x + 1)( y + 1)(z + 1) + k(2 − k)],
and hence
E = (x − 1)2 ( y − 1)2 (z − 1)2 [(x + 1)( y + 1)(z + 1) + k(2 − k)] ≥ 0,
because
p
p
p
(x + 1)( y + 1)(z + 1) + k(2 − k) ≥ (2 x)(2 y)(2 z) + k(2 − k) = (2 + k)(4 − k) ≥ 0.
The equality holds for a = b, or b = c, or c = a.
P 1.153. If a, b, c are nonnegative real numbers, no two of which are zero, then
b2
1
1
1
1
1
1
+ 2
+ 2
≥ 2
+ 2
+ 2
.
2
2
2
+ bc + c
c + ca + a
a + ab + b
2a + bc 2b + ca 2c + a b
(Vasile Cîrtoaje, 2014)
Solution. Write the inequality as follows:
‹
X
1
1
−
≥ 0,
b2 + bc + c 2 2a2 + bc
(a2 − b2 ) + (a2 − c 2 )
≥ 0,
(b2 + bc + c 2 )(2a2 + bc)
X
X
a2 − b2
b2 − a2
+
≥ 0,
(b2 + bc + c 2 )(2a2 + bc)
(c 2 + ca + a2 )(2b2 + ca)
X
c(a2 − b2 )(a − b)
(a2 + b2 + c 2 − a b − bc − ca)
≥ 0.
(b2 + bc + c 2 )(c 2 + ca + a2 )(2a2 + bc)(2b2 + ca)
X
Clearly, the last form is obvious. The equality holds for a = b = c.
P 1.154. If a, b, c are nonnegative real numbers such that a + b + c = 3, then
1
1
1
1
1
1
+
+
≥ 2
+ 2
+ 2
.
2a b + 1 2bc + 1 2ca + 1
a +2 b +2 c +2
(Vasile Cîrtoaje, 2014)
204
Vasile Cîrtoaje
Solution. Write the inequality as
X

‹
3 X
1
1
1
− ≥
−
,
2a b + 1 2
a2 + 2 2
X
X
1
a2
3
+
≥ .
2
2a b + 1
2(a + 2) 2
Let us denote
q = a b + bc + ca,
q ≤ 3.
By the Cauchy-Schwarz inequality, we have
X
and
X
1
9
9
≥P
=
2a b + 1
(2a b + 1) 2q + 3
P 2
a
a2
9
≥P
=
.
2
2
2(a + 2)
2(a + 2) 2(15 − 2q)
Therefore, it suffices to prove that
9
3
9
+
≥ .
2q + 3 2(15 − 2q) 2
This inequality is true because it reduces to the obvious inequality
(3 − q)(9 − 2q) ≥ 0.
The equality holds for a = b = c = 1.
P 1.155. If a, b, c are nonnegative real numbers such that a + b + c = 4, then
1
1
1
1
1
1
+
+
≥ 2
+ 2
+ 2
.
a b + 2 bc + 2 ca + 2
a +2 b +2 c +2
(Vasile Cîrtoaje, 2014)
First Solution (by Nguyen Van Quy). Rewrite the inequality as follows:
‹
X 2
1
1
−
−
≥ 0,
a b + 2 a2 + 2 b2 + 2
˜
X•
a(a − b)
b(b − a)
+
≥ 0,
(a b + 2)(a2 + 2) (a b + 2)(b2 + 2)
Symmetric Rational Inequalities
205
X (2 − a b)(a − b)2 (c 2 + 2)
≥ 0.
ab + 2
Without loss of generality, assume that a ≥ b ≥ c ≥ 0. Then,
bc ≤ ac ≤
a(b + c) (a + b + c)2
≤
=2
2
8
and
X (2 − a b)(a − b)2 (c 2 + 2)
ab + 2
(2 − a b)(a − b)2 (c 2 + 2) (2 − ac)(a − c)2 (b2 + 2)
+
ab + 2
ac + 2
2 2
(2 − a b)(a − b) (c + 2) (2 − ac)(a − b)2 (c 2 + 2)
≥
+
ab + 2
ab + 2
(4 − a b − ac)(a − b)2 (c 2 + 2)
≥ 0.
=
ab + 2
≥
The equality holds for a = b = c = 4/3, and also for a = 2 and b = c = 1 (or any cyclic
permutation).
Second Solution. Write the inequality as
X

‹
1
3 X
1
1
− ≥
−
,
bc + 2 2
a2 + 2 2
X
X
1
a2
3
+
≥ .
bc + 2
2(a2 + 2) 2
Assume that a ≥ b ≥ c and denote
s=
b+c
,
2
p = bc,
0≤s≤
4
, 0 ≤ p ≤ s2 .
3
By the Cauchy-Schwarz inequality, we have
b2
c2
(b + c)2
s2
+
≥
=
.
2(b2 + 2) 2(c 2 + 2) 2(b2 + 2) + 2(c 2 + 2) + 4 2s2 − p + 2
In addition,
1
1
a(b + c) + 4
2as + 4
+
=
= 2
.
ca + 2 a b + 2 (a b + 2)(ac + 2)
a p + 4as + 4
Therefore, it suffices to show that E(a, b, c) ≥ 0, where
E(a, b, c) =
1
2(as + 2)
a2
s2
3
+ 2
+
+
− .
p + 2 a p + 4as + 4 2(a2 + 2) 2s2 − p + 2 2
We will prove that
E(a, b, c) ≥ E(a, s, s) ≥ 0.
206
Vasile Cîrtoaje
We have
‹

‹
1
1
1
1
−
+ 2(as + 2) 2
−
E(a, b, c) − E(a, s, s) =
p + 2 s2 + 2
a p + 4as + 4 a2 s2 + 4as + 4

‹
1
1
+ s2
− 2
2
2s − p + 2 s + 2
2
s −p
2a2 (as + 2)(s2 − p)
=
+
(p + 2)(s2 + 2) (a2 p + 4as + 4)(a2 s2 + 4as + 4)
s2 (s2 − p)
− 2
.
(s + 2)(2s2 − p + 2)

Since s2 − p ≥ 0, it remains to show that
2a2 (as + 2)
s2
1
+
≥
,
(p + 2)(s2 + 2) (a2 p + 4as + 4)(a2 s2 + 4as + 4) (s2 + 2)(2s2 − p + 2)
which is equivalent to
p(s2 + 1) − 2
2a2 (as + 2)
≥
.
(a2 p + 4as + 4)(a2 s2 + 4as + 4) (p + 2)(s2 + 2)(2s2 − p + 2)
Since
a2 p + 4as + 4 ≤ a2 s2 + 4as + 4 = (as + 2)2
and
2s2 − p + 2 ≥ s2 + 2,
it is enough to prove that
p(s2 + 1) − 2
2a2
≥
.
(as + 2)3
(p + 2)(s2 + 2)2
In addition, since
as + 2 = (4 − 2s)s + 2 ≤ 4
and
p(s2 + 1) − 2
2(s2 + 2)
2(s2 + 2)
= s2 + 1 −
≤ s2 + 1 − 2
= s2 − 1,
p+2
p+2
s +2
it suffices to show that
a2
s2 − 1
≥ 2
,
32 (s + 2)2
which is equivalent to
(2 − s)2 (2 + s2 )2 ≥ 8(s2 − 1).
Symmetric Rational Inequalities
207
4
, we have
3

‹
4 2
4
2
2 2
2
(2 − s) (2 + s ) − 8(s − 1) ≥ 2 −
(2 + s2 )2 − 8(s2 − 1) = (s4 − 14s2 + 22)
3
9

‹2
4
4
88
16
2 2
=
(7 − s ) − 27 ≥
− 27 =
> 0.
7−
9
9
9
729
Indeed, for the non-trivial case 1 < s ≤
To end the proof, we need to show that E(a, s, s) ≥ 0. Notice that E(a, s, s) can be find
from E(a, b, c) by replacing p with s2 . We get
1
2
a2
s2
3
+
+
+
−
s2 + 2 as + 2 2(a2 + 2) s2 + 2 2
(s − 1)2 (3s − 4)2
=
≥ 0.
2(s2 + 2)(1 + 2s − s2 )(2s2 − 8s + 9)
E(a, s, s) =
P 1.156. If a, b, c are nonnegative real numbers, no two of which are zero, then
(a)
(b)
a b + bc + ca
(a − b)2 (b − c)2 (c − a)2
+
≤ 1;
a2 + b2 + c 2
(a2 + b2 )(b2 + c 2 )(c 2 + a2 )
a b + bc + ca
(a − b)2 (b − c)2 (c − a)2
+
≤ 1.
a2 + b2 + c 2
(a2 − a b + b2 )(b2 − bc + c 2 )(c 2 − ca + a2 )
(Vasile Cîrtoaje, 2014)
Solution. (a) First Solution. Consider the non-trivial case where a, b, c are distinct and
write the inequality as follows:
(a − b)2 (b − c)2 (c − a)2
(a − b)2 + (b − c)2 + (c − a)2
≤
,
(a2 + b2 )(b2 + c 2 )(c 2 + a2 )
2(a2 + b2 + c 2 )
(a2 + b2 ) + (b2 + c 2 ) + (c 2 + a2 ) (a − b)2 + (b − c)2 + (c − a)2
≤
,
(a2 + b2 )(b2 + c 2 )(c 2 + a2 )
(a − b)2 (b − c)2 (c − a)2
X
X
1
1
≤
.
(b2 + c 2 )(c 2 + a2 )
(b − c)2 (c − a)2
Since
a2 + b2 ≥ (a − b)2 ,
b2 + c 2 ≥ (b − c)2 ,
c 2 + a2 ≥ (c − a)2 ,
the conclusion follows. The equality holds for a = b = c.
208
Vasile Cîrtoaje
Second Solution. Assume that a ≥ b ≥ c. We have
(a − b)2 (b − c)2 (c − a)2
a b + bc + ca
(a − b)2 (a − c)2
a b + bc + ca
+
≤
+
a2 + b2 + c 2
(a2 + b2 )(b2 + c 2 )(c 2 + a2 )
a2 + b2 + c 2
(a2 + b2 )(a2 + c 2 )
2a b + c 2
(a − b)2 a2
≤ 2
+
a + b2 + c 2 a2 (a2 + b2 + c 2 )
2a b + c 2 + (a − b)2
= 1.
=
a2 + b2 + c 2
(b) Consider the non-trivial case where a, b, c are distinct and write the inequality
as follows:
(a − b)2 (b − c)2 (c − a)2
(a − b)2 + (b − c)2 + (c − a)2
≤
,
(a2 − a b + b2 )(b2 − bc + c 2 )(c 2 − ca + a2 )
2(a2 + b2 + c 2 )
2(a2 + b2 + c 2 )
(a − b)2 + (b − c)2 + (c − a)2
≤
,
(a2 − a b + b2 )(b2 − bc + c 2 )(c 2 − ca + a2 )
(a − b)2 (b − c)2 (c − a)2
X
1
2(a2 + b2 + c 2 )
≥
.
(a − b)2 (a − c)2
(a2 − a b + b2 )(b2 − bc + c 2 )(c 2 − ca + a2 )
Assume that a = min{a, b, c} and use the substitution
b = a + x,
c = a + y,
x, y ≥ 0.
The inequality can be written as
1
x2 y2
+
x 2 (x
1
1
+ 2
≥ 2 f (a),
2
− y)
y (x − y)2
where
f (a) =
3a2 + 2(x + y)a + x 2 + y 2
.
(a2 + x a + x 2 )(a2 + y a + y 2 )[a2 + (x + y)a + x 2 − x y + y 2 ]
We will show that
1
x2 y2
+
x 2 (x
1
1
+ 2
≥ 2 f (0) ≥ 2 f (a).
2
− y)
y (x − y)2
We have
2(x 2 + y 2 − x y)
2(x 2 + y 2 )
1
1
1
+
+
−
2
f
(0)
=
−
x 2 y 2 x 2 (x − y)2
y 2 (x − y)2
x 2 y 2 (x − y)2
x 2 y 2 (x 2 − x y + y 2 )
2
=
≥ 0.
2
2
(x − y) (x − x y + y 2 )
Symmetric Rational Inequalities
209
Also, since
(a2 + x a + x 2 )(a2 + y a + y 2 ) ≥ (x 2 + y 2 )a2 + x y(x + y)a + x 2 y 2
and
a2 + (x + y)a + x 2 − x y + y 2 ≥ x 2 − x y + y 2 ,
we get f (a) ≤ g(a), where
g(a) =
3a2 + 2(x + y)a + x 2 + y 2
.
[(x 2 + y 2 )a2 + x y(x + y)a + x 2 y 2 ](x 2 − x y + y 2 )
Therefore,
x2 + y2
− g(a)
x 2 y 2 (x 2 − x y + y 2 )
(x 4 − x 2 y 2 + y 4 )a2 + x y(x + y)(x − y)2 a
= 2 2 2
≥ 0.
x y (x − x y + y 2 )[(x 2 + y 2 )a2 + x y(x + y)a + x 2 y 2 ]
f (0) − f (a) ≥
Thus, the proof is completed. The equality holds for a = b = c.
P 1.157. If a, b, c are nonnegative real numbers, no two of which are zero, then
a2
1
1
1
45
+ 2
+ 2
≥
.
2
2
2
2
2
2
+b
b +c
c +a
8(a + b + c ) + 2(a b + bc + ca)
(Vasile Cîrtoaje, 2014)
First Solution (by Nguyen Van Quy). Multiplying by a2 +b2 +c 2 , the inequality becomes
X
a2
45(a2 + b2 + c 2 )
+
3
≥
.
b2 + c 2
8(a2 + b2 + c 2 ) + 2(a b + bc + ca)
Applying the Cauchy-Schwarz inequality, we have
X
P 2 2
a
a2
(a2 + b2 + c 2 )2
P
≥
=
.
b2 + c 2
a2 (b2 + c 2 ) 2(a2 b2 + b2 c 2 + c 2 a2 )
Therefore, it suffices to show that
(a2 + b2 + c 2 )2
45(a2 + b2 + c 2 )
+
3
≥
,
2(a2 b2 + b2 c 2 + c 2 a2 )
8(a2 + b2 + c 2 ) + 2(a b + bc + ca)
210
Vasile Cîrtoaje
which is equivalent to
45(a2 + b2 + c 2 )
(a2 + b2 + c 2 )2
−
3
≥
− 9,
a2 b2 + b2 c 2 + c 2 a2
4(a2 + b2 + c 2 ) + a b + bc + ca
9(a2 + b2 + c 2 − a b − bc − ca)
a4 + b4 + c 4 − a2 b2 − b2 c 2 − c 2 a2
.
≥
a2 b2 + b2 c 2 + c 2 a2
4(a2 + b2 + c 2 ) + a b + bc + ca
By Schur’s inequality of degree four, we have
a4 + b4 + c 4 − a2 b2 − b2 c 2 − c 2 a2 ≥ (a2 + b2 + c 2 − a b − bc − ca)(a b + bc + ca) ≥ 0.
Therefore, it suffices to show that
[4(a2 + b2 + c 2 ) + a b + bc + ca](a b + bc + ca) ≥ 9(a2 b2 + b2 c 2 + c 2 a2 ).
Since
(a b + bc + ca)2 ≥ a2 b2 + b2 c 2 + c 2 a2 ,
this inequality is true if
4(a2 + b2 + c 2 )(a b + bc + ca) ≥ 8(a2 b2 + b2 c 2 + c 2 a2 ),
which is equivalent to the obvious inequality
a b(a − b)2 + bc(b − c)2 + ca(c − a)2 + a bc(a + b + c) ≥ 0.
The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).
Second Solution. Write the inequality as f6 (a, b, c) ≥ 0, where
Y
X 2
f6 (a, b, c) = 8(a2 + b2 + c 2 ) + 2(a b + bc + ca)
(a + b2 )(a2 + c 2 ) − 45
(b2 + c 2 ).
Clearly, f6 (a, b, c) has the same highest coefficient A as
Y
Y
f (a, b, c) = −45
(b2 + c 2 ) = −45
(p2 − 2q − a2 ),
where p = a + b + c and q = a b + bc + ca; that is,
A = 45.
Since A > 0, we will apply the highest coefficient cancellation method. We have
f6 (a, 1, 1) = 4a(2a + 5)(a2 + 1)(a − 1)2 ,
f6 (0, b, c) = (b − c)2 [8(b4 + c 4 ) + 18bc(b2 + c 2 ) + 15b2 c 2 ].
Symmetric Rational Inequalities
211
Since
f6 (1, 1, 1) = f6 (0, 1, 1) = 0,
define the homogeneous function
P(a, b, c) = a bc + B(a + b + c)3 + C(a + b + c)(a b + bc + ca)
such that P(1, 1, 1) = P(0, 1, 1) = 0; that is,
1
4
P(a, b, c) = a bc + (a + b + c)3 − (a + b + c)(a b + bc + ca).
9
9
We will show that the following sharper inequality holds
f6 (a, b, c) ≥ 45P 2 (a, b, c).
Let us denote
g6 (a, b, c) = f6 (a, b, c) − 45P 2 (a, b, c).
Clearly, g6 (a, b, c) has the highest coefficient A = 0. By P 3.76-(a) in Volume 1, it suffices
to prove that g6 (a, 1, 1) ≥ 0 and g6 (0, b, c) ≥ 0 for all a, b, c ≥ 0. We have
P(a, 1, 1) =
a(a − 1)2
,
9
hence
g6 (a, 1, 1) = f6 (a, 1, 1) − 45P 2 (a, 1, 1) =
Also, we have
P(0, b, c) =
a(a − 1)2 (67a3 + 190a2 + 67a + 180)
≥ 0.
9
(b + c)(b − c)2
,
9
hence
g6 (0, b, c) = f6 (0, b, c) − 45P 2 (0, b, c)
=
(b − c)2 [67(b4 + c 4 ) + 162bc(b2 + c 2 ) + 145b2 c 2 ]
≥ 0.
9
P 1.158. If a, b, c are real numbers, no two of which are zero, then
a2 − 7bc b2 − 7ca c 2 − 7a b 9(a b + bc + ca)
+ 2
+ 2
+
≥ 0.
b2 + c 2
a + b2
a + b2
a2 + b2 + c 2
(Vasile Cîrtoaje, 2014)
212
Vasile Cîrtoaje
Solution. Let
p = a + b + c,
q = a b + bc + ca,
r = a bc.
Write the inequality as f8 (a, b, c) ≥ 0, where
X
f8 (a, b, c) =(a2 + b2 + c 2 )
(a2 − 7bc)(a2 + b2 )(a2 + c 2 )
Y
+ 9(a b + bc + ca)
(b2 + c 2 )
is a symmetric homogeneous polynomial of degree eight. Always, f8 (a, b, c) can be
written in the form
f8 (a, b, c) = A(p, q)r 2 + B(p, q)r + C(p, q),
where the highest polynomial A(p, q) has the form αp2 + βq. Since
X
f8 (a, b, c) =(p2 − 2q)
(a2 − 7bc)(p2 − 2q − c 2 )(p2 − 2q − b2 )
Y
+ 9q
(p2 − 2q − a2 ),
f8 (a, b, c) has the same highest polynomial as
X
g8 (a, b, c) =(p2 − 2q)
(a2 − 7bc)b2 c 2 + 9q(−a2 b2 c 2 )
X
€
Š
=(p2 − 2q) 3r 2 − 7
b3 c 3 − 9qr 2 ;
that is,
A(p, q) = (p2 − 2q)(3 − 21) − 9q = −9(p2 − 3q).
Since A(p, q) ≤ 0 for all real a, b, c, by Lemma below, it suffices to prove that f8 (a, 1, 1) ≥
0 for all a, b, c ≥ 0. We have
f8 (a, 1, 1) = (a2 + 1)(a − 1)2 (a + 2)2 (a2 − 2a + 3) ≥ 0.
The equality holds for a = b = c, and also for −a/2 = b = c (or any cyclic permutation).
Lemma. Let
p = a + b + c,
q = a b + bc + ca,
r = a bc,
and let f8 (a, b, c) be a symmetric homogeneous polynomial of degree eight written in the
form
f8 (a, b, c) = A(p, q)r 2 + B(p, q)r + C(p, q),
where A(p, q) ≤ 0 for all real a, b, c. The inequality f8 (a, b, c) ≥ 0 holds for all real
numbers a, b, c if and only if f8 (a, 1, 1) ≥ 0 for all real a.
Proof. For fixed p and q,
h8 (r) = A(p, q)r 2 + B(p, q)r + C(p, q)
Symmetric Rational Inequalities
213
is a concave quadratic function of r. Therefore, h8 (r) is minimal when r is minimal or
maximal; this is, according to P 2.53 in Volume 1, when f8 (a, 1, 1) ≥ 0 and f8 (a, 0, 0) ≥ 0
for all real a. Notice that the condition " f8 (a, 0, 0) ≥ 0 for all real a" is not necessary
because it follows from the condition " f8 (a, 1, 1) ≥ 0 for all real a" as follows:
f8 (a, 0, 0) = lim f8 (a, t, t) = lim t 8 f8 (a/t, 1, 1) ≥ 0.
t→0
t→0
Notice that A(p, q) is called the highest polynomial of f8 (a, b, c).
Remark. This Lemma can be extended as follow.
• The inequality f8 (a, b, c) ≥ 0 holds for all real numbers a, b, c satisfying A(p, q) ≤ 0
if and only if f8 (a, 1, 1) ≥ 0 for all real a such that A(a + 2, 2a + 1) ≤ 0.
P 1.159. If a, b, c are real numbers such that a bc 6= 0, then
(b + c)2 (c + a)2 (a + b)2
10(a + b + c)2
+
+
≥
2
+
.
a2
b2
c2
3(a2 + b2 + c 2 )
(Vasile Cîrtoaje and Michael Rozenberg, 2014)
Solution. Let
p = a + b + c,
q = a b + bc + ca,
r = a bc.
Write the inequality as f8 (a, b, c) ≥ 0, where
”X
—
f8 (a, b, c) = 3(a2 + b2 + c 2 )
b2 c 2 (b + c)2 − 2a2 b2 c 2 − 10a2 b2 c 2 (a + b + c)2 .
From
X
b2 c 2 (b + c)2 − 2a2 b2 c 2 =
X
b2 c 2 (p − a)2 − 2r 2 = p2
X
b2 c 2 − 2pqr + r 2 ,
it follows that f8 (a, b, c) has the same highest polynomial as
3(a2 + b2 + c 2 )r 2 − 10r 2 (a + b + c)2 ;
that is,
A(p, q) = 3(p2 − 2q) − 10p2 = −7p2 − 6q.
There are two cases to consider.
Case 1: A(p, q) ≤ 0. According to Remark from the preceding P 1.158, it suffices to show
that f8 (a, 1, 1) ≥ 0 for all real a such that A(a + 2, 2a + 1) ≤ 0. Indeed, we have
f8 (a, 1, 1) = 3(a2 + 2)[4 + 2a2 (a + 1)2 − 2a2 ] − 10a2 (a + 2)2
= 2(3a6 + 6a5 + a4 − 8a3 − 14a2 + 12)
= 2(a − 1)2 (3a4 + 12a3 + 22a2 + 24a + 12)
= 2(a − 1)2 [3(a + 1)4 + (2a + 3)2 ] ≥ 0.
214
Vasile Cîrtoaje
Case 2: A(p, q) > 0. We will show that there exist two real numbers B and C such that
the following sharper inequality holds:
f8 (a, b, c) ≥ A(p, q)P 2 (a, b, c),
where
P(a, b, c) = r + Bp3 + C pq.
Let us denote
g8 (a, b, c) = f8 (a, b, c) − A(p, q)P 2 (a, b, c).
We see that the highest polynomial of g8 (a, b, c) is zero. Thus, according to Remark
from P 1.158, it suffices to prove that g(a) ≥ 0 for all real a, where g(a) = g8 (a, 1, 1).
We have
g(a) = f8 (a, 1, 1) − A(a + 2, 2a + 1)P 2 (a, 1, 1),
where
f8 (a, 1, 1) = 2(3a6 + 6a5 + a4 − 8a3 − 14a2 + 12),
A(a + 2, 2a + 1) = −7(a + 2)2 − 6(2a + 1) = −7a2 − 40a − 34,
P(a, 1, 1) = a + B(a + 2)3 + C(a + 2)(2a + 1).
Since
g(−2) = 72 − 18(−2)2 = 0,
a necessary condition to have g(a) ≥ 0 in the vicinity of −2 is g 0 (−2) = 0. This condition
involves C = −1. We can check that g 00 (−2) = 0 for this value of C. Thus, a necessary
condition to have g(a) ≥ 0 in the vicinity of −2 is g 000 (−2) = 0 is necessary. This
condition involves B = 2/3. For these values of B and C, we have
•
˜2
2
2
3
g(a) = f8 (a, 1, 1) + (7a + 40a + 34) a + (a + 2) − (a + 2)(2a + 1)
3
2
4
4
3
2
= (a + 2) (14a + 52a + 117a + 154a + 113).
9
Since
14a4 + 52a3 + 117a2 + 154a + 113 = (a2 + 1)2 + 13a2 (a + 2)2 + 7(9a2 + 22a + 16) > 0,
the proof is completed. The equality holds for a = b = c.
P 1.160. If a, b, c are nonnegative real numbers, no two of which are zero, then
a2 − 4bc b2 − 4ca c 2 − 4a b 9(a b + bc + ca) 9
+ 2
+ 2
+
≥ .
b2 + c 2
a + b2
a + b2
a2 + b2 + c 2
2
(Vasile Cîrtoaje, 2014)
Symmetric Rational Inequalities
215
Solution. Let
p = a + b + c,
q = a b + bc + ca,
r = a bc.
Write the inequality as f8 (a, b, c) ≥ 0, where
X
f8 (a, b, c) =2(a2 + b2 + c 2 )
(a2 − 4bc)(a2 + b2 )(a2 + c 2 )
Y
+ 9(2a b + 2bc + 2ca − a2 − b2 − c 2 )
(b2 + c 2 )
is a symmetric homogeneous polynomial of degree eight. Always, f8 (a, b, c) can be
written in the form
f8 (a, b, c) = A(p, q)r 2 + B(p, q)r + C(p, q),
where A(p, q) = αp2 + βq is called the highest polynomial of f8 (a, b, c). Since
X
f8 (a, b, c) =2(p2 − 2q)
(a2 − 4bc)(p2 − 2q − c 2 )(p2 − 2q − b2 )
Y
+ 9(4q − p2
(p2 − 2q − a2 ),
f8 (a, b, c) has the same highest polynomial as
X
g8 (a, b, c) =2(p2 − 2q)
(a2 − 4bc)b2 c 2 + 9(4q − p2 )(−a2 b2 c 2 )
X
€
Š
=2(p2 − 2q) 3r 2 − 4
b3 c 3 − 9(4q − p2 )r 2 ;
that is,
A(p, q) = 2(p2 − 2q)(3 − 12) − 9(4q − p2 ) = −9p2 .
Since A(p, q) ≤ 0 for all a, b, c ≥ 0, by Lemma below, it suffices to prove that f8 (a, 1, 1) ≥
0 and f8 (0, b, c) ≥ 0 for all a, b, c ≥ 0. We have
f8 (a, 1, 1) = 2a(a + 4)(a2 + 1)(a − 1)4 ≥ 0
and
f8 (0, b, c) = (b2 + c 2 )(2E − 9F ),
where
E = −4b3 c 3 + (b2 + c 2 )(b4 + c 4 ),
F = b2 c 2 (b − c)2 .
Since
E ≥ −4b3 c 3 + 2bc(b4 + c 4 ) = 2bc(b2 − c 2 )2 ,
we have
2E − 9F ≥ 4bc(b2 − c 2 )2 − 9b2 c 2 (b − c)2 = bc(b − c)2 [4(b + c)2 − 9bc] ≥ 0.
Thus, the proof is completed. The equality holds for a = b = c, and also for a = 0 and
b = c (or any cyclic permutation).
216
Vasile Cîrtoaje
Lemma. Let
p = a + b + c,
q = a b + bc + ca,
r = a bc,
and let f8 (a, b, c) be a symmetric homogeneous polynomial of degree eight written in the
form
f8 (a, b, c) = A(p, q)r 2 + B(p, q)r + C(p, q),
where A(p, q) ≤ 0 for all a, b, c ≥ 0. The inequality f8 (a, b, c) ≥ 0 holds for all nonnegative
real numbers a, b, c if and only if f8 (a, 1, 1) ≥ 0 and f8 (0, b, c) ≥ 0 for all a, b, c ≥ 0.
Proof. For fixed p and q,
h8 (r) = A(p, q)r 2 + B(p, q)r + C(p, q)
is a concave quadratic function of r. Therefore, h8 (r) is minimal when r is minimal or
maximal. This is, according to P 3.57 in Volume 1, when b = c or a = 0. Thus, the
conclusion follows. Notice that A(p, q) is called the highest polynomial of f8 (a, b, c).
Remark. This Lemma can be extended as follow.
• The inequality f8 (a, b, c) ≥ 0 holds for all a, b, c ≥ 0 satisfying A(p, q) ≤ 0 if and only
if f8 (a, 1, 1) ≥ 0 and f8 (0, b, c) ≥ 0 for all a, b, c ≥ 0 such that A(a + 2, 2a + 1) ≤ 0 and
A(b + c, bc) ≤ 0.
P 1.161. If a, b, c are nonnegative real numbers, no two of which are zero, then
9(a − b)2 (b − c)2 (c − a)2
a2 + b2 + c 2
≥1+
.
a b + bc + ca
(a + b)2 (b + c)2 (c + a)2
(Vasile Cîrtoaje, 2014)
Solution. Consider the non-trivial case where a, b, c are distinct and a = min{a, b, c}.
Write the inequality as follows:
(a − b)2 + (b − c)2 + (c − a)2
9(a − b)2 (b − c)2 (c − a)2
≥
,
2(a b + bc + ca)
(a + b)2 (b + c)2 (c + a)2
(a − b)2 + (b − c)2 + (c − a)2
18(a b + bc + ca)
≥
,
2
2
2
(a − b) (b − c) (c − a)
(a + b)2 (b + c)2 (c + a)2
X
1
18(a b + bc + ca)
≥
.
2
2
(b − a) (c − a)
(a + b)2 (a + c)2 (b + c)2
Since
X
1
1
1
1
2(b2 + c 2 − bc)
≥
+
+
=
(b − a)2 (c − a)2
b2 c 2 b2 (b − c)2 c 2 (b − c)2
b2 c 2 (b − c)2
Symmetric Rational Inequalities
and
217
a b + bc + ca
1
a b + bc + ca
≤
≤
,
2
2
2
2
2
(a + b) (a + c) (b + c)
(a b + bc + ca) (b + c)
bc(b + c)2
it suffices to show that
b2 + c 2 − bc
9
≥
.
2
2
2
b c (b − c)
bc(b + c)2
Write this inequality as follows:
(b + c)2 − 3bc
9(b + c)2 − 36bc
≥
,
bc
(b + c)2
(b + c)2
36bc
− 12 +
≥ 0,
bc
(b + c)2
(b + c)4 − 12bc(b + c)2 + 36b2 c 2 ≥ 0,
[(b + c)2 − 6bc]2 ≥ 0.
Thus, the proof is completed. The equality holds for a = b = c, and also for a = 0 and
b/c + c/b = 4 (or any cyclic permutation).
P 1.162. If a, b, c are nonnegative real numbers, no two of which are zero, then
p
(a − b)2 (b − c)2 (c − a)2
a2 + b2 + c 2
≥ 1 + (1 + 2)2 2
.
a b + bc + ca
(a + b2 )(b2 + c 2 )(c 2 + a2 )
(Vasile Cîrtoaje, 2014)
p
Solution. Consider the non-trivial case where a, b, c are distinct and denote k = 1+ 2.
Write the inequality as follows:
(a − b)2 + (b − c)2 + (c − a)2
k2 (a − b)2 (b − c)2 (c − a)2
≥ 2
,
2(a b + bc + ca)
(a + b2 )(b2 + c 2 )(c 2 + a2 )
(a − b)2 + (b − c)2 + (c − a)2
2k2 (a b + bc + ca)
≥
,
(a − b)2 (b − c)2 (c − a)2
(a2 + b2 )(b2 + c 2 )(c 2 + a2 )
X
1
2k2 (a b + bc + ca)
≥
.
(b − a)2 (c − a)2
(a2 + b2 )(b2 + c 2 )(c 2 + a2 )
Assume that a = min{a, b, c}, and use the substitution
b = a + x,
c = a + y,
x, y ≥ 0.
218
Vasile Cîrtoaje
The inequality becomes
1
x2 y2
+
x 2 (x
1
1
+ 2
≥ 2k2 f (a),
2
− y)
y (x − y)2
where
f (a) =
3a2 + 2(x + y)a + x y
.
(2a2 + 2x a + x 2 )(2a2 + 2 y a + y 2 )[2a2 + 2(x + y)a + x 2 + y 2 ]
We will show that
1
x2 y2
+
x 2 (x
1
1
+ 2
≥ 2k2 f (0) ≥ 2k2 f (a).
2
− y)
y (x − y)2
We have
2(x 2 + y 2 − x y)
2k2 x y
1
1
1
2
+
+
−
2k
f
(0)
=
−
x 2 y 2 x 2 (x − y)2
y 2 (x − y)2
x 2 y 2 (x − y)2
x 2 y 2 (x 2 + y 2 )
p
2[x 2 + y 2 − (2 + 2 )x y]2
≥ 0.
= 2 2
x y (x − y)2 (x 2 − x y + y 2 )
Also, since
(2a2 + 2x a + x 2 )(2a2 + 2 y a + y 2 ) ≥ 2(x 2 + y 2 )a2 + 2x y(x + y)a + x 2 y 2
and
2a2 + 2(x + y)a + x 2 + y 2 ≥ x 2 + y 2 ,
we get f (a) ≤ g(a), where
g(a) =
3a2 + 2(x + y)a + x y
.
[2(x 2 + y 2 )a2 + 2x y(x + y)a + x 2 y 2 ](x 2 + y 2 )
Therefore,
1
− g(a)
x y(x 2 + y 2 )
(2x 2 + 2 y 2 − 3x y)a2
=
≥ 0.
x y(x 2 + y 2 )[2(x 2 + y 2 )a2 + 2x y(x + y)a + x 2 y 2 ]
f (0) − f (a) ≥
Thus, the proof is
pcompleted. The equality holds for a = b = c, and also for a = 0 and
b/c + c/b = 2 + 2 (or any cyclic permutation).
Symmetric Rational Inequalities
219
P 1.163. If a, b, c are nonnegative real numbers, no two of which are zero, then
2
2
5
5
5
2
+
+
≥
+
+
.
a+b b+c c+a
3a + b + c 3b + c + a 3c + a + b
Solution. Write the inequality as follows:
‹
X 2
5
−
≥ 0,
b + c 3a + b + c
X
2a − b − c
≥ 0,
(b + c)(3a + b + c)
X
X
a−c
a−b
+
≥ 0,
(b + c)(3a + b + c)
(b + c)(3a + b + c)
X
X
a−b
b−a
+
≥ 0,
(b + c)(3a + b + c)
(c + a)(3b + c + a)
X
(a − b)2 (a + b − c)
,
(b + c)(c + a)(3a + b + c)(3b + c + a)
X
(b − c)2 Sa ≥ 0,
where
Sa = (b + c − a)(b + c)(3a + b + c).
Assume that a ≥ b ≥ c. Since Sc > 0, it suffices to show that
(b − c)2 Sa + (a − c)2 S b ≥ 0.
Since S b ≥ 0 and (a − c)2 ≥ (b − c)2 , we have
(b − c)2 Sa + (a − c)2 S b ≥ (b − c)2 Sa + (b − c)2 S b = (b − c)2 (Sa + S b ).
Thus, it is enough to prove that Sa + S b ≥ 0, which is equivalent to
(c + a − b)(c + a)(3b + c + a) ≥ (b + c − a)(b + c)(3a + b + c).
Consider the nontrivial case where b + c − a > 0. Since c + a − b ≥ b + c − a, we only
need to show that
(c + a)(3b + c + a) ≥ (b + c)(3a + b + c).
Indeed,
(c + a)(3b + c + a) − (b + c)(3a + b + c) = (a − b)(a + b − c) ≥ 0.
Thus, the proof is completed. The equality holds for a = b = c, and also for a = 0 and
b = c (or any cyclic permutation).
220
Vasile Cîrtoaje
P 1.164. If a, b, c are real numbers, no two of which are zero, then
(a)
8b2 + 3ca
8c 2 + 3a b
8a2 + 3bc
+
+
≥ 11;
b2 + bc + c 2 c 2 + ca + a2 a2 + a b + b2
(b)
8a2 − 5bc
8b2 − 5ca
8c 2 − 5a b
+
+
≥ 9.
b2 − bc + c 2 c 2 − ca + a2 a2 − a b + b2
(Vasile Cîrtoaje, 2011)
Solution. Consider the more general inequality
a2 + mbc
b2 + mca
c 2 + ma b
3(m + 1)
+
+
≥
.
2
2
2
2
2
2
b + k bc + c
c + kca + a
a + ka b + b
k+2
Let p = a + b + c and q = a b + bc + ca. Write the inequality in the form f6 (a, b, c) ≥ 0,
where
X
f6 (a, b, c) = (k + 2)
(a2 + mbc)(a2 + ka b + b2 )(a2 + kac + c 2 )
Y
−3(m + 1)
(b2 + k bc + c 2 ).
From
f6 (a, b, c) = (k + 2)
X
(a2 + mbc)(ka b − c 2 + p2 − 2q)(kac − b2 + p2 − 2q)
Y
−3(m + 1)
(k bc − a2 + p2 − 2q).
it follows that f6 (a, b, c) has the same highest coefficient A as
(k + 2)P2 (a, b, c) − 3(m + 1)P3 (a, b, c),
where
X
(a2 + mbc)(ka b − c 2 )(kac − b2 ),
Y
P3 (a, b, c) =
(k bc − a2 ).
P2 (a, b, c) =
According to Remark 2 from the proof of P 2.75 in Volume 1,
A = (k + 2)P2 (1, 1, 1) − 3(m + 1)P3 (1, 1, 1)
= 3(k + 2)(m + 1)(k − 1)2 − 3(m + 1)(k − 1)3 = 9(m + 1)(k − 1)2 .
Also, we have
f6 (a, 1, 1) = (k + 2)(a2 + ka + 1)(a − 1)2 [a2 + (k + 2)a + 1 + 2k − 2m].
(a) For our particular case m = 3/8 and k = 1, we have A = 0. Therefore, according
to P 2.75 in Volume 1, it suffices to prove that f6 (a, 1, 1) ≥ 0 for all real a. Indeed,

‹
3 2
f6 (a, 1, 1) = 3(a2 + a + 1)(a − 1)2 a +
≥ 0.
2
Symmetric Rational Inequalities
221
Thus, the proof is completed. The equality holds for a = b = c, and also for −2a/3 =
b = c (or any cyclic permutation).
(b) For m = −5/8 and k = −1, we have A = 27/2 and
f6 (a, 1, 1) =
1 2
(a − a + 1)(a − 1)2 (2a + 1)2 .
4
Since A > 0, we will use the highest coefficient cancellation method. Define the homogeneous polynomial
P(a, b, c) = r + Bp3 + C pq,
where B and C are real constants. Since the desired inequality becomes an equality
for a = b = c = 1, and also for a = −1 and b = c = 2, determine B and C such that
P(1, 1, 1) = P(−1, 2, 2) = 0. We find
B=
when
P(a, 1, 1) =
4
,
27
C=
2
(a − 1)2 (2a + 1),
27
We will show that
f6 (a, b, c) ≥
−5
,
9
P(a, 0, 0) = Ba3 =
4 3
a .
27
27 2
P (a, b, c).
2
Let us denote
g6 (a, b, c) = f6 (a, b, c) −
27 2
P (a, b, c).
2
Since g6 (a, b, c) has the highest coefficient A = 0, it suffices to prove that g6 (a, 1, 1) ≥ 0
for all real a (see P 2.75 in Volume 1). Indeed,
g6 (a, 1, 1) = f6 (a, 1, 1) −
27 2
1
P (a, 1, 1) =
(a − 1)2 (2a + 1)2 (19a2 − 11a + 19) ≥ 0.
2
108
Thus, the proof is completed. The equality holds for a = b = c, and also for −2a = b = c
(or any cyclic permutation).
P 1.165. If a, b, c are real numbers, no two of which are zero, then
4a2 + bc
4b2 + ca
4c 2 + a b
+
+
≥ 1.
4b2 + 7bc + 4c 2 4c 2 + 7ca + 4a2 4a2 + 7a b + 4b2
(Vasile Cîrtoaje, 2011)
222
Vasile Cîrtoaje
Solution. Write the inequality as f6 (a, b, c) ≥ 0, where
X
Y
f6 (a, b, c) =
(4a2 + bc)(4a2 + 7a b + 4b2 )(4a2 + 7ac + 4c 2 ) −
(4b2 + 7bc + 4c 2 ).
Let
p = a + b + c, q = a b + bc + ca,
r = a bc.
From
f6 (a, b, c) =
X
(4a2 + bc)(7a b − 4c 2 + 4p2 − 8q)(7ac − 4b2 + 4p2 − 8q)
−
Y
(7bc − 4a2 + 4p2 − 8q),
it follows that f6 (a, b, c) has the same highest coefficient A as
P2 (a, b, c) − P3 (a, b, c),
where
X
(4a2 + bc)(7a b − 4c 2 )(7ac − 4b2 ),
Y
P3 (a, b, c) =
(7bc − 4a2 ).
P2 (a, b, c) =
According to Remark 2 from the proof of P 2.75 in Volume 1,
A = P2 (1, 1, 1) − P3 (1, 1, 1) = 135 − 27 = 108.
Since A > 0, we will apply the highest coefficient cancellation method. Define the homogeneous polynomial
P(a, b, c) = r + Bp3 + C pq,
where B and C are real constants. We will show that there are two real numbers B and
C such that the following sharper inequality holds
f6 (a, b, c) ≥ 108P 2 (a, b, c).
Let us denote
g6 (a, b, c) = f6 (a, b, c) − 108P 2 (a, b, c).
Clearly, g6 (a, b, c) has the highest coefficient A1 = 0. Then, by P 2.75 in Volume 1, it
suffices to prove that g6 (a, 1, 1) ≥ 0 for all real a.
We have
g6 (a, 1, 1) = f6 (a, 1, 1) − 108P 2 (a, 1, 1),
where
f6 (a, 1, 1) = 4(4a2 + 7a + 4)(a − 1)2 (4a2 + 15a + 16),
P(a, 1, 1) = a + B(a + 2)3 + C(a + 2)(2a + 1).
Symmetric Rational Inequalities
223
Let us denote g(a) = f6 (a, 1, 1). Since g(−2) = 0, we can have g(a) ≥ 0 in the vicinity
of a = −2 only if g 0 (−2) = 0, which involves C = −5/9. On the other hand, from
g(1) = 0, we get B = 4/27. For these values of B and C, we get
P(a, 1, 1) =
g6 (a, 1, 1) =
2(a − 1)2 (2a + 1)
,
27
4
(a − 1)2 (a + 2)2 (416a2 + 728a + 431) ≥ 0.
27
The proof is completed. The equality holds for a = b = c, and for a = 0 and b + c = 0
(or any cyclic permutation).
P 1.166. If a, b, c are real numbers, no two of which are equal, then
1
1
1
27
+
+
≥
.
2
2
2
2
2
2
(a − b)
(b − c)
(c − a)
4(a + b + c − a b − bc − ca)
First Solution. Write the inequality as follows
(a − b)2 + (b − c)2 + (a − c)2
•
(a − b)2 (b − c)2
+
+1
(a − c)2 (a − c)2
˜
1
1
27
1
+
+
≥
,
(a − b)2 (b − c)2 (a − c)2
2
(a − c)2 (a − c)2
27
+
+1 ≥
,
(a − b)2 (b − c)2
2

‹
1
1
27
(x 2 + y 2 + 1)
+
+
1
≥
,
x2
y2
2
where
x=
a−b
,
a−c
y=
b−c
,
a−c
x + y = 1.
We have
‹
1
1
27 (x + 1)2 (x − 2)2 (2x − 1)2
(x + y + 1)
+
+
1
−
=
≥ 0.
x2
y2
2
2x 2 (1 − x)2
2
2

The proof is completed. The equality holds for 2a = b + c (or any cyclic permutation).
Second Solution. Assume that a > b > c. We have
1
1
2
8
8
+
≥
≥
=
.
2
2
2
(a − b)
(b − c)
(a − b)(b − c) [(a − b) + (b − c)]
(a − c)2
224
Vasile Cîrtoaje
Therefore, it suffices to show that
27
9
,
≥
2
2
2
2
(a − c)
4(a + b + c − a b − bc − ca)
which is equivalent to
(a − 2b + c)2 ≥ 0.
Third Solution. Write the inequality as f6 (a, b, c) ≥ 0, where
X
f6 (a, b, c) = 4(a2 + b2 + c 2 − a b − bc − ca)
(a − b)2 (a − c)2 −27(a − b)2 (b − c)2 (c − a)2 .
Clearly, f6 (a, b, c) has the same highest coefficient A as
−27(a − b)2 (b − c)2 (c − a)2 ;
that is,
A = −27(−27) = 729.
Since A > 0, we will use the highest coefficient cancellation method. Define the homogeneous polynomial

‹
1
P(a, b, c) = a bc + B(a + b + c)3 − 3B +
(a + b + c)(a b + bc + ca),
9
which satisfies the property P(1, 1, 1) = 0. We will show that there is a real value of B
such that the following sharper inequality holds
f6 (a, b, c) ≥ 729P 2 (a, b, c).
Let us denote
g6 (a, b, c) = f6 (a, b, c) − 729P 2 (a, b, c).
Clearly, g6 (a, b, c) has the highest coefficient A1 = 0. Then, by P 2.75 in Volume 1, it
suffices to prove that g6 (a, 1, 1) ≥ 0 for all real a.
We have
f6 (a, 1, 1) = 4(a − 1)6
and
P(a, 1, 1) =
1
(a − 1)2 [9B(a + 2) + 2],
9
hence
g6 (a, 1, 1) = f6 (a, 1, 1)−729P 2 (a, 1, 1) = (27B +2)(a −1)4 (a +2)[(2−27B)a −54B −8].
Choosing B = −2/27, we get g6 (a, 1, 1) = 0 for all real a.
Remark. The inequality is equivalent to
(a − 2b + c)2 (b − 2c + a)2 (c − 2a + b)2 ≥ 0.
Symmetric Rational Inequalities
225
P 1.167. If a, b, c are real numbers, no two of which are zero, then
a2
1
1
1
14
+ 2
+ 2
≥
.
2
2
2
2
− ab + b
b − bc + c
c − ca + a
3(a + b2 + c 2 )
(Vasile Cîrtoaje and BJSL, 2014)
Solution. Write the inequality as f6 (a, b, c) ≥ 0, where
X
f6 (a, b, c) = 3(a2 + b2 + c 2 )
(a2 − a b + b2 )(a2 − ac + c 2 )
−14(a2 − a b + b2 )(b2 − bc + c 2 )(c 2 − ca + a2 ).
Clearly, f6 (a, b, c) has the same highest coefficient A as
−14(a2 − a b + b2 )(b2 − bc + c 2 )(c 2 − ca + a2 ),
hence as
f (a, b, c) = −14(−c 2 − a b)(−a2 − bc)(−b2 − ca);
that is, according to Remark 2 from the proof of P 2.75 in Volume 1,
A = f (1, 1, 1) = −14(−2)3 = 112.
Since A > 0, we apply the highest coefficient cancellation method. Define the homogeneous polynomial
P(a, b, c) = a bc + B(a + b + c)3 + C(a + b + c)(a b + bc + ca).
We will show that there are two real numbers B and C such that the following sharper
inequality holds
f6 (a, b, c) ≥ 112P 2 (a, b, c).
Let us denote
g6 (a, b, c) = f6 (a, b, c) − 112P 2 (a, b, c).
Clearly, g6 (a, b, c) has the highest coefficient A1 = 0. By P 2.75 in Volume 1, it suffices
to prove that g6 (a, 1, 1) ≥ 0 for all real a.
We have
g6 (a, 1, 1) = f6 (a, 1, 1) − 112P 2 (a, 1, 1),
where
f6 (a, 1, 1) = (a2 − a + 1)(3a4 − 3a3 + a2 + 8a + 4),
P(a, 1, 1) = 1 + B(a + 2)3 + C(a + 2)(2a + 1).
Let us denote g(a) = g6 (a, 1, 1). Since g(−2) = 0, we can have g(a) ≥ 0 in the vicinity
of a = −2 only if g 0 (−2) = 0, which involves C = −4/7. In addition, setting B = 9/56,
we get
1
(9a3 − 10a2 + 4a + 8),
P(a, 1, 1) =
56
226
Vasile Cîrtoaje
3 6
(a + 4a5 + 8a4 + 16a3 + 20a2 + 16a + 16)
28
3(a + 2)2 (a2 + 2)2
=
≥ 0.
28
g6 (a, 1, 1) =
The proof is completed. The equality holds for a = 0 and b + c = 0 (or any cyclic
permutation).
P 1.168. Let a, b, c be real numbers such that a b + bc + ca ≥ 0 and no two of which are
zero. Prove that
(a)
(b) i f a b ≤ 0, then
a
b
c
3
+
+
≥ ;
b+c c+a a+b
2
a
b
c
+
+
≥ 2.
b+c c+a a+b
(Vasile Cîrtoaje, 2014)
Solution. Let as show first that b + c 6= 0, c + a 6= 0 and a + b 6= 0. Indeed, if b + c = 0,
then a b + bc + ca ≥ 0 yields b = c = 0, which is not possible.
(a) Write the inequality as follows
9
X a
+1 ≥ ,
b+c
2

”X
— X 1 ‹
(b + c)
≥ 9,
b+c
‹
Xa + b a + c
+
− 2 ≥ 0,
a+c a+b
X
(b − c)2
≥ 0,
(a + b)(a + c)
X
(b − c)2
≥ 0.
a2 + (a b + bc + ca)
Clearly, the last inequality is true. The equality holds for a = b = c 6= 0.
(b) From a b + bc + ca ≥ 0, it follows that if one of a, b, c is zero, then the others
are the same sign. In this case, the desired inequality is trivial. So, due to symmetry
and homogeneity, it suffices to consider that a < 0 < b ≤ c.
Symmetric Rational Inequalities
227
First Solution. We will show that
F (a, b, c) > F (0, b, c) ≥ 2,
where
F (a, b, c) =
b
c
a
+
+
.
b+c c+a a+b
We have
F (0, b, c) =
b c
+ ≥2
c
b
and
F (a, b, c) − F (0, b, c) = a
•
˜
1
b
c
−
−
.
b + c c(c + a) b(a + b)
Since a < 0, we need to show that
c
1
b
+
>
.
c(c + a) b(a + b)
b+c
From a b + bc + ca ≥ 0, we get
c+a≥
−ca
> 0,
b
a+b≥
−a b
> 0,
c
hence
b
b
> 2,
c(c + a)
c
c
c
> 2.
b(a + b)
b
Therefore, it suffices to prove that
b
c
1
+ 2≥
.
2
c
b
b+c
Indeed, by virtue of the AM-GM inequality, we have
b
c
1
2
1
+ 2−
≥ p − p > 0.
2
c
b
b+c
bc 2 bc
This completes the proof. The equality holds for a = 0 and b = c, or b = 0 and a = c.
Second Solution. Since b + c > 0 and
(b + c)(a + b) = b2 + (a b + bc + ca) > 0, (b + c)(c + a) = c 2 + (a b + bc + ca) > 0,
we get a + b > 0 and c + a > 0. By virtue of the Cauchy-Schwarz inequality and AM-GM
inequality, we have
228
Vasile Cîrtoaje
a
b
c
a
(b + c)2
+
+
≥
+
b+c c+a a+b
b + c b(c + a) + c(a + b)
a
(b + c)2
>
+
2a + b + c (b + c)2
+ a(b + c)
2
4a
2(b + c)
>
+
= 2.
2a + b + c 2a + b + c
P 1.169. If a, b, c are nonnegative real numbers, then
a
b
c
a b + bc + ca
+
+
≥
.
7a + b + c 7b + c + a 7c + a + b
(a + b + c)2
(Vasile Cîrtoaje, 2014)
First Solution. Write the inequality as follows:
˜
X•
2a
a(b + c)
−
≥ 0,
7a + b + c (a + b + c)2
X a[(a − b) + (a − c)](a − b − c)
≥ 0,
7a + b + c
X a(a − b)(a − b − c) X a(a − c)(a − b − c)
+
≥ 0,
7a + b + c
7a + b + c
X a(a − b)(a − b − c) X b(b − a)(b − c − a)
+
≥ 0,
7a + b + c
7b + c + a
˜
•
X
a(a − b − c) b(b − c − a)
−
≥ 0,
(a − b)
7a + b + c
7b + c + a
X
(a − b)2 (a2 + b2 − c 2 + 14a b)(a + b + 7c) ≥ 0.
Since
a2 + b2 − c 2 + 14a b ≥ (a + b)2 − c 2 = (a + b + c)(a + b − c),
it suffices to show that
X
(a − b)2 (a + b − c)(a + b + 7c) ≥ 0.
Assume that a ≥ b ≥ c. It suffices to show that
(a − c)2 (a − b + c)(a + 7b + c) + (b − c)2 (−a + b + c)(7a + b + c) ≥ 0.
Symmetric Rational Inequalities
229
For the nontrivial case b > 0, we have
(a − c)2 ≥
a2
a
(b − c)2 ≥ (b − c)2 .
2
b
b
Thus, it is enough to prove that
a(a − b + c)(a + 7b + c) + b(−a + b + c)(7a + b + c) ≥ 0.
Since
a(a + 7b + c) ≥ b(7a + b + c),
we have
a(a − b + c)(a + 7b + c) + b(−a + b + c)(7a + b + c) ≥
≥ b(a − b + c)(7a + b + c) + b(−a + b + c)(7a + b + c)
= 2bc(7a + b + c) ≥ 0.
This completes the proof. The equality holds for a = b = c, and also for a = 0 and b = c
(or any cyclic permutation).
Second Solution. Assume that a ≤ b ≤ c, a + b + c = 3 and use the substitution
x=
2a + 1
,
3
y=
2b + 1
2c + 1
, z=
,
3
3
where 1/3 ≤ x ≤ y ≤ z, x + y + z = 3. We have b + c ≥ 2, y + z ≥ 2, x ≤ 1. The
inequality becomes
a
b
c
9 − a2 − b2 − c 2
+
+
≥
,
2a + 1 2b + 1 2c + 1
6
9(x 2 + y 2 + z 2 ) ≥ 4

‹
1 1 1
+ +
+ 17.
x
y z
Assume that x ≤ y ≤ z and show that
E(x, y, z) ≥ E(x, t, t) ≥ 0,
where
t = ( y + z)/2 = (3 − x)/2
and
‹
1 1 1
E(x, y, z) = 9(x + y + z ) − 4
+ +
− 17.
x
y z
2
2
2

230
Vasile Cîrtoaje
We have
1 1 2
+ −
E(x, y, z) − E(x, t, t) = 9( y + z − 2t ) − 4
y z
t
2
( y − z) [9 yz( y + z) − 8]
=
≥ 0,
2 yz( y + z)
2
2
2

‹
since
9 yz = (2b + 1)(2c + 1) ≥ 2(b + c) + 1 ≥ 5,
y + z ≥ 2.
Also,
E(x, t, t) = 9x 2 + 2t 2 − 15 −
4 8 (x − 1)2 (3x − 1)(8 − 3x)
− =
≥ 0.
x
t
2x(3 − x)
Third Solution. Write the inequality as f5 (a, b, c) ≥ 0, where f5 (a, b, c) is a symmetric
homogeneous inequality of degree five. According to P 3.68-(a) in Volume 1, it suffices
to prove the inequality for a = 0 and for b = c = 1. For a = 0, the inequality is
equivalent to
(b − c)2 (b2 + c 2 + 11bc) ≥ 0,
while, for b = c = 1, the inequality is equivalent to
a(a − 1)2 (a + 14) ≥ 0.
P 1.170. If a, b, c are the lengths of the sides of a triangle, then
b2
c2
1
a2
+
+
≥ .
2
2
2
4a + 5bc 4b + 5ca 4c + 5a b
3
(Vasile Cîrtoaje, 2009)
Solution. Write the inequality as f6 (a, b, c) ≥ 0, where
X
Y
f6 (a, b, c) = 3
a2 (4b2 + 5ca)(4c 2 + 5a b) −
(4a2 + 5bc)
X
X
= −45a2 b2 c 2 − 25a bc
a3 + 40
a3 b3 .
Since f6 (a, b, c has the highest coefficient
A = −45 − 75 + 120 = 0,
Symmetric Rational Inequalities
231
according to 2.67-(b), it suffices to prove the original inequality b = c = 1 and 0 ≤ a ≤ 2,
and for a = b + c.
Case 1: b = c = 1, 0 ≤ a ≤ 2. The original inequality reduces to
(2 − a)(a − 1)2 ≥ 0,
which is true.
Case 2: a = b + c. Using the Cauchy-Schwarz inequality
b2
c2
(b + c)2
+
≥
,
4b2 + 5ca 4c 2 + 5a b
4(b2 + c 2 ) + 5a(b + c)
it suffices to show that
a2
(b + c)2
1
+
≥ ,
4a2 + 5bc 4(b2 + c 2 ) + 5a(b + c) 3
which reduces to the obvious inequality
(b − c)2 (3b2 + 3c 2 − 4bc) ≥ 0.
The equality holds for an equilateral triangle, and for a degenerate triangle with a/2 =
b = c (or any cyclic permutation).
P 1.171. If a, b, c are the lengths of the sides of a triangle, then
7a2
1
1
1
3
+ 2
+ 2
≥
.
2
2
2
2
2
2
+b +c
7b + c + a
7c + a + b
(a + b + c)2
(Vo Quoc Ba Can, 2010)
Solution. Let p = a + b + c and q = a b + bc + ca. Write the inequality as f6 (a, b, c) ≥ 0,
where
X
Y
f6 (a, b, c) = p2
(7b2 + c 2 + a2 )(7c 2 + a2 + b2 ) − 3
(7a2 + b2 + c 2 )
X
Y
= p2
(6b2 + p2 − 2q)(6c 2 + p2 − 2q) − 3
(6a2 + p2 − 2q).
Since f6 (a, b, c has the highest coefficient
A = −3(63 ) < 0,
according to 2.67-(b), it suffices to prove the original inequality b = c = 1 and 0 ≤ a ≤ 2,
and for a = b + c.
232
Vasile Cîrtoaje
Case 1: b = c = 1, 0 ≤ a ≤ 2. The original inequality reduces to
a(8 − a)(a − 1)2 ≥ 0,
which is true.
Case 2: a = b + c. Write the inequality as
4(b2
1
1
3
1
+ 2
+ 2
≥
,
2
2
2
+ c ) + 7bc 4b + c + bc 4c + b + bc
2(b + c)2
5x + 2
3
1
+
≥
,
2
4x + 7 4x + 5x + 10 2(x + 2)
where x = b/c + c/b, x ≥ 2. This inequality is equivalent to the obvious inequality
16x 2 + 5x − 38 ≥ 0.
The equality holds for an equilateral triangle, and for a degenerate triangle with a = 0
and b = c (or any cyclic permutation).
P 1.172. Let a, b, c be the lengths of the sides of a triangle. If k > −2, then
X a(b + c) + (k + 1)bc
b2
+ k bc
+ c2
≤
3(k + 3)
.
k+2
(Vasile Cîrtoaje, 2009)
Solution. Let p = a + b + c and q = a b + bc + ca. Write the inequality as f6 (a, b, c) ≥ 0,
where
Y
f6 (a, b, c) = 3(k + 3)
(b2 + k bc + c 2 )
X
−(k + 2)
[a(b + c) + (k + 1)bc](c 2 + kca + a2 )(a2 + ka b + b2 )
Y
= 3(k + 3)
(p2 − 2q + k bc − a2 )
X
−(k + 2)
(q + k bc)(p2 − 2q + kca − b2 )(p2 − 2q + ka b − c 2 ).
Since f6 (a, b, c has the same highest coefficient A as f (a, b, c), where
Y
X
f (a, b, c) = 3(k + 3)
(k bc − a2 ) − k(k + 2)
bc(kca − b2 )(ka b − c 2 )
X
X
= 3(k + 3)[(k3 − 1)a2 b2 c 2 − k2 a bc
a3 + k
a3 b3 ]
X
X
−k(k + 2)(3k2 a2 b2 c 2 − 2ka bc
a3 +
a3 b3 ),
Symmetric Rational Inequalities
233
we get
A = 3(k + 3)(k3 − 1 − 3k2 + 3k) − k(k + 2)(3k2 − 6k + 3) = −9(k − 1)2 ≤ 0.
According to 2.67-(b), it suffices to prove the original inequality b = c = 1 and 0 ≤ a ≤
2, and for a = b + c.
Case 1: b = c = 1, 0 ≤ a ≤ 2. The original inequality reduces to
(2 − a)(a − 1)2 ≥ 0,
which is true.
Case 2: a = b + c. Write the inequality as follows
˜
X • a(b + c) + (k + 1)bc
3
,
−1 ≤
2
2
b + k bc + c
k+2
X a b + bc + ca − b2 − c 2
b2
+ k bc
+ c2
≤
3
,
k+2
3bc
bc − c 2
bc − b2
3
+
+
≤
.
2
2
2
2
2
2
b + k bc + c
b + (k + 2)(bc + c ) c + (k + 2)(bc + b )
k+2
Since
3bc
3
≤
,
b2 + k bc + c 2
k+2
it suffices to prove that
bc − b2
bc − c 2
+
≤ 0.
b2 + (k + 2)(bc + c 2 ) c 2 + (k + 2)(bc + b2 )
This reduces to the obvious inequality
(b − c)2 (b2 + bc + c 2 ) ≥ 0.
The equality holds for an equilateral triangle, and for a degenerate triangle with a/2 =
b = c (or any cyclic permutation).
P 1.173. Let a, b, c be the lengths of the sides of a triangle. If k > −2, then
X 2a2 + (4k + 9)bc
b2
+ k bc
+ c2
≤
3(4k + 11)
.
k+2
(Vasile Cîrtoaje, 2009)
234
Vasile Cîrtoaje
Solution. Let p = a + b + c and q = a b + bc + ca. Write the inequality as f6 (a, b, c) ≥ 0,
where
Y
f6 (a, b, c) = 3(4k + 11)
(b2 + k bc + c 2 )
X
−(k + 2)
[2a2 + (4k + 9)bc](c 2 + kca + a2 )(a2 + ka b + b2 )
Y
= 3(4k + 11)
(p2 − 2q + k bc − a2 )
X
−(k + 2)
[2a2 + (4k + 9)bc](p2 − 2q + kca − b2 )(p2 − 2q + ka b − c 2 ).
Since f6 (a, b, c has the same highest coefficient A as f (a, b, c), where
Y
f (a, b, c) = 3(4k + 11)
(k bc − a2 )
X
[2a2 + (4k + 9)bc](kca − b2 )(ka b − c 2 )
X
X
= 3(4k + 11)[(k3 − 1)a2 b2 c 2 − k2 a bc
a3 + k
a3 b3 ]
X
X
−(k + 2)[3(4k3 + 9k2 + 2)a2 b2 c 2 − 6k(k + 3)a bc
a3 + 9
a3 b3 ],
−(k + 2)
we get
A = 3(4k + 11)(k3 − 1 − 3k2 + 3k) − (k + 2)[3(4k3 + 9k2 + 2) − 18k(k + 3) + 27]
= −9(4k + 11)(k − 1)2 ≤ 0.
According to 2.67-(b), it suffices to prove the original inequality b = c = 1 and 0 ≤ a ≤
2, and for a = b + c.
Case 1: b = c = 1, 0 ≤ a ≤ 2. The original inequality reduces to
(2 − a)(a − 1)2 ≥ 0,
which is true.
Case 2: a = b + c. Write the inequality as follows
X 2a2 + (4k + 9)bc
3(2k + 7)
−2 ≤
,
2
2
b + k bc + c
k+2
X 2(a2 − b2 − c 2 ) + (2k + 9)bc
+ c2
≤
3(2k + 7)
,
k+2
+ k bc
•
˜
c
b
(2k + 13)bc
+
(2k
+
5)(b
+
c)
+
b2 + k bc + c 2
b2 + (k + 2)(bc + c 2 ) c 2 + (k + 2)(bc + b2 )
b2
≤
3(2k + 7)
.
k+2
Symmetric Rational Inequalities
235
Using the substitution x = b/c + c/b, x ≥ 2, the inequality can be written as
2k + 13
(2k + 5)(x + 2)(x + 2k + 3)
3(2k + 7)
+
≤
,
2
2
x +k
(k + 2)x + (k + 2)(k + 3)x + 2k + 6k + 5
k+2
(x − 2)[4(k + 2)Ax 2 + 2(k + 2)B x + C] ≥ 0,
where
A = k + 4,
B = 2k2 + 13k + 22,
C = 8k3 + 51k2 + 98k + 65.
The inequality is true since A > 0,
B = 2(k + 2)2 + 5(k + 2) + 4 > 0,
C = 8(k + 2)3 + 2k2 + (k + 1)2 > 0.
The equality holds for an equilateral triangle, and for a degenerate triangle with a/2 =
b = c (or any cyclic permutation).
P 1.174. If a ≥ b ≥ c ≥ d such that a bcd = 1, then
1
1
3
1
+
+
≥
.
p
3
1+a 1+ b 1+c
1 + a bc
(Vasile Cîrtoaje, 2008)
Solution. We can get this inequality by summing the inequalities below
1
1
2
+
≥
p ,
1+a 1+ b
1 + ab
2
1
3
+
.
≥
p
p
3
1 + c 1 + ab
1 + a bc
The first inequality is true, since

‹ 
‹
1
1
2
1
1
1
1
+
−
−
−
=
+
p
p
p
1 + a 1 + b 1 + ab
1 + a 1 + ab
1 + b 1 + ab
p
p
p
( a − b)2 ( a b − 1)
=
p
(1 + a)(1 + b)(1 + a b)
p
p
and
a b ≥ a bcd = 1. To prove the second inequality, we denote x = a b andpy =
p
3
a bc (x ≥ y ≥ 1), which yield c = y 3 /x 2 . From a bc 2 ≥ a bcd = 1, we get a bc ≥ a b,
that is, y 3 ≥ x. Since
2
3
3
x2
1
2
+
=
+
−
p −
p
3
2
3
1 + c 1 + a b 1 + a bc
x +y
1+ x 1+ y
236
Vasile Cîrtoaje
=
1
x2
−
2
3
x +y
1+ y
=
+2

1
1
−
1+ x 1+ y
‹
(x − y)2 [( y − 2)x + 2 y 2 − y]
,
(1 + x)(1 + y)(x 2 + y 3 )
we still have to show that ( y − 2)x + 2 y 2 − y ≥ 0. This is clearly true for y ≥ 2, while
for 1 ≤ y < 2, we have
( y − 2)x + 2 y 2 − y ≥ ( y − 2) y 3 + 2 y 2 − y = y( y − 1)( y 2 − y + 1) ≥ 0.
The equality holds for a = b = c.
P 1.175. Let a, b, c, d be positive real numbers such that a bcd = 1. Prove that
X
1
≤ 1.
1 + a b + bc + ca
Solution. From
p
p p
p
1
1 1 1
1
1
+ + ≥p +p +p
= d( a + b + c),
a b c
ca
bc
ab
we get
p
a b + bc + ca ≥
Therefore,
X
a+
p
p
b+
d
p
c
.
p
X
1
d
≤
p
p = 1,
p
p
1 + a b + bc + ca
a+ b+ c+ d
which is just the required inequality. The equality occurs for a = b = c = d = 1.
P 1.176. Let a, b, c, d be positive real numbers such that a bcd = 1. Prove that
1
1
1
1
+
+
+
≥ 1.
2
2
2
(1 + a)
(1 + b)
(1 + c)
(1 + d)2
(Vasile Cîrtoaje, 1995)
Symmetric Rational Inequalities
237
First Solution. The inequality follows by summing the following inequalities (see P
1.1):
1
1
1
+
≥
,
(1 + a)2 (1 + b)2
1 + ab
ab
1
1
1
=
.
+
≥
2
2
(1 + c)
(1 + d)
1 + cd
1 + ab
The equality occurs for a = b = c = d = 1.
Second Solution. Using the substitutions a = 1/x 4 , b = 1/ y 4 , c = 1/z 4 , d = 1/t 4 ,
where x, y, z, t are positive real numbers such that x yz t = 1, the inequality becomes as
follows
y6
x6
z6
t6
+
+
+

‹2 
‹2 
‹2 
‹2 ≥ 1.
1
1
1
1
x3 +
y3 +
z3 +
t3 +
x
y
z
t
By the Cauchy-Schwarz inequality, we get
P
P
X
( x 3 )2
( x 3 )2
x6
P
P
.

‹2 ≥ 
‹2 = P 6
P
x + 2 x 2 + x 2 y 2z2
1
1
x3 +
x3 +
x
x
Thus, it suffices to show that
2(x 3 y 3 + x 3 z 3 + x 3 t 3 + y 3 z 3 + y 3 t 3 + z 3 t 3 ) ≥ 2x yz t
X
x2 +
X
x 2 y 2z2.
This is true if
2(x 3 y 3 + x 3 z 3 + x 3 t 3 + y 3 z 3 + y 3 t 3 + z 3 t 3 ) ≥ 3x yz t
and
2(x 3 y 3 + x 3 z 3 + x 3 t 3 + y 3 z 3 + y 3 t 3 + z 3 t 3 ) ≥ 3
X
X
x2
x 2 y 2z2,
Write these inequalities as
X
and
x 3 ( y 3 + z 3 + t 3 − 3 yz t) ≥ 0
X
(x 3 y 3 + y 3 z 3 + z 3 x 3 − 3x 2 y 2 z 2 ) ≥ 0,
respectively. By the AM-GM inequality, we have y 3 + z 3 + t 3 ≥ 3 yz t and x 3 y 3 + y 3 z 3 +
z 3 x 3 ≥ 3x 2 y 2 z 2 . Thus the conclusion follows.
Third Solution. Using the substitutions a = yz/x 2 , b = z t/ y 2 , c = t x/z 2 , d = x y/t 2 ,
where x, y, z, t are positive real numbers, the inequality becomes
y4
x4
z4
t4
+
+
+
≥ 1.
(x 2 + yz)2 ( y 2 + z t)2 (z 2 + t x)2 (t 2 + x y)2
238
Vasile Cîrtoaje
Using the Cauchy-Schwarz inequality two times, we deduce
z4
x4
z4
x4
+
+
≥
(x 2 + yz)2 (z 2 + t x)2
(x 2 + y 2 )(x 2 + z 2 ) (z 2 + t 2 )(z 2 + x 2 )
x 2 + z2
1
x4
z4
≥
= 2
+
,
x + z2 x 2 + y 2 z2 + t 2
x 2 + y 2 + z2 + t 2
and hence
x4
z4
x 2 + z2
+
≥
.
(x 2 + yz)2 (z 2 + t x)2
x 2 + y 2 + z2 + t 2
Adding this to the similar inequality
y2 + t2
y4
t4
+
≥
,
( y 2 + z t)2 (t 2 + x y)2
x 2 + y 2 + z2 + t 2
we get the required inequality.
Fourth Solution. Using the substitutions a = x/ y, b = y/z, c = z/t, d = t/x, where
x, y, z, t are positive real numbers, the inequality can be written as
y2
z2
t2
x2
+
+
+
≥ 1.
(x + y)2 ( y + z)2 (z + t)2 (t + x)2
By the Cauchy-Schwarz inequality and the AM-GM inequality, we get
P
X
[ y( y + z)]2
y2
≥P
(x + y)2
(x + y)2 ( y + z)2
=
[(x + y)2 + ( y + z)2 + (z + t)2 + (t + x)2 ]2
≥ 1.
4[(x + y)2 + (z + t)2 ][( y + z)2 + (t + x)2 ]
Remark. The following generalization holds true (Vasile Cîrtoaje, 2005):
• Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an = 1. If k ≥
then
1
1
1
n
+
+ ··· +
≥
.
2
2
2
(1 + ka1 )
(1 + ka2 )
(1 + kan )
(1 + k)2
P 1.177. Let a, b, c, d 6=
p
n − 1,
1
be positive real numbers such that a bcd = 1. Prove that
3
1
1
1
1
+
+
+
≥ 1.
2
2
2
(3a − 1)
(3b − 1)
(3c − 1)
(3d − 1)2
(Vasile Cîrtoaje, 2006)
Symmetric Rational Inequalities
239
First Solution. It suffices to show that
1
a−3
≥
.
(3a − 1)2
a−3 + b−3 + c −3 + d −3
This inequality is equivalent to
6a−2 + b−3 + c −3 + d −3 ≥ 9a−1 ,
which follows from the AM-GM inequality, as follows
6a−2 + b−3 + c −3 + d −3 ≥ 9
p
9
a−12 b−3 c −3 d −3 = 9a−1 .
The equality occurs for a = b = c = d = 1.
Second Solution. Let a ≤ b ≤ c ≤ d. If a < 1/3, then
1
> 1,
(3a − 1)2
and the desired inequality is clearly true. Otherwise, if 1/3 < a ≤ b ≤ c ≤ d, we have
4a3 − (3a − 1)2 = (a − 1)2 (4a − 1) ≥ 0.
Therefore, using this result and the AM-GM inequality, we get
X
v
t
1
1
1X 1
4
≥
≥
= 1.
2
3
3
3
(3a − 1)
4
a
a b c3 d 3
Third Solution. We have
1
a(a − 1)2 (a + 2)(a2 + 3)
1
−
=
≥ 0.
(3a − 1)2 (a3 + 1)2
(3a − 1)2 (a3 + 1)2
Therefore,
X
X
1
1
≥
,
2
3
(3a − 1)
(a + 1)2
and it suffices to prove that
X
(a3
1
≥ 1.
+ 1)2
This inequality is an immediate consequence of the inequality in P 1.176.
240
Vasile Cîrtoaje
P 1.178. Let a, b, c, d be positive real numbers such that a bcd = 1. Prove that
1
1
1
1
+
+
+
≥ 1.
1 + a + a2 + a3 1 + b + b2 + b3 1 + c + c 2 + c 3 1 + d + d 2 + d 3
(Vasile Cîrtoaje, 1999)
First Solution. We get the desired inequality by summing the inequalities
1
1
1
+
≥
,
2
3
2
3
1+a+a +a
1+ b+ b + b
1 + (a b)3/2
1
1
1
.
+
≥
2
3
2
3
1+c+c +c
1+d +d +d
1 + (cd)3/2
Thus, it suffices to show that
1
1
1
+
≥
,
1 + x2 + x4 + x6 1 + y2 + y4 + y6
1 + x3 y3
where x and y are positive real numbers. Putting p = x y and s = x 2 + x y + y 2 , this
inequality becomes
p3 (x 6 + y 6 ) + p2 (p − 1)(x 4 + y 4 ) − p2 (p2 − p + 1)(x 2 + y 2 ) − p6 − p4 + 2p3 − p2 + 1 ≥ 0,
p3 (x 3 − y 3 )2 + p2 (p − 1)(x 2 − y 2 )2 − p2 (p2 − p + 1)(x − y)2 + p6 − p4 − p2 + 1 ≥ 0.
p3 s2 (x − y)2 + p2 (p − 1)(s + p)2 (x − y)2 − p2 (p2 − p + 1)(x − y)2 + p6 − p4 − p2 + 1,
p2 (s + 1)(ps − 1)(x − y)2 + (p2 − 1)(p4 − 1) ≥ 0.
If ps − 1 ≥ 0, then this inequality is clearly true. Consider further that ps < 1. From
ps < 1 and s ≥ 3p, we get p2 < 1/3. Write the desired inequality in the form
(1 − p2 )(1 − p4 ) ≥ p2 (1 + s)(1 − ps)(x − y)2 .
Since
p(x − y)2 = p(s − 3p) < 1 − 3p2 < 1 − p2 ,
it suffices to show that
1 − p4 ≥ p(1 + s)(1 − ps).
Indeed,
4p(1 + s)(1 − ps) ≤ [p(1 + s) + (1 − ps)]2 = (1 + p)2 < 2(1 + p2 ) < 4(1 − p4 ).
The equality occurs for a = b = c = d = 1.
Second Solution. Assume that a ≥ b ≥ c ≥ d, and write the inequality as
X
1
≥ 1.
(1 + a)(1 + a2 )
Symmetric Rational Inequalities
241
Since
1
1
1
1
1
1
≤
≤
,
≤
≤
,
2
2
1+a
1+ b
1+c
1+a
1+ b
1 + c2
by Chebyshev’s inequality, it suffices to prove that

‹
‹
1
1
1
1
1
1
1
1
+
+
+
+
+
≥ 1.
3 1+a 1+ b 1+c
1 + a2 1 + b2 1 + c 2
(1 + d)(1 + d 2 )
In addition, from the inequality in P 1.174, we have
p
3
1
1
1
3
3 d
+
+
≥
=
p
p
3
3
1+a 1+ b 1+c
1 + a bc
d +1
and
p
3
3 d2
1
1
3
1
=
+
+
≥
.
p
p
3
3
1 + a2 1 + b2 1 + c 2
1 + a2 b2 c 2
d2 + 1
Thus, it suffices to prove that
(1 +
Putting x =
p
3
p
3
3d
d)(1 +
p
3
d2
)
+
1
≥ 1.
(1 + d)(1 + d 2 )
d, this inequality becomes as follows
3x 3
1
+
≥ 1,
(1 + x)(1 + x 2 ) (1 + x 3 )(1 + x 6 )
3x 3 (1 − x + x 2 )(1 − x 2 + x 4 ) + 1 ≥ (1 + x 3 )(1 + x 6 ),
x 3 (2 − 3x + 2x 3 − 3x 5 + 2x 6 ) ≥ 0,
x 3 (1 − x)2 (2 + x + x 3 + 2x 4 ) ≥ 0.
Remark. The following generalization holds true (Vasile Cîrtoaje, 2004):
• If a1 , a2 , . . . , an are positive real numbers such that a1 a2 · · · an = 1, then
1
1 + a1 + · · · + a1n−1
+
1
1 + a2 + · · · + a2n−1
+ ··· +
1
≥ 1.
1 + an + · · · + ann−1
P 1.179. Let a, b, c, d be positive real numbers such that a bcd = 1. Prove that
1
1
1
1
+
+
+
≥ 1.
1 + a + 2a2 1 + b + 2b2 1 + c + 2c 2 1 + d + 2d 2
(Vasile Cîrtoaje, 2006)
242
Vasile Cîrtoaje
Solution. We will show that
1
1
≥
,
2
k
1 + a + 2a
1 + a + a2k + a3k
where k = 5/6. Then, it suffices to show that
X
1
≥ 1,
1 + a k + a2k + a3k
which immediately follows from the inequality in P 1.178. Setting a = x 6 , x > 0, the
claimed inequality can be written as
2x 12
1
1
≥
,
6
5
+ x + 1 1 + x + x 10 + x 15
which is equivalent to
x 10 + x 5 + 1 ≥ 2x 7 + x.
We can prove it by summing the AM-GM inequalities
x 5 + 4 ≥ 5x
and
5x 10 + 4x 5 + 1 ≥ 10x 7 .
This completes the proof. The equality occurs for a = b = c = d = 1.
Remark. The inequalities in P 1.176, P 1.178 and P 1.179 are particular cases of the
following more general inequality (Vasile Cîrtoaje, 2009):
• Let a1 , a2 , . . . , an (n ≥ 4) be positive real numbers such that a1 a2 · · · an = 1. If p, q, r
are nonnegative real numbers satisfying p + q + r = n − 1, then
i=n
X
1
i=1
1 + pai + qai2 + r ai3
≥ 1.
P 1.180. Let a, b, c, d be positive real numbers such that a bcd = 1. Prove that
1 1 1 1
9
25
+ + + +
≥
.
a b c d a+b+c+d
4
Symmetric Rational Inequalities
243
Solution (by Vo Quoc Ba Can). Replacing a, b, c, d by a4 , b4 , c 4 , d 4 , respectively, the inequality becomes as follows:
1
1
1
9
25
1
+ 4+ 4+ 4+ 4
≥
,
4
4
4
4
a
b
c
d
a +b +c +d
4a bcd
1
1
1
1
4
9
9
+ 4+ 4+ 4−
≥
− 4
,
4
4
a
b
c
d
a bcd
4a bcd a + b + c 4 + d 4
9(a4 + b4 + c 4 + d 4 − 4a bcd)
1
1
1
1
4
≥
.
+
+
+
−
a4 b4 c 4 d 4 a bcd
4a bcd(a4 + b4 + c 4 + d 4 )
Using the identities
a4 + b4 + c 4 + d 4 − 4a bcd = (a2 − b2 )2 + (c 2 − d 2 )2 + 2(a b − cd)2 ,
1
1
1
1
4
(a2 − b2 )2 (c 2 − d 2 )2 2(a b − cd)2
+
+
+
−
=
+
+ 2 2 2 2 ,
a4 b4 c 4 d 4 a bcd
a4 b4
c4 d 4
a b c d
the inequality can be written as
(a2 − b2 )2 (c 2 − d 2 )2 2(a b − cd)2
9[(a2 − b2 )2 + (c 2 − d 2 )2 + 2(a b − cd)2 ]
+
+
≥
,
a4 b4
c4 d 4
a2 b2 c 2 d 2
4a bcd(a4 + b4 + c 4 + d 4 )
4
4
4
4
4
4
4
4
2
2 2 4cd(a + b + c + d )
2
2 2 4a b(a + b + c + d )
(a − b )
− 9 + (c − d )
−9
a3 b3
c3 d 3
4
4
4
4
2 4(a + b + c + d )
+2(a b − cd)
− 9 ≥ 0.
a bcd
By the AM-GM inequality, we have
a4 + b4 + c 4 + d 4 ≥ 4a bcd.
Therefore, it suffices to show that
4
4
4
4
4
4
4
4
2
2 2 4cd(a + b + c + d )
2
2 2 4a b(a + b + c + d )
(a − b )
− 9 + (c − d )
− 9 ≥ 0.
a3 b3
c3 d 3
Without loss of generality, assume that a ≥ c ≥ d ≥ b. Since
(a2 − b2 )2 ≥ (c 2 − d 2 )2
and
4cd(a4 + b4 + c 4 + d 4 ) 4(a4 + b4 + c 4 + d 4 ) 4(a4 + 3b4 )
≥
≥
> 9,
a3 b3
a3 b
a3 b
it is enough to prove that
4cd(a4 + b4 + c 4 + d 4 )
4a b(a4 + b4 + c 4 + d 4 )
−9 +
− 9 ≥ 0,
a3 b3
c3 d 3
244
Vasile Cîrtoaje
which is equivalent to
2(a4 + b4 + c 4 + d 4 )

cd
ab
+ 3 3
3
3
a b
c d
‹
≥ 9.
Indeed, by the AM-GM inequality,

‹

‹
cd
ab
2
4
4
4
4
2(a + b + c + d ) 3 3 + 3 3 ≥ 8a bcd
= 16 > 9.
a b
c d
a bcd
The equality occurs for a = b = c = d = 1.
P 1.181. If a, b, c, d are real numbers such that a + b + c + d = 0, then
(a − 1)2 (b − 1)2 (c − 1)2 (d − 1)2
+
+ 2
+
≤ 4.
3a2 + 1
3b2 + 1
3c + 1
3d 2 + 1
Solution. Since
4−
3(a − 1)2
(3a + 1)2
=
,
3a2 + 1
3a2 + 1
we can write the inequality as
X (3a + 1)2
3a2 + 1
≥ 4.
On the other hand, since
4a2 = 3a2 + (b + c + d)2 ≤ 3a2 + 3(b2 + c 2 + d 2 ) = 3(a2 + b2 + c 2 + d 2 ),
9 2
9(a2 + b2 + c 2 + d 2 ) + 4
(a + b2 + c 2 + d 2 ) + 1 =
,
4
4
P
X (3a + 1)2
4 (3a + 1)2
≥
= 4.
3a2 + 1
9(a2 + b2 + c 2 + d 2 ) + 4
3a2 + 1 ≤
we have
The equality holds for a = b = c = d = 0, and also for a = 1 and b = c = d = −1/3 (or
any cyclic permutation).
Remark. The following generalization is also true.
• If a1 , a2 , . . . , an are real numbers such that a1 + a2 + · · · + an = 0, then
(a1 − 1)2
(n − 1)a12 + 1
+
(a2 − 1)2
(n − 1)a22 + 1
+ ··· +
(an − 1)2
≤ n,
(n − 1)an2 + 1
with equality for a1 = a2 = · · · = an = 0, and also for a1 = 1 and a2 = a3 = · · · = an =
−1/(n − 1 (or any cyclic permutation).
Symmetric Rational Inequalities
245
P 1.182. If a, b, c, d ≥ −5 such that a + b + c + d = 4, then
1− b
1−c
1−d
1−a
+
+
+
≥ 0.
2
2
2
(1 + a)
(1 + b)
(1 + c)
(1 + d)2
Solution. Assume that a ≤ b ≤ c ≤ d. We show first that x ∈ [−5, −1) ∪ (−1, ∞)
involves
1− x
−1
,
≥
2
(1 + x)
8
and x ∈ [−5, −1) ∪ (−1, 1/3] involves
3
1− x
≥ .
2
(1 + x)
8
Indeed, we have
1
(x − 3)2
1− x
+
=
≥0
(1 + x)2 8 8(1 + x)2
and
1− x
3 (5 + x)(1 − 3x)
− =
≥ 0.
(1 + x)2 8
8(1 + x)2
Then, if a ≤ 1/3, then
1−a
1− b
1−c
1−d
3 1 1 1
+
+
+
≥ − − − = 0.
2
2
2
2
(1 + a)
(1 + b)
(1 + c)
(1 + d)
8 8 8 8
Assume now that 1/3 ≤ a ≤ b ≤ c ≤ d. Since
1−a ≥1− b ≥1−c ≥1−d
and
1
1
1
1
≥
≥
≥
,
(1 + a)2
(1 + b)2
(1 + c)2
(1 + d)2
by Chebyshev’s inequality, we have
1− b
1−c
1−d
1−a
+
+
+
≥
2
2
2
(1 + a)
(1 + b)
(1 + c)
(1 + d)2
˜
— •X
1 ”X
1
≥
(1 − a)
= 0.
4
(1 + a)2
The equality holds for a = b = c = d = 1, and also for a = −5 and b = c = d = 3 (or
any cyclic permutation).
246
Vasile Cîrtoaje
P 1.183. Let a1 , a2 , . . . , an be positive real numbers such that a1 + a2 + · · · + an = n. Prove
that
X
1
1
≤ .
2
2
2
2
(n + 1)a1 + a2 + · · · + an
(Vasile Cîrtoaje, 2008)
First Solution. By the Cauchy-Schwarz inequality, we have
n2
X
(n + 1)a12 + a22 + · · · + an2
=
(a1 + a2 + · · · + an )2
X
2a12 + (a12 + a22 ) + · · · + (a12 + an2 )
X 1
an2
a22
≤
+
+ ··· + 2
2 a12 + a22
a1 + an2
=
n2
n n(n − 1)
+
= ,
2
2
2
from which the conclusion follows. The equality holds for a1 = a2 = · · · = an = 1.
Second Solution. Write the inequality as
X
a12 + a22 + · · · + an2
≤
(n + 1)a12 + a22 + · · · + an2
Since
a12 + a22 + · · · + an2
(n + 1)a12 + a22 + · · · + an2
=1−
a12 + a22 + · · · + an2
2
.
na12
(n + 1)a12 + a22 + · · · + an2
,
we need to prove that
a12
X
(n + 1)a12 + a22 + · · · + an2
+
a12 + a22 + · · · + an2
2n
≥ 1.
By the Cauchy-Schwarz inequality, we have
X
a12
(a1 + a2 + · · · + an )2
P
≥
(n + 1)a12 + a22 + · · · + an2
[(n + 1)a12 + a22 + · · · + an2 ]
n
=
.
2
2
2(a1 + a2 + · · · + an2 )
Then, it suffices to prove that
n
a12 + a22 + · · · + an2
+
a12 + a22 + · · · + an2
n
which follows immediately from the AM-GM inequality.
≥ 2,
Symmetric Rational Inequalities
247
P 1.184. Let a1 , a2 , . . . , an be real numbers such that a1 + a2 + · · · + an = 0. Prove that
(a1 + 1)2
a12 + n − 1
+
(a2 + 1)2
a22 + n − 1
+ ··· +
(an + 1)2
n
≥
.
2
an + n − 1
n−1
(Vasile Cîrtoaje, 2010)
Solution. Without loss of generality, assume that an2 = max{a12 , a22 , · · · , an2 }. Since
(an + 1)2
(n − 1 − an )2
n
=
,
−
an2 + n − 1
n − 1 (n − 1)(an2 + n − 1)
we can write the inequality as
n−1
X
(ai + 1)2
i=1
ai2 + n − 1
≥
(n − 1 − an )2
.
(n − 1)(an2 + n − 1)
From the Cauchy-Schwarz inequality
™ – n−1
– n−1
™ – n−1
™2
X (a + 1)2
X
X
i
(ai2 + n − 1)
≥
(ai + 1) ,
2
a
+
n−1
i=1 i
i=1
i=1
we get
n−1
X
(ai + 1)2
i=1
(n − 1 − an )2
.
≥
P
n−1 2
2
ai2 + n − 1
i=1 ai + (n − 1)
Thus, it suffices to show that
n−1
X
ai2 + (n − 1)2 ≤ (n − 1)(an2 + n − 1),
i=1
which is clearly true. The proof is completed. The equality holds for
· · · = an (or any cyclic permutation).
−a1
= a2 = a3 =
n−1
P 1.185. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an = 1. Prove that
1
1
1
+
+ ··· +
≥ 1.
1 + (n − 1)a1 1 + (n − 1)a2
1 + (n − 1)an
(Vasile Cîrtoaje, 1991)
248
Vasile Cîrtoaje
First Solution. Let k = (n − 1)/n. We can get the required inequality by summing the
inequalities below for i = 1, 2, · · · , n:
−k
ai
1
.
≥ −k
−k
1 + (n − 1)ai
a1 + a2 + · · · + an−k
This inequality is equivalent to
−k
−k
a1−k + · · · + ai−1
+ ai+1
+ · · · + an−k ≥ (n − 1)ai1−k ,
which follows from the AM-GM inequality. The equality holds for a1 = a2 = · · · = an =
1.
Second Solution. Using the substitutions ai = 1/x i for all i, the inequality becomes
xn
x1
x2
+
+ ··· +
≥ 1,
x1 + n − 1 x2 + n − 1
xn + n − 1
where x 1 , x 2 , · · · , x n are positive real numbers such that x 1 x 2 · · · x n = 1. By the CauchySchwarz inequality, we have
Pp 2
X
(
x1)
xi
≥P
.
xi + n − 1
(x 1 + n − 1)
Thus, we still have to prove that
Xp
X
(
x 1 )2 ≥
x 1 + n(n − 1),
which reduces to
X
p
xi x j ≥
1≤i< j≤n
n(n − 1)
.
2
Since x 1 x 2 · · · x n = 1, this inequality follows from the AM-GM inequality.
Third Solution. For the sake of contradiction, assume that the required inequality is not
true. Then, it suffices to show that the hypothesis a1 a2 · · · an = 1 does not hold. More
precisely, we will prove that
1
1
1
+
+ ··· +
<1
1 + (n − 1)a1 1 + (n − 1)a2
1 + (n − 1)an
involves a1 a2 · · · an > 1. Let x i =
ai =
1
, 0 < x i < 1, for i = 1, 2, · · · , n. Since
1 + (n − 1)ai
1 − xi
for all i, we need to show that
(n − 1)x i
x1 + x2 + · · · + x n < 1
Symmetric Rational Inequalities
249
implies
(1 − x 1 )(1 − x 2 ) · · · (1 − x n ) > (n − 1)n x 1 x 2 · · · x n .
Using the AM-GM inequality, we have
1 − xi >
X
vY
u
t
xk.
x k ≥ (n − 1) n−1
k6=i
k6=i
Multiplying the inequalities
vY
u
t
1 − x i > (n − 1) n−1
xk.
k6=i
for i = 1, 2, · · · , n, the conclusion follows.
Remark. The inequality in P 1.185 is a particular case of the following more general
results (Vasile Cîrtoaje, 2005):
• Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an = 1. If 0 < k ≤ n − 1
and p ≥ n1/k − 1, then
1
1
1
n
+
+ ··· +
≥
.
k
k
k
(1 + pa1 )
(1 + pa2 )
(1 + pan )
(1 + p)k
1
• Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an = 1. If k ≥
and
n−1
n 1/k
− 1, then
0<p≤
n−1
1
1
1
n
+
+ ··· +
≤
.
k
k
k
(1 + pa1 )
(1 + pa2 )
(1 + pan )
(1 + p)k
P 1.186. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an = 1. Prove that
1
1 − a1 + na12
+
1
1 − a2 + na22
+ ··· +
1
≥ 1.
1 − an + nan2
(Vasile Cîrtoaje, 2009)
Solution. First, we show that
1
1
≥
,
2
1 − x + nx
1 + (n − 1)x k
250
Vasile Cîrtoaje
where x > 0 and k = 2 +
1
. Write the inequality as
n−1
(n − 1)x k + x ≥ nx 2 .
We can get this inequality using the AM-GM inequality as follows
p
n
(n − 1)x k + x ≥ n x (n−1)k x = nx 2 .
Thus, it suffices to show that
1
1 + (n − 1)a1k
+
1
1 + (n − 1)a2k 2
+ ··· +
1
≥ 1,
1 + (n − 1)ank
which follows immediately from the inequality in the preceding P 1.185. The equality
holds for a1 = a2 = · · · = an = 1.
Remark 1. Similarly, we can prove the following more general statement.
• Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an = 1. If p and q are
p
real numbers such that p + q = n − 1 and n − 1 ≤ q ≤ ( n + 1)2 , then
1
1+
pa1 + qa12
+
1
1+
pa2 + qa22
+ ··· +
1
≥ 1.
1 + pan + qan2
Remark 2. We can extend the inequality in Remark 1 as follows (Vasile Cîrtoaje, 2009).
• Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an = 1. If p and q are
p
real numbers such that p + q = n − 1 and 0 ≤ q ≤ ( n + 1)2 , then
1
1+
pa1 + qa12
+
1
1+
pa2 + qa22
+ ··· +
1
≥ 1.
1 + pan + qan2
P 1.187. Let a1 , a2 , . . . , an be positive real numbers such that
a1 , a2 , . . . , an ≥
k(n − k − 1)
,
kn − k − 1
k>1
and
a1 a2 · · · an = 1.
Prove that
1
1
1
n
+
+ ··· +
≤
.
a1 + k a2 + k
an + k
1+k
(Vasile Cîrtoaje, 2005)
Symmetric Rational Inequalities
251
Solution. We use the induction on n. Let
En (a1 , a2 , . . . , an ) =
1
1
n
1
+
+ ··· +
−
.
a1 + k a2 + k
an + k 1 + k
For n = 2, we have
p
p
(1 − k)( a1 − a2 )2
E2 (a1 , a2 ) =
≤ 0.
(1 + k)(a1 + k)(a2 + k)
Assume that the inequality is true for n−1 numbers (n ≥ 3), and prove that En (a1 , a2 , . . . , an ) ≥
0 for a1 a2 · · · an = 1 and a1 , a2 , . . . , an ≥ pn , where
pn =
k(n − k − 1)
.
kn − k − 1
Due to symmetry, we may assume that a1 ≥ 1 and a2 ≤ 1. There are two cases to
consider.
Case 1: a1 a2 ≤ k2 . Since a1 a2 ≥ a2 and pn−1 < pn , from a1 , a2 , . . . , an ≥ pn it follows
that
a1 a2 , a3 , · · · , an > pn−1 .
Then, by the inductive hypothesis, we have En−1 (a1 a2 , a2 , . . . , an ) ≤ 0, and it suffices to
show that
En (a1 , a2 , . . . , an ) ≤ En−1 (a1 a2 , a2 , . . . , an ).
This is equivalent to
1
1
1
1
+
−
−
≤ 0,
a1 + k a2 + k a1 a2 + k 1 + k
which reduces to the obvious inequality
(a1 − 1)(1 − a2 )(a1 a2 − k2 ) ≤ 0.
Case 2: a1 a2 ≥ k2 . Since
a1 + a2 + 2k
a1 + a2 + 2k
1
1
1
+
=
≤ 2
=
a1 + k a2 + k
a1 a2 + k(a1 + a2 ) + k2
k + k(a1 + a2 ) + k2
k
and
1
1
n−2
kn − k − 1
+ ··· +
≤
=
,
a3 + k
an + k
pn + k
k(k + 1)
we have
En (a1 , a2 , . . . , an ) ≤
1 kn − k − 1
n
+
−
= 0.
k
k(k + 1)
1+k
Thus, the proof is completed. The equality holds for a1 = a2 = · · · = an = 1.
252
Vasile Cîrtoaje
Remark. For k = n − 1, we get the inequality in P 1.185. Also, for k → ∞, we get the
known inequalities
1<
1
1
1
+
+ ··· +
< n − 1,
1 + a1 1 + a2
1 + an
which holds for all positive numbers a1 , a2 , . . . , an satisfying a1 a2 · · · an = 1.
P 1.188. Let a1 , a2 , . . . , an be positive real numbers such that
a1 ≥ 1 ≥ a2 ≥ · · · ≥ an ,
Prove that
1 − a1
3 + a12
+
1 − a2
3 + a22
a1 a2 · · · an = 1.
+ ··· +
1 − an
≥ 0.
3 + an2
(Vasile Cîrtoaje, 2013)
Solution. For n = 2, we have
1 − a1
3 + a12
+
1 − a2
3 + a22
=
(a1 − 1)4
(3 + a12 )(3a12 + 1)
≥ 0.
For n ≥ 3, we will use the induction method and the inequality
1− y
1− xy
1− x
+
≥
,
2
2
3+ x
3+ y
3 + x2 y2
which holds for all x, y ∈ [0, 1]. Indeed, we can write this inequality as
(1 − x)(1 − y)(3 + x y)(3 − x y − x 2 y − x y 2 ) ≥ 0,
which is obviously true. Based on the inequality
1 − an−1
2
3 + an−1
+
1 − an−1 an
1 − an
≥
,
2
3 + an2
3 + an−1
an2
it suffices to show that
1 − a1
3 + a12
+ ··· +
1 − an−2
2
3 + an−2
+
1 − an−1 an
2
3 + an−1
an2
≥ 0.
Since
a1 ≥ 1 ≥ a2 ≥ · · · an−2 ≥ an−1 an ,
this inequality follows from the hypothesis induction. Thus, the proof is completed. The
equality holds for a1 = a2 = · · · = an = 1.
Symmetric Rational Inequalities
253
P 1.189. If a1 , a2 , . . . , an ≥ 0, then
1
1
n
1
+
+ ··· +
≥
.
1 + na1 1 + na2
1 + nan
n + a1 a2 · · · an
(Vasile Cîrtoaje, 2013)
Solution. If one of a1 , a2 , . . . , an is zero, the inequality is obvious. Consider further that
a1 , a2 , . . . , an > 0 and let
p
r = n a1 a2 · · · an .
By the Cauchy-Schwarz inequality, we have
Pp
Pp
X
(
(
a2 a3 · · · an )2
a2 a3 · · · an )2
1
=P
≥P
.
1 + na1
(1 + na1 )a2 a3 · · · an
a2 a3 · · · an + n2 r n
Therefore, it suffices to show that
X
Xp
a2 a3 · · · an )2 ≥ n
a2 a3 · · · an + n3 r n .
(n + r n )(
By the AM-GM inequality, we have
Xp
X
(
a2 a3 · · · an )2 ≥
a2 a3 · · · an + n(n − 1)r n−1 .
Thus, it is enough to prove that
X
X
(n + r n )[
a2 a3 · · · an + n(n − 1)r n−1 ] ≥ n
a2 a3 · · · an + n3 r n ,
which is equivalent to
X
rn
a2 a3 · · · an + n(n − 1)r 2n−1 + n2 (n − 1)r n−1 ≥ n3 r n .
Also, by the AM-GM inequality,
X
a2 a3 · · · an ≥ nr n−1 ,
and it suffices to show the inequality
nr 2n−1 + n(n − 1)r 2n−1 + n2 (n − 1)r n−1 ≥ n3 r n ,
which can be rewritten as
n2 r n−1 (r n − nr + n − 1) ≥ 0.
Indeed, by the AM-GM inequality, we get
r n − nr + n − 1 = r n + 1 + · · · + 1 − nr ≥ n
The equality holds for a1 = a2 = · · · = an = 1.
p
n
r n · 1 · · · 1 − nr = 0.
254
Vasile Cîrtoaje
P 1.190. If a1 , a2 , . . . , an are positive real numbers, then
bn
an
b1 b2
a1 a2
+
+ ··· +
≥
+
+ ··· + ,
a1 a2
an
b1 b2
bn
where
bi =
Solution. Let
1 X
aj,
n − 1 j6=i
i = 1, 2, · · · , n.
a1 + a2 + · · · + an
,
n
1
1
1
A=
+
+ ··· + .
a1 a2
an
a=
Using the Cauchy-Schwarz inequality, we have
(n − 1)2
1
1
1
1
≤
+
+ ··· +
= A− ,
a2 + a3 + · · · + an
a2 a3
an
a1
n−1
1
≤ A− ,
b1
a1
Aa1 − 1
a1
≤
,
b1
n−1
n
X
a
i=1
n
i
bi
≤
n
X
a
i=1
Since
i
bi
A X
n
ai −
,
n − 1 i=1
n−1
≤
naA
n
−
.
n−1 n−1
n
n
X
bi
1 X na − ai
naA
n
=
=
−
,
a
n
−
1
a
n
−
1
n
−
1
i
i
i=1
i=1
the conclusion follows. The equality holds for a1 = a2 = · · · = an .
P 1.191. If a1 , a2 , . . . , an are positive real numbers such that
a1 + a2 + · · · + an =
then
1
1
1
+
+ ··· + ,
a1 a2
an
Symmetric Rational Inequalities
255
(a)
1
1
1
+
+ ··· +
≥ 1;
1 + (n − 1)a1 1 + (n − 1)a2
1 + (n − 1)an
(b)
1
1
1
+
+ ··· +
≤ 1.
n − 1 + a1 n − 1 + a2
n − 1 + an
(Vasile Cîrtoaje, 1996)
Solution. (a) We use the contradiction method. So, assume that
1
1
1
+
+ ··· +
< 1,
1 + (n − 1)a1 1 + (n − 1)a2
1 + (n − 1)an
and show that
a1 + a2 + · · · + an >
1
1
1
+
+ ··· + .
a1 a2
an
Using the substitution
xi =
1
,
1 + (n − 1)ai
i = 1, 2, · · · , n,
the hypothesis inequality becomes
x 1 + x 2 + · · · + x n < 1.
This inequality involves
1 − x i > (n − 1)bi ,
bi =
1 X
x j,
n − 1 j6=i
i = 1, 2, · · · , n.
Using this result and the inequality from the preceding P 1.190, we get
a1 + a2 + · · · + an =
n
n
n
X
X
1 − xi
bi X x i
>
≥
.
(n − 1)x i
x
b
i=1
i=1 i
i=1 i
Thus, it suffices to show that
n
X
xi
1
1
1
≥
+
+ ··· + .
b
a1 a2
an
i=1 i
Indeed, we have
n
n
n
X
x i X (n − 1)x i X 1
>
=
.
b
1 − xi
a
i=1 i
i=1
i=1 i
The proof is completed. The equality holds for a1 = a2 = · · · = an = 1.
(b) The desired inequality follows from the inequality in (a) by replacing a1 , a2 , . . . , an
with 1/a1 , 1/a2 , . . . , 1/an , respectively.
256
Vasile Cîrtoaje
Chapter 2
Symmetric Nonrational Inequalities
2.1
Applications
2.1. If a, b, c are nonnegative real numbers, then
Xp
Æ
a2 − a b + b2 ≤ 6(a2 + b2 + c 2 ) − 3(a b + bc + ca).
2.2. If a, b, c are nonnegative real numbers, then
p
a2 − a b + b2 +
p
b2 − bc + c 2 +
p
c 2 − ca + a2 ≤ 3
v
t a2 + b2 + c 2
2
.
2.3. If a, b, c are nonnegative real numbers, then
v
v
v
t
t
t
p
2
2
2
a2 + b2 − a b + b2 + c 2 − bc + c 2 + a2 − ca ≥ 2 a2 + b2 + c 2 .
3
3
3
2.4. If a, b, c are nonnegative real numbers, then
Xp
Æ
a2 + a b + b2 ≥ 4(a2 + b2 + c 2 ) + 5(a b + bc + ca).
2.5. If a, b, c are nonnegative real numbers, then
Xp
Æ
a2 + a b + b2 ≤ 5(a2 + b2 + c 2 ) + 4(a b + bc + ca).
257
258
Vasile Cîrtoaje
2.6. If a, b, c are nonnegative real numbers, then
Xp
p
p
a2 + a b + b2 ≤ 2 a2 + b2 + c 2 + a b + bc + ca.
2.7. If a, b, c are nonnegative real numbers, then
p
p
p
p
p
a2 + 2bc + b2 + 2ca + c 2 + 2a b ≤ a2 + b2 + c 2 + 2 a b + bc + ca.
2.8. If a, b, c are nonnegative real numbers, then
p
1
a2 + 2bc
+p
1
b2 + 2ca
+p
1
c 2 + 2a b
≥p
1
a2 + b2 + c 2
+p
2
a b + bc + ca
.
2.9. If a, b, c are positive real numbers, then
p
p
p
p
p
2a2 + bc + 2b2 + ca + 2c 2 + a b ≤ 2 a2 + b2 + c 2 + a b + bc + ca.
2.10. Let a, b, c be nonnegative real numbers such that a + b + c = 3. If k =
then
XÆ
p
a(a + k b)(a + kc) ≤ 3 3.
p
3 − 1,
2.11. If a, b, c are nonnegative real numbers such that a + b + c = 3, then
XÆ
a(2a + b)(2a + c) ≥ 9.
2.12. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that
Æ
Æ
Æ
b2 + c 2 + a(b + c) + c 2 + a2 + b(c + a) + a2 + b2 + c(a + b) ≥ 6.
2.13. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that
(a)
p
(b)
p
p
p
p
3a2 + a bc + 3b2 + a bc + 3c 2 + a bc ≥ 3 3 + a bc.
a(3a2 + a bc) +
p
b(3b2 + a bc) +
p
c(3c 2 + a bc) ≥ 6;
Symmetric Nonrational Inequalities
259
2.14. Let a, b, c be positive real numbers such that a b + bc + ca = 3. Prove that
Æ
Æ
Æ
a (a + 2b)(a + 2c) + b (b + 2c)(b + 2a) + c (c + 2a)(c + 2b) ≥ 9.
2.15. Let a, b, c be nonnegative real numbers such that a + b + c = 1. Prove that
Æ
Æ
Æ
p
a + (b − c)2 + b + (c − a)2 + c + (a − b)2 ≥ 3.
2.16. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
v
v
v
t a(b + c) t b(c + a) t c(a + b)
+
+
≥ 2.
a2 + bc
b2 + ca
c2 + a b
2.17. Let a, b, c be positive real numbers such that a bc = 1. Prove that
p
3
1
a2
+ 25a + 1
+p
3
1
b2
+ 25b + 1
+p
3
1
c2
+ 25c + 1
≥ 1.
2.18. If a, b, c are nonnegative real numbers, then
p
a2 + bc +
p
b2 + ca +
p
c2 + a b ≤
3
(a + b + c).
2
2.19. If a, b, c are nonnegative real numbers, then
p
p
p
p
a2 + 9bc + b2 + 9ca + c 2 + 9a b ≥ 5 a b + bc + ca.
2.20. If a, b, c are nonnegative real numbers, then
XÆ
(a2 + 4bc)(b2 + 4ca) ≥ 5(a b + ac + bc).
2.21. If a, b, c are nonnegative real numbers, then
XÆ
(a2 + 9bc)(b2 + 9ca) ≥ 7(a b + ac + bc).
260
Vasile Cîrtoaje
2.22. If a, b, c are nonnegative real numbers, then
XÆ
(a2 + b2 )(b2 + c 2 ) ≤ (a + b + c)2 .
2.23. If a, b, c are nonnegative real numbers, then
XÆ
(a2 + a b + b2 )(b2 + bc + c 2 ) ≥ (a + b + c)2 .
2.24. If a, b, c are nonnegative real numbers, then
XÆ
(a2 + 7a b + b2 )(b2 + 7bc + c 2 ) ≥ 7(a b + ac + bc).
2.25. If a, b, c are nonnegative real numbers, then
v
‹
‹
X t
7
7
13
2
2
2
2
a + ab + b
b + bc + c ≤
(a + b + c)2 .
9
9
12
2.26. If a, b, c are nonnegative real numbers, then
v
‹
‹
X t
1
1
61
2
2
2
2
a + ab + b
b + bc + c ≤
(a + b + c)2 .
3
3
60
2.27. If a, b, c are nonnegative real numbers, then
a
b
c
+p
+p
≥ 1.
p
2
2
2
2
2
4b + bc + 4c
4c + ca + 4a
4a + a b + 4b2
2.28. If a, b, c are nonnegative real numbers, then
p
a
b2
+ bc
+ c2
+p
b
c2
+ ca + a2
+p
c
a2
+ ab +
b2
≥p
a+b+c
a b + bc + ca
2.29. If a, b, c are nonnegative real numbers, then
p
a
a2
+ 2bc
+p
b
b2
+ 2ca
+p
c
c2
+ 2a b
≤p
a+b+c
a b + bc + ca
.
.
Symmetric Nonrational Inequalities
261
2.30. If a, b, c are nonnegative real numbers, then
p
p
p
a3 + b3 + c 3 + 3a bc ≥ a2 a2 + 3bc + b2 b2 + 3ca + c 2 c 2 + 3a b.
2.31. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
a
b
c
+p
+p
≤ 1.
p
4a2 + 5bc
4b2 + 5ca
4c 2 + 5a b
2.32. Let a, b, c be nonnegative real numbers. Prove that
p
p
p
a 4a2 + 5bc + b 4b2 + 5ca + c 4c 2 + 5a b ≥ (a + b + c)2 .
2.33. Let a, b, c be nonnegative real numbers. Prove that
p
p
p
a a2 + 3bc + b b2 + 3ca + c c 2 + 3a b ≥ 2(a b + bc + ca).
2.34. Let a, b, c be nonnegative real numbers. Prove that
p
p
p
a a2 + 8bc + b b2 + 8ca + c c 2 + 8a b ≤ (a + b + c)2 .
2.35. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
p
a2 + 2bc
b2 + bc + c 2
+p
b2 + 2ca
c 2 + ca + a2
+p
c 2 + 2a b
a2 + a b + b2
≥3
p
a b + bc + ca.
2.36. Let a, b, c be nonnegative real numbers, no two of which are zero. If k ≥ 1, then
a k+1
b k+1
c k+1
ak + bk + c k
+
+
≤
.
2a2 + bc 2b2 + ca 2c 2 + a b
a+b+c
2.37. If a, b, c are positive real numbers, then
a2 − bc
b2 − ca
c2 − a b
+p
+p
≥ 0;
p
3a2 + 2bc
3b2 + 2ca
3c 2 + 2a b
(a)
(b)
a2 − bc
p
8a2 + (b + c)2
+p
b2 − ca
8b2 + (c + a)2
+p
c2 − a b
8c 2 + (a + b)2
≥ 0.
262
Vasile Cîrtoaje
p
2.38. Let a, b, c be positive real numbers. If 0 ≤ k ≤ 1 + 2 2, then
p
a2 − bc
ka2 + b2 + c 2
+p
b2 − ca
k b2 + c 2 + a2
+p
c2 − a b
kc 2 + a2 + b2
≥ 0.
2.39. If a, b, c are nonnegative real numbers, then
p
p
p
(a2 − bc) b + c + (b2 − ca) c + a + (c 2 − a b) a + b ≥ 0.
2.40. If a, b, c are nonnegative real numbers, then
p
p
p
(a2 − bc) a2 + 4bc + (b2 − ca) b2 + 4ca + (c 2 − a b) c 2 + 4a b ≥ 0.
2.41. If a, b, c are nonnegative real numbers, then
v
t
v
v
t
t
a3
b3
c3
+
+
≥ 1.
a3 + (b + c)3
b3 + (c + a)3
c 3 + (a + b)3
2.42. If a, b, c are positive real numbers, then
v
t
1 1 1
(a + b + c)
+ +
a b c

‹
≥1+
v
u
t
1+
v
t
(a2 + b2 + c 2 )

‹
1
1
1
+
+
.
a2 b2 c 2
2.43. If a, b, c are positive real numbers, then
5+
v
t
2(a2
+
b2
+ c2)

‹

‹
1
1
1
1 1 1
+
+
− 2 ≥ (a + b + c)
+ +
.
a2 b2 c 2
a b c
2.44. If a, b, c are real numbers, then
Æ
2(1 + a bc) + 2(1 + a2 )(1 + b2 )(1 + c 2 ) ≥ (1 + a)(1 + b)(1 + c).
Symmetric Nonrational Inequalities
263
2.45. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
v
t a2 + bc
b2
+ c2
+
v
t b2 + ca
c2
+ a2
+
v
t c2 + a b
a2
+
1
≥2+ p .
2
b2
2.46. If a, b, c are nonnegative real numbers, then
Æ
a(2a + b + c) +
Æ
b(2b + c + a) +
Æ
c(2c + a + b) ≥
Æ
12(a b + bc + ca).
2.47. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that
a
Æ
(4a + 5b)(4a + 5c) + b
Æ
(4b + 5c)(4b + 5a) + c
Æ
(4c + 5a)(4c + 5b) ≥ 27.
2.48. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. Prove that
a
Æ
(a + 3b)(a + 3c) + b
Æ
(b + 3c)(b + 3a) + c
Æ
(c + 3a)(c + 3b) ≥ 12.
2.49. Let a, b, c be nonnegative real numbers such that a2 + b2 + c 2 = 3. Prove that
p
2 + 7a b +
p
2 + 7bc +
p
2 + 7ca ≥ 3
Æ
3(a b + bc + ca).
2.50. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. Prove that
Pp
(a)
P
(b)
(c)
P
a(b + c)(a2 + bc) ≥ 6;
p
p
a(b + c) a2 + 2bc ≥ 6 3;
p
a(b + c) (a + 2b)(a + 2c) ≥ 18.
2.51. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. Prove that
a
p
p
p
bc + 3 + b ca + 3 + c a b + 3 ≥ 6.
264
Vasile Cîrtoaje
2.52. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that
(a)
P
p
(b + c) b2 + c 2 + 7bc ≥ 18;
(b)
p
P
p
(b + c) b2 + c 2 + 10bc ≤ 12 3.
2.53. Let a, b, c be nonnegative real numbers such then a + b + c = 2. Prove that
p
p
p
p
a + 4bc + b + 4ca + c + 4a b ≥ 4 a b + bc + ca.
2.54. If a, b, c are nonnegative real numbers, then
p
p
p
p
a2 + b2 + 7a b + b2 + c 2 + 7bc + c 2 + a2 + 7ca ≥ 5 a b + bc + ca.
2.55. If a, b, c are nonnegative real numbers, then
p
p
p
Æ
a2 + b2 + 5a b + b2 + c 2 + 5bc + c 2 + a2 + 5ca ≥ 21(a b + bc + ca).
2.56. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. Prove that
v
t2
p
p
p
a a2 + 5 + b b2 + 5 + c c 2 + 5 ≥
(a + b + c)2 .
3
2.57. Let a, b, c be nonnegative real numbers such that a2 + b2 + c 2 = 1. Prove that
p
p
p
a 2 + 3bc + b 2 + 3ca + c 2 + 3a b ≥ (a + b + c)2 .
2.58. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that
(a)
(b)
a
a
v
t 2a + bc
3
v
t a(1 + b + c)
3
+b
+b
v
t 2b + ca
3
+c
v
t b(1 + c + a)
3
v
t 2c + a b
+c
3
≥ 3;
v
t c(1 + a + b)
3
≥ 3.
Symmetric Nonrational Inequalities
265
2.59. If a, b, c are nonnegative real numbers such that a + b + c = 3, then
Æ
8(a2 + bc) + 9 +
Æ
8(b2 + ca) + 9 +
Æ
8(c 2 + a b) + 9 ≥ 15.
2.60. Let a, b, c be nonnegative real numbers such that a + b + c = 3. If k ≥
p
a2 + bc + k +
p
b2 + ca + k +
p
9
, then
8
p
c 2 + a b + k ≥ 3 2 + k.
2.61. If a, b, c are nonnegative real numbers such that a + b + c = 3, then
p
a3 + 2bc +
p
b3 + 2ca +
p
p
c 3 + 2a b ≥ 3 3.
2.62. If a, b, c are positive real numbers, then
p
a2 + bc
+
b+c
p
b2 + ca
+
c+a
p
p
c2 + a b
3 2
≥
.
a+b
2
2.63. If a, b, c are nonnegative real numbers, no two of which are zero,then
p
bc + 4a(b + c)
+
b+c
p
ca + 4b(c + a)
+
c+a
p
a b + 4c(a + b) 9
≥ .
a+b
2
2.64. If a, b, c are nonnegative real numbers, no two of which are zero,then
p
p
p
a a2 + 3bc b b2 + 3ca c c 2 + 3a b
+
+
≥ a + b + c.
b+c
c+a
a+b
2.65. If a, b, c are nonnegative real numbers, no two of which are zero,then
v
t
v
v
t
t
2a(b + c)
2b(c + a)
2c(a + b)
+
+
≥ 2.
(2b + c)(b + 2c)
(2c + a)(c + 2a)
(2a + b)(a + 2b)
266
Vasile Cîrtoaje
2.66. If a, b, c are nonnegative real numbers such that a b + bc + ca = 3, then
v
v
v
v
s
s
t ab
t bc
t ab
t bc
ca
ca
+
+
≤
1
≤
+
+
.
3a2 + 6
3b2 + 6
3c 2 + 6
6a2 + 3
6b2 + 3
6c 2 + 3
2.67. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. If k > 1, than
a k (b + c) + b k (c + a) + c k (a + b) ≥ 6.
2.68. Let a, b, c be nonnegative real numbers such that a + b + c = 2. If 2 ≤ k ≤ 3, than
a k (b + c) + b k (c + a) + c k (a + b) ≤ 2.
2.69. Let a, b, c be nonnegative real numbers, no two of which are zero. If m > n ≥ 0,
than
bm + c m
c m + am
am + bm
(b
+
c
−
2a)
+
(c
+
a
−
2b)
+
(a + b − 2c) ≥ 0.
bn + c n
c n + an
an + bn
2.70. Let a, b, c be positive real numbers such that a bc = 1. Prove that
p
p
p
a2 − a + 1 + a2 − a + 1 + a2 − a + 1 ≥ a + b + c.
2.71. Let a, b, c be positive real numbers such that a bc = 1. Prove that
p
p
p
16a2 + 9 + 16b2 + 9 + 16b2 + 9 ≥ 4(a + b + c) + 3.
2.72. Let a, b, c be positive real numbers such that a bc = 1. Prove that
p
p
p
25a2 + 144 + 25b2 + 144 + 25c 2 + 144 ≤ 5(a + b + c) + 24.
2.73. If a, b are positive real numbers such that a b + bc + ca = 3, then
(a)
(b)
p
a2 + 3 +
p
b2 + 3 +
p
b2 + 3 ≥ a + b + c + 3;
p
p
p
p
a + b + b + c + c + a ≥ 4(a + b + c) + 6.
Symmetric Nonrational Inequalities
267
2.74. If a, b, c are nonnegative real numbers such that a + b + c = 3, then
Æ
Æ
Æ
(5a2 + 3)(5b2 + 3) + (5b2 + 3)(5c 2 + 3) + (5c 2 + 3)(5a2 + 3) ≥ 24.
2.75. If a, b, c are nonnegative real numbers such that a + b + c = 3, then
v
t 4(a2 + b2 + c 2 ) + 42
p
p
p
.
a2 + 1 + b2 + 1 + c 2 + 1 ≥
3
2.76. If a, b, c are nonnegative real numbers such that a + b + c = 3, then
p
p
p
p
(a)
a2 + 3 + b2 + 3 + c 2 + 3 ≥ 2(a2 + b2 + c 2 ) + 30;
p
p
p
p
(b)
3a2 + 1 + 3b2 + 1 + 3c 2 + 1 ≥ 2(a2 + b2 + c 2 ) + 30.
2.77. If a, b, c are nonnegative real numbers such that a + b + c = 3, then
Æ
Æ
Æ
(32a2 + 3)(32b2 + 3) + (32b2 + 3)(32c 2 + 3) + (32c 2 + 3)(32a2 + 3) ≤ 105.
2.78. If a, b, c are positive real numbers, then
b+c
c + a
a + b
≥ 2.
+
−
3
−
3
+
−
3
a
c
b
2.79. If a, b, c are real numbers such that a bc 6= 0, then
b + c c + a a + b +
≥ 2.
+
a b
c 2.80. Let a, b, c be nonnegative real numbers, no two of which are zero, and let
x=
2a
,
b+c
y=
2b
2c
, z=
.
c+a
a+b
Prove that
(a)
(b)
p
yz + z x ≥ 6;
p
p
p
p
x + y + z ≥ 8 + x yz.
x + y +z+
p
xy+
p
268
Vasile Cîrtoaje
2.81. Let a, b, c be nonnegative real numbers, no two of which are zero, and let
x=
2a
,
b+c
y=
2b
2c
, z=
.
c+a
a+b
Prove that
p
1 + 24x +
p
1 + 24 y +
p
1 + 24z ≥ 15.
2.82. If a, b, c are positive real numbers, then
v
v
v
t
t
t
7a
7b
7c
+
+
≤ 3.
a + 3b + 3c
b + 3c + 3a
c + 3a + 3b
2.83. If a, b, c are positive real numbers such that a + b + c = 3, then
Æ
Æ
Æ
p
3
3
3
3
a2 (b2 + c 2 ) + b2 (c 2 + a2 ) + c 2 (a2 + b2 ) ≤ 3 2.
2.84. If a, b, c are nonnegative real numbers, no two of which are zero, then
1
1
1
1
2
+
+
≥
+p
.
a+b b+c c+a
a+b+c
a b + bc + ca
2.85. If a, b ≥ 1, then
1
1
1
1
+ ≥p
+p
.
p
2
3a + 1
3a b + 1
3b + 1
2.86. Let a, b, c be positive real numbers such that a ≥ 1 ≥ b ≥ c and a bc = 1. Prove
that
1
1
1
3
+p
≥ .
+p
p
3a + 1
3c + 1 2
3b + 1
1
2.87. Let a, b, c be positive real numbers such that a + b + c = 3. If k ≥ p , then
2
(a bc)k (a2 + b2 + c 2 ) ≤ 3.
Symmetric Nonrational Inequalities
269
2.88. Let p and q be nonnegative real numbers such that p2 ≥ 3q, and let
v
v
t 2p + w
t 2p − 2w
+2
,
g(p, q) =
3
3
s
 s
2p + 2w
2p − w

+2
,

3
3
h(p, q) =
p
p
p

 p + p + q,
where w =
p
p2 ≤ 4q
p2 ≥ 4q
,
p2 − 3q. If a, b, c are nonnegative real numbers such that
a + b + c = p,
a b + bc + ca = q,
then
p
(a)
with equality for a =
(b)
a+b+
p
b+c+
p
c + a ≥ g(p, q),
p−w
p + 2w
and b = c =
(or any cyclic permutation);
3
3
p
p
p
a + b + b + c + c + a ≤ h(p, q),
p − 2w
p+w
and b = c =
(or any cyclic permutation) - when
3
3
p2 ≤ 4q, and for a = 0, b+c = p and bc = q (or any cyclic permutation) - when p2 ≥ 4q.
with equality for a =
2.89. Let a, b, c, d be nonnegative real numbers such that a2 + b2 + c 2 + d 2 = 1. Prove
that
p
p
p
p
p
p
p
p
1 − a + 1 − b + 1 − c + 1 − d ≥ a + b + c + d.
2.90. Let a, b, c, d be positive real numbers. Prove that
p
A + 2 ≥ B + 4,
where
‹
1 1 1 1
A = (a + b + c + d)
+ + +
− 16,
a b c d

‹
1
1
1
1
2
2
2
2
B = (a + b + c + d ) 2 + 2 + 2 + 2 − 16.
a
b
c
d

270
Vasile Cîrtoaje
2.91. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = 1.
Prove that
p
p
p
3a1 + 1 + 3a2 + 1 + · · · + 3an + 1 ≥ n + 1.
2.92. Let 0 ≤ a < b and a1 , a2 , . . . , an ∈ [a, b]. Prove that
€p
p
p Š2
b− a .
a1 + a2 · · · + an − n n a1 a2 · · · an ≤ (n − 1)
2.93. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an = 1. Prove that
1
p
1 + (n2
− 1)a1
+p
1
1 + (n2 − 1)a2
+ ··· + p
1
1 + (n2 − 1)an
≥ 1.
2.94. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an = 1. Prove that
n
X
i=1
1
1+
p
1 + 4n(n − 1)ai
≥
1
.
2
2.95. If f is a convex function on a real interval I and a1 , a2 , . . . , an ∈ I, then
a + a + ··· + a 1
2
n
f (a1 ) + f (a2 ) + · · · + f (an ) + n(n − 2) f
≥
n
≥ (n − 1)[ f (b1 ) + f (b2 ) + · · · + f (bn )],
where
bi =
1 X
aj,
n − 1 j6=i
i = 1, 2, · · · , n.
2.96. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an = 1. Prove that
n
X
i=1
1
n−1+
p
(n − 1)2
+ 4nai
≤
1
.
2
2.97. If a1 , a2 , . . . , an are positive real numbers such that a1 a2 · · · an = 1, then
v
u 2
t a1 + a22 + · · · + an2
a1 + a2 + · · · + an ≥ n − 1 +
.
n
Symmetric Nonrational Inequalities
271
2.98. If a1 , a2 , . . . , an are positive real numbers such that a1 a2 · · · an = 1, then
q
Æ
(n − 1)(a12 + a22 + · · · + an2 ) + n − n(n − 1) ≥ a1 + a2 + · · · + an .
2.99. Let a1 , a2 , . . . , an (n ≥ 3) be positive real numbers such that a1 a2 · · · an = 1. If
0<p≤
then
1
p
1 + pa1
+p
1
1 + pa2
2n − 1
,
(n − 1)2
+p
1
1 + pan
≤p
n
1+p
.
2.100. If a1 , a2 , . . . , an (n ≥ 3) are positive real numbers such that a1 a2 · · · an = 1, then
n Ç
X
‚
(n − 1)2 ai4
+ 2n − 1 ≥
i=1
n
X
ai
n
X
Œ2
ai
;
i=1
q
‚
(n − 1)2 ai2 + 2n − 1 ≥
n
X
Œ2
ai
.
i=1
i=1
2.101. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an ≥ 1. If k > 1,
then
X
a1k
≥ 1.
a1k + a2 + · · · + an
2.102. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an ≥ 1. If
−2
≤ k < 1,
n−2
then
X
a1k
a1k + a2 + · · · + an
≤ 1.
2.103. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an ≥ 1. If k > 1,
then
X
a1
≤ 1.
k
a1 + a2 + · · · + an
272
Vasile Cîrtoaje
2.104. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an ≥ 1. If
−1 −
then
2
≤ k < 1,
n−2
a1
X
a1k
+ a2 + · · · + an
≥ 1.
2.105. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an = 1. If k ≥ 0,
then
X
1
≤ 1.
k
a1 + a2 + · · · + an
2.106. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an ≥ n.
If 1 < k ≤ n + 1, then
a1
a1k
+ a2 + · · · + an
+
a2
a1 + a2k
+ · · · + an
+ ··· +
an
≤ 1.
a1 + a2 + · · · + ank
2.107. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an ≤ n.
If 0 ≤ k < 1, then
1
a1k
+ a2 + · · · + an
+
1
a1 + a2k
+ · · · + an
+ ··· +
1
≥ 1.
a1 + a2 + · · · + ank
2.108. Let a1 , a2 , . . . , an be positive real numbers. If k > 1, then
X ak + ak + · · · + ak
2
3
n
a2 + a3 + · · · + an
≤
n(a1k + a2k + · · · + ank )
a1 + a2 + · · · + an
.
Symmetric Nonrational Inequalities
2.2
273
Solutions
P 2.1. If a, b, c are nonnegative real numbers, then
Xp
Æ
a2 − a b + b2 ≤ 6(a2 + b2 + c 2 ) − 3(a b + bc + ca).
Solution. By squaring, the inequality becomes as follows
XÆ
2(a b + bc + ca) + 2
(a2 − a b + b2 )(a2 − ac + c 2 ) ≤ 4(a2 + b2 + c 2 ),
X €p
a2 − a b + b2 −
p
a2 − ac + c 2
Š2
≥ 0.
The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).
P 2.2. If a, b, c are nonnegative real numbers, then
p
a2 − a b + b2 +
p
b2 − bc + c 2 +
p
c 2 − ca + a2 ≤ 3
v
t a2 + b2 + c 2
2
Solution (by Nguyen Van Quy). Assume that c = min{a, b, c}. Since
b2 − bc + c 2 ≤ b2
and
c 2 − ca + a2 ≤ a2 ,
it suffices to show that
p
a2 − a b + b2 + a + b ≤ 3
v
t a2 + b2 + c 2
2
.
Using the Cauchy-Schwarz inequality, we have
p
a2
− ab +
=
b2
+a+b≤
v•
t
(a2
− ab +
b2 ) +
˜
(a + b)2
(1 + k)
k
v
t (1 + k)[(1 + k)(a2 + b2 ) + (2 − k)a b]
k
.
.
274
Vasile Cîrtoaje
Choosing k = 2, we get
p
a2 − a b + b2 + a + b ≤ 3
v
t a2 + b2
2
v
t a2 + b2 + c 2
≤3
= 3.
2
The equality holds for a = b and c = 0 (or any cyclic permutation).
P 2.3. If a, b, c are nonnegative real numbers, then
v
t
a2
+
b2
v
v
t
t
p
2
2
2
2
2
− a b + b + c − bc + c 2 + a2 − ca ≥ 2 a2 + b2 + c 2 .
3
3
3
(Vasile Cîrtoaje, 2012)
First Solution. By squaring, the inequality becomes
XÆ
2
(3a2 + 3b2 − 2a b)(3a2 + 3c 2 − 2ac) ≥ 6(a2 + b2 + c 2 ) + 2(a b + bc + ca),
6(a2 + b2 + c 2 − a b − bc − ca) ≥
X €p
3a2 + 3b2 − 2a b −
p
3a2 + 3c 2 − 2ac
Š2
,
X
X
(b − c)2 (3b + 3c − 2a)2
(b − c)2 ≥
2 ,
p
p
3a2 + 3b2 − 2a b + 3a2 + 3c 2 − 2ac


2
X
(3b + 3c − 2a)
(b − c)2 1 − €p
Š2  .
p
2
2
2
2
9a + 9b − 6a b + 9a + 9c − 6ac
3
Since
Æ
9a2 + 9b2 − 6a b = (3b − a)2 + 8a2 ≥ |3b − a|,
Æ
p
9a2 + 9c 2 − 6ac = (3c − a)2 + 8a2 ≥ |3c − a|,
p
it suffices to show that

‹2 X
|3b + 3c − 2a|
2
(b − c) 1 −
≥ 0.
|3b − a| + |3c − a|
This is true since
|3b + 3c − 2a| = |(3b − a) + (3c − a)| ≤ |3b − a| + |3c − a|.
The equality holds for a = b = c, and also for b = c = 0 (or any cyclic permutation).
Symmetric Nonrational Inequalities
275
Second Solution. Assume that a ≥ b ≥ c. Write the inequality as
Æ
Æ
Æ
(a + b)2 + 2(a − b)2 + (b + c)2 + 2(b − c)2 + (a + c)2 + 2(a − c)2 ≥
Æ
≥ 2 3(a2 + b2 + c 2 ).
By Minkowski’s inequality, it suffices to show that
Æ
Æ
[(a + b) + (b + c) + (a + c)]2 + 2[(a − b) + (b − c) + (a − c)]2 ≥ 2 3(a2 + b2 + c 2 ),
which is equivalent to
Æ
(a + b + c)2 + 2(a − c)2 ≥
Æ
3(a2 + b2 + c 2 ).
By squaring, the inequality turns into
(a − b)(b − c) ≥ 0.
P 2.4. If a, b, c are nonnegative real numbers, then
Xp
Æ
a2 + a b + b2 ≥ 4(a2 + b2 + c 2 ) + 5(a b + bc + ca).
(Vasile Cîrtoaje, 2009)
First Solution. By squaring, the inequality becomes
XÆ
(a2 + a b + b2 )(a2 + ac + c 2 ) ≥ (a + b + c)2 .
Using the Cauchy-Schwarz inequality, we get
v
•
‹
˜

XÆ
Xu
t
b 2 3b2 c 2 3c 2
2
2
2
2
(a + a b + b )(a + ac + c ) =
a+
+
a+
+
2
4
2
4
≥
X •
‹
˜
b c 3bc
a+
a+
+
= (a + b + c)2 .
2
2
4
The equality holds for a = b = c, and also for b = c = 0 (or any cyclic permutation).
Second Solution. Assume that a ≥ b ≥ c. By Minkowski’s inequality, we get
Xp
XÆ
2
a2 + a b + b2 =
3(a + b)2 + (a − b)2
≥
Æ
3[(a + b) + (b + c) + (c + a)]2 + [(a − b) + (b − c) + (a − c)]2
276
Vasile Cîrtoaje
=2
Æ
3(a + b + c)2 + (a − c)2 .
Therefore, it suffices to show that
3(a + b + c)2 + (a − c)2 ≥ 4(a2 + b2 + c 2 ) + 5(a b + bc + ca),
which is equivalent to the obvious inequality
(a − b)(b − c) ≥ 0.
Remark. Similarly, we can prove the following generalization.
• Let a, b, c be nonnegative real numbers such that a b + bc + ca = 4. If |k| ≤ 2, then
Xp
p
a2 + ka b + b2 ≥ 2 a2 + b2 + c 2 + 3k + 2.
For k = −2/3 and k = 1, we get the inequalities in P 2.3 and P 2.4, respectively. For
k = −1 and k = 0, we get the inequalities
Xp
p
a2 − a b + b2 ≥ 2 a2 + b2 + c 2 − 1,
Xp
a2 + b2 ≥ 2
p
a2 + b2 + c 2 + 2.
P 2.5. If a, b, c are nonnegative real numbers, then
Xp
Æ
a2 + a b + b2 ≤ 5(a2 + b2 + c 2 ) + 4(a b + bc + ca).
(Michael Rozenberg, 2008)
First Solution (by Vo Quoc Ba Can). Using the Cauchy-Schwarz inequality, we have
€X p
Š2 ”X
— X b2 + bc + c 2
2
2
≤
(b + c)
b + bc + c
b+c
X 2
X
b + bc + c 2
a 2
= 2(a + b + c)
=2
1+
(b + bc + c 2 )
b+c
b+c
= 4(a2 + b2 + c 2 ) + 2(a b + bc + ca) +
X 2a(b2 + bc + c 2 )
b+c

bc
= 4(a + b + c ) + 2(a b + bc + ca) +
2a b + c −
b+c
X 1
= 4(a2 + b2 + c 2 ) + 6(a b + bc + ca) − 2a bc
.
b+c
2
2
2
X
‹
Symmetric Nonrational Inequalities
277
Thus, it suffices to prove that
4(a2 + b2 + c 2 ) + 6(a b + bc + ca) − 2a bc
X
1
≤ 5(a2 + b2 + c 2 ) + 4(a b + bc + ca),
b+c
which is equivalent to Schur’s inequality
2(a b + bc + ca) ≤ a2 + b2 + c 2 + 2a bc
X
1
.
b+c
We can prove this inequality by writing it as follows:
‹
X 
bc
,
a a+
(a + b + c)2 ≤ 2
b+c
X a
(a + b + c)2 ≤ 2(a b + bc + ca)
,
b+c
”X
—X a
(a + b + c)2 ≤
a(b + c)
.
b+c
Clearly, the last inequality follows from the Cauchy-Schwarz inequality. The equality
holds for a = b = c.
Second Solution. Let us denote
p
p
A = b2 + bc + c 2 , B = c 2 + ca + a2 ,
C=
p
a2 + a b + b2 .
Without loss of generality, assume that a ≥ b ≥ c. For b = c = 0, the inequality is clearly
true. Consider further b > 0. By squaring, the inequality becomes
X
X
X
2
BC ≤ 3
a2 + 3
a b,
X
a2 −
X
ab ≤
X
(B − C)2 ,
X
X (b − c)2
(b − c)2 ≤ 2(a + b + c)2
.
(B + C)2
Since
(B + C)2 ≤ 2(B 2 + C 2 ) = 2(2a2 + b2 + c 2 + ca + a b),
it suffices to show that
X
X
(b − c)2 ≤ (a + b + c)2
(b − c)2
,
2a2 + b2 + c 2 + ca + a b
which is equivalent to
X
(b − c)2 Sa ≥ 0,
278
Vasile Cîrtoaje
where
Sa =
−a2 + a b + 2bc + ca
(a + b + c)2
−
1
=
,
2a2 + b2 + c 2 + ca + a b
2a2 + b2 + c 2 + ca + a b
Sb =
−b2 + bc + 2ca + a b
≥ 0,
2b2 + c 2 + a2 + a b + bc
Sc =
−c 2 + ca + 2a b + bc
≥ 0.
2c 2 + a2 + b2 + bc + ca
According to
X
a2
(b − c)2 Sa ≥ (b − c)2 Sa + (a − c)2 S b ≥ (b − c)2 Sa + 2 (b − c)2 S b
b

‹
Sb
a
2
2
2 Sa
≥ (b − c) Sa + (b − c) S b = a(b − c)
+
,
b
a
b
it suffices to prove that
Sa S b
+
≥ 0,
a
b
which is equivalent to
a2 − a b − 2bc − ca
−b2 + bc + 2ca + a b
≥
.
b(2b2 + c 2 + a2 + a b + bc)
a(2a2 + b2 + c 2 + ca + a b)
Consider the non-trivial case a2 − a b − 2bc − ca ≥ 0. Since
(2a2 + b2 + c 2 + ca + a b) − (2b2 + c 2 + a2 + a b + bc) = (a − b)(a + b + c) ≥ 0,
it suffices to show that
−b2 + bc + 2ca + a b
a2 − a b − 2bc − ca
≥
.
b
a
Indeed,
a(−b2 + bc + 2ca + a b) − b(a2 − a b − 2bc − ca) = 2c(a2 + a b + b2 ) > 0.
P 2.6. If a, b, c are nonnegative real numbers, then
Xp
p
p
a2 + a b + b2 ≤ 2 a2 + b2 + c 2 + a b + bc + ca.
(Vasile Cîrtoaje, 2010)
Symmetric Nonrational Inequalities
279
First Solution (by Nguyen Van Quy). Assume that a = max{a, b, c}. Since
p
p
Æ
a2 + a b + b2 + c 2 + ca + a2 ≤ 2[(a2 + a b + b2 ) + (c 2 + ca + a2 )],
it suffices to show that
p
p
p
p
2 A + b2 + bc + c 2 ≤ 2 X + Y ,
where
1
A = a2 + (b2 + c 2 + a b + ac),
2
X = a2 + b2 + c 2 ,
Y = a b + bc + ca.
Write the desired inequality as follows
p
p
p
p
2( A − X ) ≤ Y − b2 + bc + c 2 ,
2(A − X )
Y − (b2 + bc + c 2 )
,
p
p
p ≤p
A+ X
Y + b2 + bc + c 2
b(a − b) + c(a − c)
b(a − b) + c(a − c)
≤p
.
p
p
p
A+ X
Y + b2 + bc + c 2
Since b(a − b) + c(a − c) ≥ 0, we need to show that
p
p
p
p
A + X ≥ Y + b2 + bc + c 2 .
This inequality is true because X ≥ Y and
p
p
A ≥ b2 + bc + c 2 .
Indeed,
2(A − b2 − bc − c 2 ) = 2a2 + (b + c)a − (b + c)2 = (2a − b − c)(a + b + c) ≥ 0.
The equality holds for a = b = c, and also for b = c = 0 (or any cyclic permutation).
Second Solution. In the first solution of P 2.5, we have shown that
X
€X p
Š2
b2 + bc + c 2 ≤ 4(a2 + b2 + c 2 ) + 6(a b + bc + ca) − 2a bc
1
.
b+c
Thus, it suffices to prove that
4(a2 + b2 +c 2 )+6(a b+ bc +ca)−2a bc
X
€ p
Š2
p
1
≤ 2 a2 + b2 + c 2 + a b + bc + ca ,
b+c
which is equivalent to
2a bc
X
Æ
1
+ 4 (a2 + b2 + c 2 )(a b + bc + ca) ≥ 5(a b + bc + ca).
b+c
280
Vasile Cîrtoaje
Since
X
1
9
9
≥P
=
,
b+c
(b + c) 2(a + b + c)
it is enough to prove that
Æ
9a bc
+ 4 (a2 + b2 + c 2 )(a b + bc + ca) ≥ 5(a b + bc + ca),
a+b+c
which can be written as
Æ
9a bc
+ 4 q(p2 − 2q) ≥ 5q,
p
where
p = a + b + c, q = a b + bc + ca.
p
For p2 ≥ 4q, this inequality is true because 4 q(p2 − 2q) ≥ 5q. Consider further
3q ≤ p2 ≤ 4q.
By Schur’s inequality of third degree,
9a bc
≥ 4q − p2 .
p
Therefore, it suffices to show that
(4q − p2 ) + 4
Æ
q(p2 − 2q) ≥ 5q,
which is
Æ
4 q(p2 − 2q) ≥ p2 + q.
Indeed,
16q(p2 − 2q) − (p2 + q)2 = (p2 − 3q)(11q − p2 ) ≥ 0.
Third Solution. Let us denote
p
p
p
A = b2 + bc + c 2 , B = c 2 + ca + a2 , C = a2 + a b + b2 ,
p
p
X = a2 + b2 + c 2 , Y = a b + bc + ca.
By squaring, the inequality becomes
X
X
2
BC ≤ 2
a2 + 4X Y,
X
(B − C)2 ≥ 2(X − Y )2 ,
P
2
X (b − c)2
(b − c)2
2
2(a + b + c)
≥
.
(B + C)2
(X + Y )2
Symmetric Nonrational Inequalities
281
Since
B + C ≤ (c + a) + (a + b) = 2a + b + c,
it suffices to show that
2
2(a + b + c)
X
(b − c)2
≥
(2a + b + c)2
P
2
(b − c)2
(X + Y )2
.
According to the Cauchy-Schwarz inequality, we have
X
P
2
(b − c)2
(b − c)2
≥P
.
(2a + b + c)2
(b − c)2 (2a + b + c)2
Therefore, it is enough to prove that
2(a + b + c)2
1
P
≥
,
(X + Y )2
(b − c)2 (2a + b + c)2
which is
2(a + b + c)2 (X + Y )2 ≥
X
(b − c)2 (2a + b + c)2 .
We see that
(a + b + c)2 (X + Y )2 ≥
=
and
€X
a2 + 2
X
ab
Š €X
a2 +
X
ab
Š
Š2
€X Š €X Š
€X Š2
a2 + 3
ab
a2 + 2
ab
X
X
X
≥
a4 + 3
a b(a2 + b2 ) + 4
a2 b2
€X
X
X
(b − c)2 (2a + b + c)2 =
(b − c)2 [4a2 + 4a(b + c) + (b + c)2 ]
X
X
X
=4
a2 (b − c)2 + 4
a(b − c)(b2 − c 2 ) +
(b2 − c 2 )2
X
X
X
≤8
a2 b2 + 4
a(b3 + c 3 ) + 2
a4 .
Thus, it suffices to show that
X
X
X
X
X
X
a4 + 3
a b(a2 + b2 ) + 4
a2 b2 ≥ 4
a2 b2 + 2
a(b3 + c 3 ) +
a4 ,
which is equivalent to the obvious inequality
X
a b(a2 + b2 ) ≥ 0.
282
Vasile Cîrtoaje
P 2.7. If a, b, c are nonnegative real numbers, then
p
p
p
p
p
a2 + 2bc + b2 + 2ca + c 2 + 2a b ≤ a2 + b2 + c 2 + 2 a b + bc + ca.
(Vasile Cîrtoaje and Nguyen Van Quy, 1989)
Solution (by Nguyen Van Quy). Let
p
X = a2 + b2 + c 2 ,
Y=
p
a b + bc + ca.
Consider the nontrivial case when no two of a, b, c are zero (Y 6= 0) and write the
inequality as
X€
Š
p
X − a2 + 2bc ≥ 2(X − Y ),
P
X
(b − c)2
(b − c)2
≥
.
p
X +Y
X + a2 + 2bc
By the Cauchy-Schwarz inequality, we have
X
P
2
(b − c)2
(b − c)2
≥P
.
p
p
X + a2 + 2bc
(b − c)2 X + a2 + 2bc
Therefore, it suffices to show that
P
(b − c)2
1
,
≥
p
P
2
2
X
+
Y
(b − c) X + a + 2bc
which is equivalent to
X
€
Š
p
(b − c)2 Y − a2 + 2bc ≥ 0.
From
€
Y−
p
a2 + 2bc
Š2
≥ 0.
we get
Y−
Thus,
p
a2 + 2bc ≥
Y 2 − (a2 + 2bc) (a − b)(c − a)
=
.
2Y
2Y
X
€
Š X (b − c)2 (a − b)(c − a)
p
(b − c)2 Y − a2 + 2bc ≥
2Y
X
(a − b)(b − c)(c − a)
=
(b − c) = 0.
2Y
The equality holds for a = b, or b = c, or c = a.
Symmetric Nonrational Inequalities
283
P 2.8. If a, b, c are nonnegative real numbers, then
p
1
a2 + 2bc
+p
1
b2 + 2ca
+p
1
c 2 + 2a b
≥p
1
a2 + b2 + c 2
+p
2
a b + bc + ca
.
(Vasile Cîrtoaje, 1989)
Solution . Let
X=
p
a2 + b2 + c 2 ,
Y=
p
a b + bc + ca.
Consider the nontrivial case when Y > 0 and write the inequality as
‹

‹
X
1
1
1
1
−
−
≥2
,
p
Y
X
a2 + 2bc X
P
X
(b − c)2
(b − c)2
.
≥
p
p
Y (X + Y )
a2 + 2bc X + a2 + 2bc
By the Cauchy-Schwarz inequality, we have
X
P
2
(b − c)2
(b − c)2
≥P
.
p
p
p
p
a2 + 2bc X + a2 + 2bc
(b − c)2 a2 + 2bc X + a2 + 2bc
Therefore, it suffices to show that
P
(b − c)2
1
,
≥
p
p
P
2
2
2
Y
(X
+ Y)
(b − c) a + 2bc X + a + 2bc
which is equivalent to
X
p
(b − c)2 [X Y − X a2 + 2bc + (a − b)(c − a)] ≥ 0.
Since
X
X
(b − c)2 (a − b)(c − a) = (a − b)(b − c)(c − a)
(b − c) = 0,
the inequality becomes
X
€
Š
p
(b − c)2 X Y − a2 + 2bc ≥ 0,
X
€
Š
p
(b − c)2 Y − a2 + 2bc ≥ 0.
We have proved this inequality at the preceding problem P 2.7. The equality holds for
a = b, or b = c, or c = a.
284
Vasile Cîrtoaje
P 2.9. If a, b, c are positive real numbers, then
p
p
p
p
p
2a2 + bc + 2b2 + ca + 2c 2 + a b ≤ 2 a2 + b2 + c 2 + a b + bc + ca.
Solution. We will apply Lemma below for
X = 2a2 + bc, Y = 2b2 + ca, Z = 2c 2 + a b
and
A = B = a2 + b2 + c 2 , C = a2 + b2 + c 2 .
We have
X + Y + Z = A+ B + C
and
A = B ≥ C.
Without loss of generality, assume that
a ≥ b ≥ c.
By Lemma, it suffices to show that
max{X , Y, Z} ≥ A,
min{X , Y, Z} ≤ C.
Indeed, we have
max{X , Y, Z} − A ≥ X − A = (a2 − b2 ) + c(b − c) ≥ 0,
min{X , Y, Z} − C ≤ Z − C = c(2c − a − b) ≤ 0.
Equality holds for a = b = c.
Lemma. If X , Y, Z and A, B, C are positive real numbers such that
X + Y + Z = A + B + C,
max{X , Y, Z} ≥ max{A, B, C},
then
p
X+
p
Y+
p
Z≤
min{X , Y, Z} ≤ min{A, B, C},
p
p
p
A + B + C.
Proof. On the assumption that X ≥ Y ≥ Z and A ≥ B ≥ C, we have
X ≥ A,
Z ≤ C,
and hence
p
p
p
p
p
p
p
p
p
p
p
p
X + Y + Z − A− B − C = ( X − A ) + ( Y − B ) + ( Z − C )
Symmetric Nonrational Inequalities
285
X −A Y −B Z −C
X −A Y −B Z −C
p + p + p ≤ p + p + p
2 C
2 C
2 A
2 B
2 B
2 B

‹
1
C −Z Z −C
1
≤ 0.
= p + p = (C − Z) p − p
2 C
2 B
2 B 2 C
≤
Remark. This Lemma is a particular case of Karamata’s inequality.
P 2.10. Let a, b, c be nonnegative real numbers such that a + b + c = 3. If k =
then
XÆ
p
a(a + k b)(a + kc) ≤ 3 3.
p
3 − 1,
Solution. By the Cauchy-Schwarz inequality, we have
XÆ
a(a + k b)(a + kc) ≤
r€X Š ”X
—
a
(a + k b)(a + kc) .
Thus, it suffices to show that
rX
(a + k b)(a + kc) ≤ a + b + c,
which is an identity. The equality holds for a = b = c = 1, and also for a = 3 and
b = c = 0 (or any cyclic permutation).
P 2.11. If a, b, c are nonnegative real numbers such that a + b + c = 3, then
XÆ
a(2a + b)(2a + c) ≥ 9.
Solution. Write the inequality as follows
X Ӯ
—
Æ
a(2a + b)(2a + c) − a 3(a + b + c) ≥ 0,
X
(a − b)(a − c)Ea ≥ 0,
where
p
Ea = p
a
(2a + b)(2a + c) +
p
3a(a + b + c)
.
286
Vasile Cîrtoaje
Assume that a ≥ b ≥ c. Since (c − a)(c − b)Ec ≥ 0, it suffices to show that
(a − c)Ea ≥ (b − c)E b ,
which is equivalent to
Æ
Æ
Æ
(a − b) 3a b(a + b + c) + (a − c) a(2b + c)(2b + a) ≥ (b − c) b(2a + b)(2a + c).
This is true if
Æ
Æ
(a − c) a(2b + c)(2b + a) ≥ (b − c) b(2a + b)(2a + c).
For the non-trivial case b > c, we have
p
a
a
a−c
≥ ≥p .
b−c
b
b
Therefore, it is enough to show that
a2 (2b + c)(2b + a) ≥ b2 (2a + b)(2a + c).
Write this inequality as
a2 (2a b + 2bc + ca) ≥ b2 (2a b + bc + 2ca).
It is true if
a(2a b + 2bc + ca) ≥ b(2a b + bc + 2ca).
Indeed,
a(2a b + 2bc + ca) − b(2a b + bc + 2ca) = (a − b)(2a b + bc + ca) ≥ 0.
The equality holds for a = b = c = 1, and also for a = 0 and b = c = 3/2 (or any cyclic
permutation).
P 2.12. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that
Æ
b2 + c 2 + a(b + c) +
Æ
c 2 + a2 + b(c + a) +
Æ
a2 + b2 + c(a + b) ≥ 6.
Symmetric Nonrational Inequalities
287
Solution. Denote
A = b2 + c 2 + a(b + c),
B = c 2 + a2 + b(c + a),
C = a2 + b2 + c(a + b).
First Solution. Write the inequality in the homogeneous form
p
p
p
A + B + C ≥ 2(a + b + c).
By squaring, the inequality becomes
Xp
X
X
2
BC ≥ 2
a2 + 6
bc,
X
X p
p 2
(a − b)2 ≥
B− C ,
X
(b − c)2 Sa ≥ 0,
where
(b + c − a)2
Sa = 1 − p
.
p
( B + C)2
Since
Sa ≥ 1 −
a(a + 3b + 3c)
(b + c − a)2
=
≥ 0, S b ≥ 0, Sc ≥ 0,
B+C
B+C
the conclusion follows. The equality holds for a = b = c = 1, and also for a = 3 and
b = c = 0 (or any cyclic permutation).
Second Solution. Write the original inequality as follows
Xp
( A − b − c) ≥ 0,
X c(a − b) + b(a − c)
≥ 0,
p
A+ b + c
X c(a − b)
X c(b − a)
+
≥ 0,
p
p
A+ b + c
B+c+a
X c(a − b)[a − b − (pA − pB)]
≥ 0.
p
p
( A + b + c)( B + c + a)
It suffices to show that
p
p
(a − b)[a − b + ( B − A)] ≥ 0.
Indeed,

‹
p
p
a+b−c
2
(a − b)[a − b + ( B − A)] = (a − b) 1 + p
≥ 0,
p
B+ A
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Vasile Cîrtoaje
because, for the nontrivial case a + b − c < 0, we have
a+b−c
a+b−c
> 0.
1+ p
p >1+
c+c
B+ A
Open Generalization. Let a, b, c be nonnegative real numbers. If 0 < k ≤
XÆ
16
, then
9
(b + c)2 + k(a b − 2bc + ca) ≥ 2(a + b + c).
16
, then the equality holds for a = b = c = 1, for a = 0 and b = c (or
9
any cyclic permutation), and for b = c = 0 (or any cyclic permutation).
Notice that if k =
P 2.13. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that
(a)
p
(b)
p
p
p
p
3a2 + a bc + 3b2 + a bc + 3c 2 + a bc ≥ 3 3 + a bc.
a(3a2 + a bc) +
p
b(3b2 + a bc) +
p
c(3c 2 + a bc) ≥ 6;
(Lorian Saceanu, 2015)
Solution. (a) Write the inequality in the homogeneous form
X Æ
3
a (a + b)(a + c) ≥ 2(a + b + c)2 .
First Solution. Write the inequality as
X
a2 −
X
ab ≥
Š2
p
3 X €p
a+b− a+c ,
a
2
X
X
(b − c)2 ≥ 3
p
a(b − c)2
2 ,
p
a+b+ a+c
X
(b − c)2 Sa ≥ 0,
where
Sa = 1 − p
Since
Sa ≥ 1 − p
3a
2 .
p
a+b+ a+c
3a
p 2 > 0, S b > 0, Sc > 0,
a+ a
the inequality is true. The equality holds for a = b = c = 1.
Symmetric Nonrational Inequalities
289
Second Solution. By Hölder’s inequality, we have
—2
”X Æ
a (a + b)(a + c) ≥ P
P
( a)3
27
=P
.
a
a
(a + b)(a + c)
(a + b)(a + c)
Therefore, it suffices to show that
X
a
3
≤ .
(a + b)(a + c) 4
This inequality has the homogeneous form
X
a
9
≤
,
(a + b)(a + c) 4(a + b + c)
which is equivalent to the obvious inequality
X
a(b − c)2 ≥ 0.
(b) By squaring, the inequality becomes
3
X
a2 + 2
XÆ
(3b2 + a bc)(3c 2 + a bc) ≥ 27 + 6a bc.
By the Cauchy-Schwarz inequality, we have
Æ
(3b2 + a bc)(3c 2 + a bc) ≥ 3bc + a bc.
Therefore, it suffices to show that
X
X
a2 + 6
bc + 6a bc ≥ 27 + 6a bc,
3
which is an identity. The equality holds for a = b = c = 1, and also for a = 0, or b = 0,
or c = 0.
P 2.14. Let a, b, c be positive real numbers such that a b + bc + ca = 3. Prove that
a
Æ
(a + 2b)(a + 2c) + b
Æ
(b + 2c)(b + 2a) + c
Æ
(c + 2a)(c + 2b) ≥ 9.
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Vasile Cîrtoaje
First Solution. Write the inequality as follows:
X Æ
a (a + 2b)(a + 2c) ≥ 3(a b + bc + ca),
Š2
p
1 X €p
a
a + 2b − a + 2c ,
2
X
X
a(b − c)2
(b − c)2 ≥ 4
2 ,
p
p
a + 2b + a + 2c
X
(b − c)2 Sa ≥ 0,
X
a2 −
X
ab ≥
where
Sa = 1 − p
Since
Sa > 1 − p
4a
2 .
p
a + 2b + a + 2c
4a
p 2 = 0, S b > 0, Sc > 0,
a+ a
the inequality is true. The equality holds for a = b = c = 1.
Second Solution. We use the AM-GM inequality to get
X 2a(a + 2b)(a + 2c)
X Æ
X 2a(a + 2b)(a + 2c)
≥
a (a + 2b)(a + 2c) =
p
(a + 2b) + (a + 2c)
2 (a + 2b)(a + 2c)
X
1
=
a(a + 2b)(a + 2c).
a+b+c
Thus, it suffices to show that
X
a(a + 2b)(a + 2c) ≥ 9(a + b + c).
Write this inequality in the homogeneous form
X
a(a + 2b)(a + 2c) ≥ 3(a + b + c)(a b + bc + ca),
which is equivalent to Schur’s inequality of degree three
a3 + b3 + c 3 + 3a bc ≥ a b(a + b) + bc(b + c) + ca(c + a).
P 2.15. Let a, b, c be nonnegative real numbers such that a + b + c = 1. Prove that
Æ
Æ
Æ
p
a + (b − c)2 + b + (c − a)2 + c + (a − b)2 ≥ 3.
(Phan Thanh Nam, 2007)
Symmetric Nonrational Inequalities
291
Solution. By squaring, the inequality becomes
XÆ
[a + (b − c)2 ][b + (a − c)2 ] ≥ 3(a b + bc + ca).
Applying the Cauchy-Schwarz inequality, it suffices to show that
X
Xp
ab +
(b − c)(a − c) ≥ 3(a b + bc + ca).
This is equivalent to the homogeneous inequality
€X Š €X p Š X
ab +
a2 ≥ 4(a b + bc + ca).
a
Making the substitution x =
€X
p
p
a, y = b, z = c, the inequality turns into
X
Š €X Š X
x2
xy +
x4 ≥ 4
x 2 y 2,
p
which is equivalent to
X
X
X
X
x4 +
x y(x 2 + y 2 ) + x yz
x ≥4
x 2 y 2.
Since
4
X
x2 y2 ≤ 2
X
x y(x 2 + y 2 ),
it suffices to show that
X
x 4 + x yz
X
x≥
X
x y(x 2 + y 2 ),
which is just Schur’s inequality of degree four. The equality holds for a = b = c =
and for a = 0 and b = c =
1
(or any cyclic permutation).
2
1
,
3
P 2.16. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
v
v
v
t a(b + c) t b(c + a) t c(a + b)
+
+
≥ 2.
a2 + bc
b2 + ca
c2 + a b
(Vasile Cîrtoaje, 2006)
Solution. Using the AM-GM inequality gives
v
t a(b + c)
a(b + c)
2a(b + c)
2a(b + c)
=p
≥ 2
=
.
2
a2 + bc
(a
+
bc)
+
(a
b
+
ac)
(a
+ b)(a + c)
(a + bc)(ab + ac)
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Vasile Cîrtoaje
Therefore, it suffices to show that
b(c + a)
c(a + b)
a(b + c)
+
+
≥ 1,
(a + b)(a + c) (b + c)(b + a) (c + a)(c + b)
which is equivalent to
a(b + c)2 + b(c + a)2 + c(a + b)2 ≥ (a + b)(b + c)(c + a),
4a bc ≥ 0.
The equality holds for a = 0 and b = c (or any cyclic permutation).
P 2.17. Let a, b, c be positive real numbers such that a bc = 1. Prove that
p
3
1
a2
+ 25a + 1
+p
3
1
b2
+ 25b + 1
+p
3
1
c2
+ 25c + 1
≥ 1.
Solution. Replacing a, b, c by a3 , b3 , c 3 , respectively, we need to show that a bc = 1
yields
1
1
1
+p
+p
≥ 1.
p
3
3
3
a6 + 25a3 + 1
b6 + 25b3 + 1
c 6 + 25c 3 + 1
We first show that
1
1
≥ 2
.
p
3
6
3
a +a+1
a + 25a + 1
This is equivalent to
(a2 + a + 1)3 ≥ a6 + 25a3 + 1,
which is true since
(a2 + a + 1)3 − (a6 + 25a3 + 1) = 3a(a − 1)2 (a2 + 4a + 1) ≥ 0.
Therefore, it suffices to prove that
1
1
1
+
+
≥ 1.
a2 + a + 1 b2 + b + 1 b2 + b + 1
Putting
a=
yz
,
x2
b=
we need to show that
X
zx
,
y2
c=
xy
,
z2
x, y, z > 0
x4
≥ 1.
x 4 + x 2 yz + y 2 z 2
Symmetric Nonrational Inequalities
293
Indeed, the Cauchy-Schwarz inequality gives
X
P 2 2
P 4
P
x
x + 2 x2 y2
x4
P
P
=P
≥P
x 4 + x 2 yz + y 2 z 2
(x 4 + x 2 yz + y 2 z 2 )
x 4 + x yz x + x 2 y 2
P 2 2
P
x y − x yz x
P
P
=1+ P
≥ 1.
x 4 + x yz x + x 2 y 2
The equality holds for a = b = c = 1.
P 2.18. If a, b, c are nonnegative real numbers, then
p
a2 + bc +
p
b2 + ca +
p
c2 + a b ≤
3
(a + b + c).
2
(Pham Kim Hung, 2005)
Solution. Without loss of generality, assume that a ≥ b ≥ c. Since the equality occurs
for a = b and c = 0, we use the inequalities
p
c
a2 + bc ≤ a +
2
and
p
b2 + ca +
p
c2 + a b ≤
Æ
2(b2 + ca) + 2(c 2 + a b).
Thus, it suffices to prove that
Æ
2(b2 + ca) + 2(c 2 + a b) ≤
a + 3b + 2c
.
2
By squaring, this inequality becomes
a2 + b2 − 4c 2 − 2a b + 12bc − 4ca ≥ 0,
(a − b − 2c)2 + 8c(b − c) ≥ 0.
The equality holds for a = b and c = 0 (or any cyclic permutation).
P 2.19. If a, b, c are nonnegative real numbers, then
p
p
p
p
a2 + 9bc + b2 + 9ca + c 2 + 9a b ≥ 5 a b + bc + ca.
(Vasile Cîrtoaje, 2012)
294
Vasile Cîrtoaje
First Solution (by Nguyen Van Quy). Assume that c = min{a, b, c}. Since the equality
occurs for a = b and c = 0, we use the inequality
p
p
c 2 + 9a b ≥ 3 a b.
Also, by Minkowski’s inequality, we have
p
a2 + 9bc +
p
s
b2 + 9ca ≥
(a + b)2 + 9c
€p
a+
p Š2
b .
Therefore, it suffices to show that
s
€p
p Š2
p
p
(a + b)2 + 9c
a + b ≥ 5 a b + bc + ca − 3 a b.
By squaring, this inequality becomes
p
Æ
(a + b)2 + 18c a b + 30 a b(a b + bc + ca) ≥ 34a b + 16c(a + b).
Since
c(a + b)
a b(a b + bc + ca) − a b +
3
•
˜2
=
c(a + b)(3a b − ac − bc)
≥ 0,
9
p
it suffices to show that f (c) ≥ 0 for 0 ≤ c ≤ a b, where
p
f (c) = (a + b)2 + 18c a b + [30a b + 10c(a + b)] − 34a b − 16c(a + b)
p
= (a + b)2 − 4a b + 6c(3 a b − a − b).
p
Since f (c) is a linear function, we only need to prove that f (0) ≥ 0 and f ( a b) ≥ 0.
We have
f (0) = (a − b)2 ≥ 0,
p
p
p
f ( a b) = (a + b)2 + 14a b − 6(a + b) a b ≥ (a + b)2 + 9a b − 6(a + b) a b
€
p Š2
= a + b − 3 a b ≥ 0.
The equality holds for a = b and c = 0 (or any cyclic permutation).
Second Solution. Assume that c = min{a, b, c}. By squaring, the inequality becomes
X
XÆ
X
a2 + 2
(a2 + 9bc)(b2 + 9ca) ≥ 16
a b,
X
a2 + 2
Æ
(a2 + 9bc)(b2 + 9ca) + 2
p
c 2 + 9a b
€p
a2 + 9bc +
p
The Cauchy-Schwarz inequality gives
p
Æ
(a2 + 9bc)(b2 + 9ca) ≥ a b + 9c a b.
X
Š
b2 + 9ca ≥ 16
a b.
Symmetric Nonrational Inequalities
295
In addition, Minkowski’s inequality gives
s
€p
p Š2
p
p
a2 + 9bc + b2 + 9ca ≥ (a + b)2 + 9c
a + b ≥ a + b + 4c,
hence
p
c 2 + 9a b
€p
a2 + 9bc +
p
Š
p
b2 + 9ca ≥ 3 a b (a + b + 4c).
Therefore, it suffices to show that f (c) ≥ 0, where
X
p
p
f (c) = a2 + b2 + 2(a b + 9c a b) + 6 a b (a + b + 4c) − 16
ab
p
p
= a2 + b2 − 14a b + 6(a + b) a b + c[42 a b − 16(a + b)].
Since fp(c) is a linear function and 0 ≤ c ≤
and f ( a b) ≥ 0. We have
f (0) = (a − b)2 + 6
p
p
a b, it is sufficient to show that f (0) ≥ 0
ab
€p
a−
p Š2
b ≥0
and
p
p
p
f ( a b) = a2 + b2 + 28a b − 10(a + b) a b ≥ (a + b)2 + 25a b − 10(a + b) a b
€
p Š2
= a + b − 5 a b ≥ 0.
P 2.20. If a, b, c are nonnegative real numbers, then
XÆ
(a2 + 4bc)(b2 + 4ca) ≥ 5(a b + ac + bc).
(Vasile Cîrtoaje, 2012)
First Solution (by Michael Rozenberg). Assume that a ≥ b ≥ c. For b = c = 0, the
inequality is trivial. Assume further that b > 0 and write the inequality as follows:
X Æ
[ (b2 + 4ca)(c 2 + 4a b) − bc − 2a(b + c)] ≥ 0,
X (b2 + 4ca)(c 2 + 4a b) − (bc + 2a b + 2ac)2
≥ 0,
p
(b2 + 4ca)(c 2 + 4a b) + bc + 2a(b + c)
X
(b − c)2 Sa ≥ 0,
where
Sa =
Æ
a(b + c − a)
, A = (b2 + 4ca)(c 2 + 4a b) + bc + 2a(b + c),
A
296
Vasile Cîrtoaje
Sb =
b(c + a − b)
,
B
B=
Æ
(c 2 + 4a b)(a2 + 4bc) + ca + 2b(c + a),
Æ
c(a + b − c)
, C = (a2 + 4bc)(b2 + 4ac) + a b + 2c(a + b).
C
Since S b ≥ 0 and Sc ≥ 0, we have
Sc =
X
a2
(b − c)2 Sa ≥ (b − c)2 Sa + (a − c)2 S b ≥ (b − c)2 Sa + 2 (b − c)2 S b
b

‹
bSa aS b
a
= (b − c)2
+
.
b
a
b
Thus, it suffices to prove that
bSa aS b
+
≥ 0,
a
b
which is equivalent to
b(b + c − a) a(c + a − b)
+
≥ 0.
A
B
Since
b(b + c − a) a(c + a − b)
b(b − a) a(a − b) (a − b)(aA − bB)
+
≥
+
=
,
A
B
A
B
AB
it is enough to show that
aA − bB ≥ 0.
Indeed,
aA − bB =
p
” p
—
p
c 2 + 4a b a b2 + 4ca − b a2 + 4bc + 2(a − b)(a b + bc + ca)
p
4c(a3 − b3 ) c 2 + 4a b
= p
+ 2(a − b)(a b + bc + ca) ≥ 0.
p
a b2 + 4ca + b a2 + 4bc
The equality holds for a = b = c, and also for a = b and c = 0 (or any cyclic permutation).
Second Solution (by Nguyen Van Quy). Write the inequality as
€p
p
a2 + 4bc +
a2 + 4bc +
p
p
b2 + 4ca +
b2 + 4ca +
p
p
c 2 + 4a b
Š2
c 2 + 4a b ≥
≥ a2 + b2 + c 2 + 14(a b + bc + ca),
Æ
a2 + b2 + c 2 + 14(a b + bc + ca).
Assume that c = min{a, b, c}. For t = 2c, the inequality (b) in Lemma below becomes
p
a2 + 4bc +
p
b2 + 4ca ≥
Æ
(a + b)2 + 8(a + b)c.
Symmetric Nonrational Inequalities
297
Thus, it suffices to show that
p
Æ
Æ
(a + b)2 + 8(a + b)c + c 2 + 4a b ≥ a2 + b2 + c 2 + 14(a b + bc + ca).
By squaring, this inequality becomes
Æ
[(a + b)2 + 8(a + b)c] (c 2 + 4a b) ≥ 4a b + 3(a + b)c,
2(a + b)c 3 − 2(a + b)2 c 2 + 2a b(a + b)c + a b(a + b)2 − 4a2 b2 ≥ 0,
2(a + b)(a − c)(b − c)c + a b(a − b)2 ≥ 0.
Lemma. Let a,b and t be nonnegative numbers such that
t ≤ 2(a + b).
Then,
p
(a)
(b)
p
(a2 + 2bt)(b2 + 2at) ≥ a b + (a + b)t;
a2 + 2bt +
p
b2 + 2at ≥
p
(a + b)2 + 4(a + b)t.
Proof. (a) By squaring, the inequality becomes
(a − b)2 t[2(a + b) − t] ≥ 0,
which is clearly true.
(b) By squaring, this inequality turns into the inequality in (a).
P 2.21. If a, b, c are nonnegative real numbers, then
XÆ
(a2 + 9bc)(b2 + 9ca) ≥ 7(a b + ac + bc).
(Vasile Cîrtoaje, 2012)
Solution (by Nguyen Van Quy). We see that the equality holds for a = b and c = 0.
Without loss of generality, assume that c = min{a, b, c}. For t = 4c, the inequality (a)
in Lemma from the preceding P 2.20 becomes
Æ
(a2 + 8bc)(b2 + 8ca) ≥ a b + 4(a + b)c.
Thus, we have
Æ
(a2 + 9bc)(b2 + 9ca) ≥ a b + 4(a + b)c,
298
Vasile Cîrtoaje
and also,
p
c 2 + 9a b
€p
a2 + 9bc +
p
Š
p
Æ
4
b2 + 9ca ≥ 3 a b · 2 (a2 + 9bc)(b2 + 9ca)
p
Æ
Æ
≥ 6 a b · a b + 4(a + b)c = 3 4a2 b2 + 16a bc(a + b)
Æ
≥ 3 4a2 b2 + 4a bc(a + b) + c 2 (a + b)2 = 3(2a b + bc + ca).
Therefore,
XÆ
(a2 + 9bc)(b2 + 9ca) ≥ (a b + 4bc + 4ca) + 3(2a b + bc + ca)
= 7(a b + bc + ca).
The equality holds for a = b and c = 0 (or any cyclic permutation).
P 2.22. If a, b, c are nonnegative real numbers, then
Æ
Æ
Æ
(a2 + b2 )(b2 + c 2 ) + (b2 + c 2 )(c 2 + a2 ) + (c 2 + a2 )(a2 + b2 ) ≤ (a + b + c)2 .
(Vasile Cîrtoaje, 2007)
Solution. Without loss of generality, assume that a = min{a, b, c}. Let us denote
y=
a
+ b,
2
z=
a
+ c.
2
Since
a2 + b2 ≤ y 2 ,
b2 + c 2 ≤ y 2 + z 2 ,
c 2 + a2 ≤ z 2 ,
it suffices to prove that
Æ
yz + ( y + z) y 2 + z 2 ≤ ( y + z)2 .
This is true since
Æ
y 2 + yz + z 2 − ( y + z) y 2 + z 2 =
y 2z2
≥ 0.
p
y 2 + yz + z 2 + ( y + z) y 2 + z 2
The equality holds for a = b = 0 (or any cyclic permutation).
P 2.23. If a, b, c are nonnegative real numbers, then
XÆ
(a2 + a b + b2 )(b2 + bc + c 2 ) ≥ (a + b + c)2 .
Symmetric Nonrational Inequalities
299
Solution. By the Cauchy-Schwarz inequality, we have

‹
c 2 3c 2
b 2 3b2 2
2
2
2
+
a+
+
(a + a b + b )(a + ac + c ) = a +
2
4
2
4

‹
c 3bc
b
a(b + c)
a+
≥ a+
+
= a2 +
+ bc.
2
2
4
2
Then,
˜
XÆ
X•
a(b + c)
2
2
2
2
2
(a + a b + b )(a + ac + c ) ≥
a +
+ bc = (a + b + c)2 .
2
The equality holds for a = b = c, and also for b = c = 0 (or any cyclic permutation).
P 2.24. If a, b, c are nonnegative real numbers, then
XÆ
(a2 + 7a b + b2 )(b2 + 7bc + c 2 ) ≥ 7(a b + ac + bc).
(Vasile Cîrtoaje, 2012)
First Solution. Without loss of generality, assume that c = min{a, b, c}. We see that the
equality holds for a = b and c = 0. Since
Æ
(a2 + 7ac + c 2 )(b2 + 7bc + c 2 ) ≥ (a + 2c)(b + 2c) ≥ a b + 2c(a + b),
it suffices to show that
€p
Š
p
p
a2 + 7a b + b2
a2 + 7ac + b2 + 7bc ≥ 6a b + 5c(a + b).
By Minkowski’s inequality, we have
p
a2 + 7ac +
≥
p
v
t
s
b2 + 7bc ≥
(a + b)2 + 7c
(a + b)2 + 7c(a + b) +
€p
a+
p Š2
b
28a bc
.
a+b
Therefore, it suffices to show that
˜
•
28a bc
2
2
2
(a + 7a b + b ) (a + b) + 7c(a + b) +
≥ (6a b + 5bc + 5ca)2 .
a+b
Due to homogeneity, we may assume that a + b = 1. Let us denote d = a b, d ≤
Since
c≤
2a b
= 2d,
a+b
1
.
4
300
Vasile Cîrtoaje
we need to show that f (c) ≥ 0 for 0 ≤ c ≤ 2d ≤
1
, where
2
f (c) = (1 + 5d)(1 + 7c + 28cd) − (6d + 5c)2 .
Since f (c) is concave, it suffices to show that f (0) ≥ 0 and f (2d) ≥ 0. Indeed,
f (0) = 1 + 5d − 36d 2 = (1 − 4d)(1 + 9d) ≥ 0
and
f (2d) = (1 + 5d)(1 + 14d + 56d 2 ) − 256d 2 ≥ (1 + 4d)(1 + 14d + 56d 2 ) − 256d 2
= (1 − 4d)(1 + 22d − 56d 2 ) ≥ d(1 − 4d)(22 − 56d) ≥ 0.
The equality holds for a = b and c = 0 (or any cyclic permutation).
Second Solution. We will use the inequality
Æ
x 2 + 7x y + y 2 ≥ x + y +
2x y
,
x+y
x, y ≥ 0,
which, by squaring, reduces to
x y(x − y)2 ≥ 0.
We have
XÆ
(a2
+ 7a b +
≥
Since
X
b2 )(a2
a2 + 3
+ 7ac
X
ab +
+ c2)
≥
X
2a b
a+b+
a+b
‹
2ac
a+c+
a+c
X 2a2 b X 2a2 c X 2a bc
+
+
.
a+b
a+c
a+b
X 2a2 b X 2a2 c X 2a2 b X 2b2 a
X
+
=
+
=2
ab
a+b
a+c
a+b
b+a
and
X 2a bc
18a bc
9a bc
≥P
,
=
a+b
a+b+c
(a + b)
it suffices to show that
X
a2 +
X
9a bc
≥2
a b,
a+b+c
which is just Schur’s inequality of degree three.
‹
Symmetric Nonrational Inequalities
301
P 2.25. If a, b, c are nonnegative real numbers, then
v
‹
‹
X t
7
7
13
a2 + a b + b2
b2 + bc + c 2 ≤
(a + b + c)2 .
9
9
12
(Vasile Cîrtoaje, 2012)
Solution (by Nguyen Van Quy). Without loss of generality, assume that c = min{a, b, c}.
It is easy to see that the equality holds for a = b = 1 and c = 0. By the AM-GM inequality,
the following inequality holds for any k > 0:
‚v
Œ
v
v
t
t
t
7
7
7
2
2
2
2
2
2
12 a + a b + b
a + ac + c + b + bc + c ≤
9
9
9
‚v
Œ2
v

‹
t
t
36 2 7
7
7
≤
a + a b + b2 + k
a2 + ac + c 2 + b2 + bc + c 2 .
k
9
9
9
We can use this inequality to prove the original inequality only if
‚v
Œ2
v

‹
t
t
7
7
36 2 7
2
a + ab + b = k
a2 + ac + c 2 + b2 + bc + c 2
k
9
9
9
for a = b = 1 and c = 0. This necessary condition if satisfied for k = 5. Therefore, it
suffices to show that
v
‹
‹
‹

t
7
7
36 2 7
2
2
2
2
2
a + ac + c
b + bc + c +
a + ab + b
12
9
9
5
9
‚v
Œ2
v
t
t
7
7
+5
a2 + ac + c 2 + b2 + bc + c 2 ≤ 13(a + b + c)2 .
9
9
which is equivalent to
v
‹
‹
t
7
7
4(a + b)2 + 94a b
199c(a + b)
2
2
2
2
b + bc + c ≤
+ 3c 2 +
.
22
a + ac + c
9
9
5
9
Since
2
v
t
7
a2 + ac + c 2
9
‹
v
‹
‹
‹
t
16
16
7
2
2
2
2
b + bc + c ≤ 2
a +
ac
b +
bc
9
9
9
v 
‹ 
‹
t
16
16
=2 a b+
c ·b a+
c
9
9

‹

‹
16
16
≤a b+
c +b a+
c
9
9
16c(a + b)
= 2a b +
,
9
302
Vasile Cîrtoaje
we only need to prove that
˜
•
4(a2 + b2 ) + 102a b
199c(a + b)
8c(a + b)
≤
+ 3c 2 +
.
22 a b +
9
5
9
This reduces to the obvious inequality
4(a − b)2 23c(a + b)
+
+ 3c 2 ≥ 0.
5
9
Thus, the proof is completed. The equality holds for a = b and c = 0 (or any cyclic
permutation).
P 2.26. If a, b, c are nonnegative real numbers, then
v
‹
‹
X t
1
1
61
2
2
2
2
a + ab + b
b + bc + c ≤
(a + b + c)2 .
3
3
60
(Vasile Cîrtoaje, 2012)
Solution (by Nguyen Van Quy). Without loss of generality, assume that c = min{a, b, c}.
It is easy to see that the equality holds for c = 0 and 11(a2 + b2 ) = 38a b. By the AM-GM
inequality, the following inequality holds for any k > 0:
v
v
‚v
Œ
t
t
t
1
1
1
60 a2 + a b + b2
a2 + ac + c 2 + b2 + bc + c 2 ≤
3
3
3
v
‚v
Œ2
‹

t
t
1
1
36 2 1
2
2
2
2
2
a + a b + b + 25k
a + ac + c + b + bc + c
.
≤
k
3
3
3
We can use this inequality to prove the original inequality only if the equality
v
Œ2
‚v

‹
t
t
36 2 1
1
1
2
a + a b + b = 25k
a2 + ac + c 2 + b2 + bc + c 2
k
3
3
3
holds for c = 0 and 11(a2 + b2 ) = 38a b. This necessary condition if satisfied for k = 1.
Therefore, it suffices to show that
v
‹
‹

‹
t
1
1
1
2
2
2
2
2
2
60
a + ab + b
b + bc + c + 36 a + a b + b
3
3
3
+25
‚v
t
v
Œ2
t
1
1
a2 + ac + c 2 + b2 + bc + c 2 ≤ 61(a + b + c)2 ,
3
3
Symmetric Nonrational Inequalities
303
which is equivalent to
v
‹
‹
t
1
31c(a + b)
1
2
2
2
2
b + bc + c ≤ 10a b + c 2 +
.
10
a + ac + c
3
3
3
Since
2
v
t
1
a2 + ac + c 2
3
‹
v
‹
‹
‹
t
1
4
4
b2 + bc + c 2 ≤ 2
a2 + ac
b2 + bc
3
3
3
v 
‹

‹
t
4
4
=2 a b+ c ·b a+ c
3
3

‹

‹
4
4
≤a b+ c +b a+ c
3
3
4c(a + b)
= 2a b +
,
3
we only need to prove that
•
˜
2c(a + b)
31c(a + b)
10 a b +
≤ 10a b + c 2 +
.
3
3
This reduces to the obvious inequality
3c 2 + 11c(a + b) ≥ 0.
Thus, the proof is completed. The equality holds for 11(a2 + b2 ) = 38a b and c = 0 (or
any cyclic permutation).
P 2.27. If a, b, c are nonnegative real numbers, then
a
b
c
+p
+p
≥ 1.
p
2
2
2
2
2
4b + bc + 4c
4c + ca + 4a
4a + a b + 4b2
(Pham Kim Hung, 2006)
Solution. By Hölder’s inequality, we have
P 3
P 3
P
‹2
X
a
a + 3 a b(a + b) + 6a bc
a
P
≥P
=
.
p
a(4b2 + bc + 4c 2 )
4 a b(a + b) + 3a bc
4b2 + bc + 4c 2
Thus, it suffices to show that
X
a3 + 3a bc ≥
X
a b(a + b),
which is Schur’s inequality of degree three. The equality holds for a = b = c, and also
for a = 0 and b = c (or any cyclic permutation).
304
Vasile Cîrtoaje
P 2.28. If a, b, c are nonnegative real numbers, then
p
a
b2
+ bc
+ c2
+p
b
c2
+ ca + a2
+p
c
a2
+ ab +
b2
≥p
a+b+c
a b + bc + ca
.
Solution. By Hölder’s inequality, we have
X
p
‹2
a
b2 + bc + c 2
P 2
a
= P
,
≥P
a(b2 + bc + c 2 )
ab
P 3
a
from which the desired inequality follows. The equality holds for a = b = c, and also
for a = 0 and b = c (or any cyclic permutation).
P 2.29. If a, b, c are nonnegative real numbers, then
p
a
a2
+ 2bc
+p
b
b2
+ 2ca
+p
c
c2
+ 2a b
≤p
a+b+c
a b + bc + ca
.
(Ho Phu Thai, 2007)
Solution. Without loss of generality, assume that a ≥ b ≥ c.
First Solution. Since
p
c
c2
+ 2a b
≤p
c
a b + bc + ca
,
it suffices to show that
p
a
a2
+ 2bc
+p
b
b2
+ 2ca
≤p
a+b
a b + bc + ca
,
which is equivalent to
p
p
p
p
a( a2 + 2bc − a b + bc + ca)
b( a b + bc + ca − b2 + 2ca)
≥
.
p
p
a2 + 2bc
b2 + 2ca
Since
p
a2 + 2bc −
and
p
p
a
a2 + 2bc
a b + bc + ca ≥ 0
≥p
b
b2 + 2ca
,
it suffices to show that
p
p
p
p
a2 + 2bc − a b + bc + ca ≥ a b + bc + ca − b2 + 2ca,
Symmetric Nonrational Inequalities
which is equivalent to
p
a2 + 2bc +
305
p
b2 + 2ca ≥ 2
p
a b + bc + ca.
Using the AM-GM inequality, it suffices to show that
(a2 + 2bc)(b2 + 2ca) ≥ (a b + bc + ca)2 ,
which is equivalent to the obvious inequality
c(a − b)2 (2a + 2b − c) ≥ 0.
The equality holds for a = b = c, and also for a = b and c = 0 (or any cyclic permutation).
Second Solution. By the Cauchy-Schwarz inequality, we have
X
p
‹2
a
a2 + 2bc
≤
€X Š X
a
a
.
a2 + 2bc
Thus, it suffices to prove that
X
a2
a+b+c
a
≤
.
+ 2bc
a b + bc + ca
This is equivalent to
X 
a
1
1
− 2
a b + bc + ca a + 2bc
X a(a − b)(a − c)
a2 + 2bc
‹
≥ 0,
≥ 0.
Assuming that a ≥ b ≥ c, we get
X a(a − b)(a − c)
a2 + 2bc
=
≥
a(a − b)(a − c) b(b − c)(b − a)
+
a2 + 2bc
b2 + 2ca
c(a − b)2 [2a(a − c) + 2b(b − c) + 3a b]
≥ 0.
(a2 + 2bc)(b2 + 2ca)
P 2.30. If a, b, c are nonnegative real numbers, then
p
p
p
a3 + b3 + c 3 + 3a bc ≥ a2 a2 + 3bc + b2 b2 + 3ca + c 2 c 2 + 3a b.
(Vo Quoc Ba Can, 2008)
306
Vasile Cîrtoaje
Solution. If a = 0, then the inequality is an identity. Consider further that a, b, c > 0,
and write the inequality as follows:
X
p
a2 ( a2 + 3bc − a) ≤ 3a bc,
X
3a2 bc
≤ 3a bc,
a2 + 3bc + a
X
1
≤ 1.
Ç
+
1
1 + 3bc
2
a
p
Let us denote
x=Ç
1
1+
3bc
a2
,
+1
y=q
1
1+
3ca
b2
+1
, z=Ç
1
1+
3ab
c2
,
+1
which implies
bc
1 − 2x
=
,
2
a
3x 2
1 − 2y
ca
=
,
2
b
3 y2
ab
1 − 2z
1
=
, 0 < x, y, z < ,
2
2
c
3z
2
and
(1 − 2x)(1 − 2 y)(1 − 2z) = 27x 2 y 2 z 2 .
We need to prove that
x + y +z ≤1
1
such that (1 − 2x)(1 − 2 y)(1 − 2z) = 27x 2 y 2 z 2 . We will use the
2
1
contradiction method. Assume that x + y + z > 1 for 0 < x, y, z < , and show that
2
(1 − 2x)(1 − 2 y)(1 − 2z) < 27x 2 y 2 z 2 . We have
for 0 < x, y, z <
(1 − 2x)(1 − 2 y)(1 − 2z) < (x + y + z − 2x)(x + y + z − 2 y)(x + y + z − 2z)
< ( y + z − x)(z + x − y)(x + y − z)(x + y + z)3
≤ 3( y + z − x)(z + x − y)(x + y − z)(x + y + z)(x 2 + y 2 + z 2 )
= 3(2x 2 y 2 + 2 y 2 z 2 + 2z 2 x 2 − x 4 − y 4 − z 4 )(x 2 + y 2 + z 2 ).
Therefore, it suffices to show that
(2x 2 y 2 + 2 y 2 z 2 + 2z 2 x 2 − x 4 − y 4 − z 4 )(x 2 + y 2 + z 2 ) ≤ 9x 2 y 2 z 2 ,
which is equivalent to
x 6 + y 6 + z 5 + 9x 2 y 2 z 2 ≥
X
y 2 z 2 ( y 2 + z 2 ).
Clearly, this is just Schur’s inequality of degree three applied to x 2 , y 2 , z 2 . So, the proof
is completed. The equality holds for a = b = c, and also for a = 0 or b = 0 or c = 0.
Symmetric Nonrational Inequalities
307
P 2.31. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
b
c
a
+p
+p
≤ 1.
p
4a2 + 5bc
4b2 + 5ca
4c 2 + 5a b
(Vasile Cîrtoaje, 2004)
First Solution (by Vo Quoc Ba Can). If one of a, b, c is zero, then the inequality becomes
an equality. Consider next that a, b, c > 0 and denote

‹
a
b
c
1
x=p
, y=p
, z=p
, x, y, z ∈ 0,
.
2
4a2 + 5bc
4b2 + 5ca
4c 2 + 5a b
We have
bc
1 − 4x 2
=
,
a2
5x 2
1 − 4 y2
ca
=
,
b2
5 y2
ab
1 − 4z 2
=
,
c2
5z 2
and
(1 − 4x 2 )(1 − 4 y 2 )(1 − 4z 2 ) = 125x 2 y 2 z 2 .
For the sake of contradiction, assume that x + y + z > 1. Using the AM-GM inequality
and the Cauchy-Schwarz inequality, we have
1 Y
1 Y
(1 − 4x 2 ) <
[(x + y + z)2 − 4x 2 ]
125
125
Y
1 Y
=
(3x + y + z) ·
( y + z − x)
125
x + y + z 3 Y
≤
( y + z − x)
3
Y
1
≤ (x 2 + y 2 + z 2 )(x + y + z)
( y + z − x)
9
1
= (x 2 + y 2 + z 2 )[2(x 2 y 2 + y 2 z 2 + z 2 x 2 ) − x 4 − y 4 − z 4 ],
9
x 2 y 2z2 =
and hence
9x 2 y 2 z 2 < (x 2 + y 2 + z 2 )[2(x 2 y 2 + y 2 z 2 + z 2 x 2 ) − x 4 − y 4 − z 4 ],
X
x 6 + y 6 + z 6 + 3x 2 y 2 z 2 <
x 2 y 2 (x 2 + y 2 ).
The last inequality contradicts Schur’s inequality
X
x 6 + y 6 + z 6 + 3x 2 y 2 z 2 ≥
x 2 y 2 (x 2 + y 2 ).
Thus, the proof is completed. The equality holds for a = b = c, and also for a = 0 and
b = c (or any cyclic permutation).
308
Vasile Cîrtoaje
Second Solution. In the nontrivial case when a, b, c > 0, setting x =
z=
ca
bc
, y = 2 and
2
a
b
ab
(x yz = 1), the desired inequality becomes E(x, y, z) ≤ 1, where
c2
1
1
1
+p
.
E(x, y, z) = p
+p
4 + 5x
4 + 5z
4 + 5y
Without loss of generality, we may assume that x ≥ y ≥ z, x ≥ 1, yz ≤ 1. We will prove
that
p
p
E(x, y, z) ≤ E(x, yz, yz) ≤ 1.
The left inequality has the form
1
p
1
1
≤p
+p
p .
4 + 5z
4 + 5y
4 + 5 yz
For the nontrivial case y 6= z, consider y > z and denote
y +z
p
, p = yz,
2
Æ
q = (4 + 5 y)(4 + 5z).
s=
We have s > p, p ≤ 1 and
Æ
Æ
q = 16 + 40s + 25p2 > 16 + 40p + 25p2 = 4 + 5p.
By squaring, the desired inequality becomes in succession as follows:
1
2
4
1
+
+ ≤
,
4 + 5 y 4 + 5z q
4 + 5p
1
1
2
2
2
+
−
≤
− ,
4 + 5 y 4 + 5z 4 + 5p
4 + 5p q
2(q − 4 − 5p)
8 + 10s
2
−
≤
,
2
q
4 + 5p
q(4 + 5p)
(s − p)(5p − 4)
8(s − p)
≤
,
2
q (4 + 5p)
q(4 + 5p)(q + 4 + 5p)
5p − 4
8
≤
,
q
q + 4 + 5p
25p2 − 16 ≤ (12 − 5p)q.
The last inequality is true since
(12 − 5p)q − 25p2 + 16 > (12 − 5p)(4 + 5p) − 25p2 + 16 = 2(8 − 5p)(4 + 5p) > 0.
Symmetric Nonrational Inequalities
309
In order to prove the right inequality, namely
1
2
≤ 1,
+p
p
p
4 + 5x
4 + 5 yz
let us denote
p
p
4 + 5 yz = 3t, t ∈ (2/3, 1]. Since
x=
25
1
=
,
2
yz
(9t − 4)2
the inequality becomes
9t 2 − 4
2
+
≤ 1,
p
3 36t 4 − 32t 2 + 21 3t
€p
Š
(2 − 3t)
36t 4 − 32t 2 + 21 − 3t 2 − 2t ≤ 0.
Since 2 − 3t < 0, we still have to show that
p
36t 4 − 32t 2 + 21 ≥ 3t 2 + 2t.
Indeed,
p
3(t − 1)2 (9t 2 + 14t + 7)
36t 4 − 32t 2 + 21 − 3t 2 − 2t = p
≥ 0.
36t 4 − 32t 2 + 21 + 3t 2 + 2t
P 2.32. Let a, b, c be nonnegative real numbers. Prove that
p
p
p
a 4a2 + 5bc + b 4b2 + 5ca + c 4c 2 + 5a b ≥ (a + b + c)2 .
(Vasile Cîrtoaje, 2004)
First Solution. Write the inequality as
X €p
Š
a
4a2 + 5bc − 2a ≥ 2(a b + bc + ca) − a2 − b2 − c 2 ,
1
≥ 2(a b + bc + ca) − a2 − b2 − c 2 .
p
2
4a + 5bc + 2a
Writing Schur’s inequality
X
a3 + b3 + c 3 + 3a bc ≥
a b(a2 + b2 )
5a bc
in the form
X
9a bc
≥ 2(a b + bc + ca) − a2 − b2 − c 2 ,
a+b+c
310
Vasile Cîrtoaje
it suffices to prove that
X
5
9
≥
.
p
2
a+b+c
4a + 5bc + 2a
Let p = a + b + c and q = a b + bc + ca. By the AM-GM inequality, we have
p
p
2
(16a2 + 20bc)(3b + 3c)2
(16a2 + 20bc) + (3b + 3c)2
4a2 + 5bc =
≤
12(b + c)
12(b + c)
16a2 + 16bc + 10(b + c)2
8a2 + 5b2 + 5c 2 + 18bc
=
,
12(b + c)
6(b + c)
≤
hence
X
5
5
≥
p
2
2
2
8a + 5b + 5c 2 + 18bc
4a + 5bc + 2a
+ 2a
6(b + c)
X
X
30(b + c)
30(b + c)
=
.
=
2
2
2
2
8a + 5b + 5c + 12a b + 18bc + 12ac
5p + 2q + 3a2 + 6bc
X
Thus, it suffices to show that
X
30(b + c)
9
≥ .
5p2 + 2q + 3a2 + 6bc
p
By the Cauchy-Schwarz inequality, we get
X
P
30[ (b + c)]2
30(b + c)
≥P
5p2 + 2q + 3a2 + 6bc
(b + c)(5p2 + 2q + 3a2 + 6bc)
=
120p2
120p2
P
=
.
10p3 + 4pq + 9 bc(b + c) 10p3 + 13pq − 27a bc
Therefore, it is enough to show that
120p2
9
≥ ,
3
10p + 13pq − 27a bc
p
which is equivalent to
10p3 + 81a bc ≥ 39pq.
From Schur’s inequality p3 + 9a bc ≥ 4pq and the known inequality pq ≥ 9a bc, we have
10p3 + 81a bc − 39pq = 10(p3 + 9a bc − 4pq) + pq − 9a bc ≥ 0.
This completes the proof. The equality holds for a = b = c, and also for a = 0 and b = c
(or any cyclic permutation).
Symmetric Nonrational Inequalities
311
Second Solution. By the Cauchy-Schwarz inequality, we have
‹
Š X
€X p
a
2
≥ (a + b + c)2 .
a 4a + 5bc
p
4a2 + 5bc
From this inequality and the inequality in P 2.31, namely
X
a
≤ 1,
p
2
4a + 5bc
the desired inequality follows.
Remark. Using the same way as in the second solution, we can prove the following
inequalities for a, b, c > 0 satisfying a bc = 1:
p
p
p
a 4a2 + 5 + b 4b2 + 5 + c 4c 2 + 5 ≥ (a + b + c)2 ;
p
p
p
4a4 + 5 + 4b4 + 5 + 4c 4 + 5 ≥ (a + b + c)2 .
The first inequality follows from the the Cauchy-Schwarz inequality
‹
€X p
Š X
a
2
a 4a + 5
≥ (a + b + c)2
p
4a2 + 5
and the inequality
X
a
≤ 1,
p
4a2 + 5
a bc = 1,
which follows from the inequality in P 2.31 by replacing bc/a2 , ca/b2 , a b/c 2 with 1/x 2 ,
1/ y 2 , 1/z 2 , respectively.
The second inequality follows from the the Cauchy-Schwarz inequality
2
€X p
Š X
a
4a4 + 5
≥ (a + b + c)2
p
4
4a + 5
and the inequality
X
a2
≤ 1,
p
4a4 + 5
a bc = 1.
P 2.33. Let a, b, c be nonnegative real numbers. Prove that
p
p
p
a a2 + 3bc + b b2 + 3ca + c c 2 + 3a b ≥ 2(a b + bc + ca).
(Vasile Cîrtoaje, 2005)
312
Vasile Cîrtoaje
First Solution (by Vo Quoc Ba Can). Using the AM-GM inequality yields
X a(b + c)(a2 + 3bc)
X p
a a2 + 3bc =
p
(b + c)2 (a2 + 3bc)
X 2a(b + c)(a2 + 3bc)
≥
.
(b + c)2 + (a2 + 3bc)
Thus, it suffices to prove the inequality
X 2a(b + c)(a2 + 3bc)
a2 + b2 + c 2 + 5bc
≥
X
a(b + c),
which can be written as follows
X a(b + c)(a2 − b2 − c 2 + bc)
a2 + b2 + c 2 + 5bc
X a3 (b + c) − a(b3 + c 3 )
a2 + b2 + c 2 + 5bc
≥ 0,
≥ 0,
X a b(a2 − b2 ) − ac(c 2 − a2 )
≥ 0,
a2 + b2 + c 2 + 5bc
X
X
a b(a2 − b2 )
ba(a2 − b2 )
−
≥ 0,
a2 + b2 + c 2 + 5bc
b2 + c 2 + a2 + 5ca
X
5a bc(a2 − b2 )(a − b)
≥ 0.
(a2 + b2 + c 2 + 5bc)(a2 + b2 + c 2 + 5ac)
Since the last inequality is clearly true, the proof is completed. The equality holds a =
b = c, and also for a = 0 and b = c (or any cyclic permutation).
Second Solution. Write the inequality as
X p
(a a2 + 3bc − a2 ) ≥ 2(a b + bc + ca) − a2 − b2 − c 2 .
Due to homogeneity, we may assume that a + b + c = 3. By the AM-GM inequality, we
have
a
p
a2 + 3bc − a2 = p
3a bc
a2
12a bc
= p
+ 3bc + a
2 4(a2 + 3bc) + 4a
12a bc
≥
.
4 + a2 + 3bc + 4a
Thus, it suffices to show that
12a bc
X
1
≥ 2(a b + bc + ca) − a2 − b2 − c 2 .
4 + a2 + 3bc + 4a
Symmetric Nonrational Inequalities
313
By the Cauchy-Schwarz inequality, we have
X
9
1
≥P
2
2
4 + a + 3bc + 4a
(4 + a + 3bc + 4a)
9
P
P
=
2
24 + a + 3 a b
27
P
P
= P
2
8( a) + 3 a2 + 9 a b
P
9 a
3
P
P
=
≥ P .
2
11( a) + 3 a b
4 a
Then, it remains to show that
9a bc
≥ 2(a b + bc + ca) − a2 − b2 − c 2 ,
a+b+c
which is equivalent to Schur’s inequality of degree three
X
X
a3 + 3a bc ≥
a b(a + b).
P 2.34. Let a, b, c be nonnegative real numbers. Prove that
p
p
p
a a2 + 8bc + b b2 + 8ca + c c 2 + 8a b ≤ (a + b + c)2 .
Solution. Multiplying by a + b + c, the inequality becomes
X Æ
a (a + b + c)2 (a2 + 8bc) ≤ (a + b + c)3 .
Since
2
Æ
(a + b + c)2 (a2 + 8bc) ≤ (a + b + c)2 + (a2 + 8bc),
it suffices to show that
X
a[(a + b + c)2 + (a2 + 8bc)] ≤ 2(a + b + c)3 ,
which can be written as
a3 + b3 + c 3 + 24a bc ≤ (a + b + c)3 .
This inequality is equivalent to
a(b − c)2 + b(c − a)2 + c(a − b)2 ≥ 0.
The equality holds for a = b = c, and also for b = c = 0 (or any cyclic permutation).
314
Vasile Cîrtoaje
P 2.35. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
p
a2 + 2bc
b2 + bc + c 2
+p
b2 + 2ca
c 2 + ca + a2
+p
c 2 + 2a b
a2 + a b + b2
≥3
p
a b + bc + ca.
(Michael Rozenberg and Marius Stanean, 2011)
Solution. By the AM-GM inequality, we have
X
p
2(a2 + 2bc) a b + bc + ca
=
p
p
b2 + bc + c 2
2 (b2 + bc + c 2 )(a b + bc + ca)
X
p
2(a2 + 2bc)
≥ ab + bc + ca
(b2 + bc + c 2 ) + (a b + bc + ca)
X
p
2(a2 + 2bc)
= ab + bc + ca
.
(b + c)(a + b + c)
a2 + 2bc
X
Thus, it suffices to show that
3
a2 + 2bc b2 + 2ca c 2 + 2a b
+
+
≥ (a + b + c).
b+c
c+a
a+b
2
This inequality is equivalent to
a4 + b4 + c 4 + a bc(a + b + c) ≥
1X
a b(a + b)2 .
2
We can prove this inequality by summing Schur’s inequality of fourth degree
X
a4 + b4 + c 4 + a bc(a + b + c) ≥
a b(a2 + b2 )
and the obvious inequality
X
a b(a2 + b2 ) ≥
1X
a b(a + b)2 .
2
The equality holds for a = b = c.
P 2.36. Let a, b, c be nonnegative real numbers, no two of which are zero. If k ≥ 1, then
a k+1
b k+1
c k+1
ak + bk + c k
+
+
≤
.
2a2 + bc 2b2 + ca 2c 2 + a b
a+b+c
(Vasile Cîrtoaje and Vo Quoc Ba Can, 2011)
Symmetric Nonrational Inequalities
315
Solution. Write the inequality as follows
X
a k+1
ak
− 2
a + b + c 2a + bc
X a k (a − b)(a − c)
2a2 + bc
≥ 0,
≥ 0.
Assume that a ≥ b ≥ c. Since (c − a)(c − b) ≥ 0, it suffices to show that
a k (a − b)(a − c) b k (b − a)(b − c)
+
≥ 0.
2a2 + bc
2b2 + ca
This is true if
a k (a − c) b k (b − c)
−
≥ 0,
2a2 + bc
2b2 + ca
which is equivalent to
a k (a − c)(2b2 + ca) ≥ b k (b − c)(2a2 + bc).
Since a k /b k ≥ a/b, it remains to show that
a(a − c)(2b2 + ca) ≥ b(b − c)(2a2 + bc),
which is equivalent to the obvious inequality
(a − b)c[a2 + 3a b + b2 − c(a + b)] ≥ 0.
The equality holds for a = b = c, and also for a = b and c = 0 (or any cyclic permutation).
P 2.37. If a, b, c are positive real numbers, then
a2 − bc
b2 − ca
c2 − a b
+p
+p
≥ 0;
p
3a2 + 2bc
3b2 + 2ca
3c 2 + 2a b
(a)
(b)
a2 − bc
p
8a2 + (b + c)2
+p
b2 − ca
8b2 + (c + a)2
+p
c2 − a b
8c 2 + (a + b)2
≥ 0.
(Vasile Cîrtoaje, 2006)
316
Vasile Cîrtoaje
Solution. (a) Let
p
A=
3a2 + 2bc,
B=
p
3b2 + 2ca,
C=
p
3c 2 + 2a b.
We have
2
X a2 − bc
A
=
=
X (a − b)(a + c) + (a − c)(a + b)
X (a − b)(a + c)
A
+
X (b − a)(b + c)
A
B
‹
X
a+c b+c
=
(a − b)
−
A
B
X a − b (a + c)2 B 2 − (b + c)2 A2
·
=
AB
(a + c)B + (b + c)A
X c(a − b)2 2(a − b)2 + c(a + b + 2c)
·
≥ 0.
=
AB
(a + c)B + (b + c)A
The equality holds for a = b = c.
(b) Let
A=
Æ
8a2 + (b + c)2 ,
B=
Æ
8b2 + (c + a)2 ,
C=
Æ
8c 2 + (a + b)2 b.
As we have shown before,
2
hence
2
X a2 − bc
A
X a2 − bc
A
=
=
X a − b (a + c)2 B 2 − (b + c)2 A2
·
,
AB
(a + c)B + (b + c)A
X (a − b)2
AB
·
C1
≥ 0,
(a + c)B + (b + c)A
since
C1 = [(a + c) + (b + c)][(a + c)2 + (b + c)2 ] − 8ac(b + c) − 8bc(a + c)
≥ [(a + c) + (b + c)](4ac + 4bc) − 8ac(b + c) − 8bc(a + c)
= 4c(a − b)2 ≥ 0.
The equality holds for a = b = c.
p
P 2.38. Let a, b, c be positive real numbers. If 0 ≤ k ≤ 1 + 2 2, then
p
a2 − bc
ka2 + b2 + c 2
+p
b2 − ca
k b2 + c 2 + a2
+p
c2 − a b
kc 2 + a2 + b2
≥ 0.
Symmetric Nonrational Inequalities
317
Solution. Let
A=
p
ka2 + b2 + c 2 ,
B=
p
k b2 + c 2 + a2 ,
C=
p
kc 2 + a2 + b2 .
As we have shown at the preceding problem,
2
therefore
2
X a2 − bc
A
X a2 − bc
A
=
=
X a − b (a + c)2 B 2 − (b + c)2 A2
·
,
AB
(a + c)B + (b + c)A
X (a − b)2
AB
·
C1
≥ 0,
(a + c)B + (b + c)A
where
C1 = (a2 + b2 + c 2 )(a + b + 2c) − (k − 1)c(2a b + bc + ca)
p
≥ (a2 + b2 + c 2 )(a + b + 2c) − 2 2 c(2a b + bc + ca).
Putting a + b = 2x, we have a2 + b2 ≥ 2x 2 , a b ≤ x 2 , and hence
p
p
C1 ≥ (2x 2 + c 2 )(2x + 2c) − 2 2 c(2x 2 + 2c x) = 2(x + c)(x 2 − c)2 ≥ 0.
The equality holds for a = b = c.
P 2.39. If a, b, c are nonnegative real numbers, then
p
p
p
(a2 − bc) b + c + (b2 − ca) c + a + (c 2 − a b) a + b ≥ 0.
First Solution. Let us denote
v
tb+c
x=
,
2
y=
s
v
ta + b
c+a
, z=
,
2
2
hence
a = y 2 + z2 − x 2,
b = z2 + x 2 − y 2,
c = x 2 + y 2 − z2.
The inequality turns into
x y(x 3 + y 3 ) + yz( y 3 + z 3 ) + z x(z 3 + x 3 ) ≥ x 2 y 2 (x + y) + y 2 z 2 ( y + z) + z 2 x 2 (z + x),
which is equivalent to the obvious inequality
x y(x + y)(x − y)2 + yz( y + z)( y − z)2 + z x(z + x)(z − x)2 ≥ 0.
318
Vasile Cîrtoaje
The equality holds for a = b = c, and also for b = c = 0 (or any cyclic permutation).
Second Solution. Write the inequality as
A(a2 − bc) + B(b2 − ca) + C(c 2 − a b) ≥ 0,
where
A=
p
b + c,
B=
p
c + a,
C=
p
a + b.
We have
2
X
A(a2 − bc) =
X
=
X
A[(a − b)(a + c) + (a − c)(a + b)]
X
A(a − b)(a + c) +
B(b − a)(b + c)
X
(a − b)[A(a + c) − B(b + c)]
X
A2 (a + c)2 − B 2 (b + c)2
=
(a − b) ·
A(a + c) + B(b + c)
X (a − b)2 (a + c)(b + c)
=
≥ 0.
A(a + c) + B(b + c)
=
P 2.40. If a, b, c are nonnegative real numbers, then
p
p
p
(a2 − bc) a2 + 4bc + (b2 − ca) b2 + 4ca + (c 2 − a b) c 2 + 4a b ≥ 0.
(Vasile Cîrtoaje, 2005)
Solution. If two of a, b, c are zero, then the inequality is clearly true. Otherwise, write
the inequality as
AX + BY + C Z ≥ 0,
where
p
A=
a2 + 4bc
,
b+c
X = (a2 − bc)(b + c),
p
B=
b2 + 4ca
,
c+a
p
C=
Y = (b2 − bc)(b + c),
c 2 + 4a b
,
a+b
X = (c 2 − a b)(a + b).
Without loss of generality, assume that a ≥ b ≥ c. We have X ≥ 0, Z ≤ 0 and
X + Y + Z = 0.
In addition,
X − Y = a b(a − b) + 2(a2 − b2 )c + (a − b)c 2 ≥ 0
Symmetric Nonrational Inequalities
319
and
a4 − b4 + 2(a3 − c 3 )c + (a2 − c 2 )c 2 + 4a bc(a − b) − 4(a − b)c 3
(b + c)2 (c + a)2
A2 − B 2 =
≥
4a bc(a − b) − 4(a − b)c 3
4c(a − b)(a b − c 2 )
=
≥ 0.
(b + c)2 (c + a)2
(b + c)2 (c + a)2
Since
2(AX + BY + C Z) = (A − B)(X − Y ) − (A + B − 2C)Z,
it suffices to show that
A + B − 2C ≥ 0.
This is true if AB ≥ C 2 . Using the Cauchy-Schwarz inequality gives
p
p
a b + 4c a b
a b + 2c a b + 2c 2
AB ≥
≥
.
(b + c)(c + a)
(b + c)(c + a)
Thus, we need to show that
(a + b)2 (a b + 2c
p
a b + 2c 2 ) ≥ (b + c)(c + a)(c 2 + 4a b).
Write this inequality as
a b(a − b)2 + 2c
p
a b(a + b)
€p
a−
p Š2
b + c 2 [2(a + b)2 − 5a b − c(a + b) − c 2 ] ≥ 0.
It is true since
2(a + b)2 − 5a b − c(a + b) − c 2 = a(2a − b − c) + b(b − c) + b2 − c 2 ≥ 0.
The equality holds for a = b = c, and also for a = b and c = 0 (or any cyclic permutation).
P 2.41. If a, b, c are nonnegative real numbers, then
v
t
v
v
t
t
a3
b3
c3
+
+
≥ 1.
a3 + (b + c)3
b3 + (c + a)3
c 3 + (a + b)3
320
Vasile Cîrtoaje
Solution. For a = 0, the inequality reduces to the obvious inequality
p
p
p
b3 + c 3 ≥ b3 + c 3 .
For a, b, c > 0, write the inequality as
v
Xu
u
u
t
1
1+

b+c
a
‹3 ≥ 1.
For any x ≥ 0, we have
p
1 + x3 =
Æ
(1 + x)(1 − x + x 2 ≤
(1 + x) + (1 − x + x 2 )
1
= 1 + x 2.
2
2
Therefore, we get
v
Xu
u
u
t
≥
1
1+

b+c
a
‹3 ≥
1
X
=
X
X
1

‹
1 b+c 2
1+
2
a
a2
= 1.
a2 + b2 + c 2
b +c
a2
The equality holds for a = b = c, and also for b = c = 0 (or any cyclic permutation).
1+
2
2
P 2.42. If a, b, c are positive real numbers, then
v
v
v
u
‹
‹


t
t
t
1
1
1 1 1
1
(a + b + c)
+ +
≥ 1 + 1 + (a2 + b2 + c 2 ) 2 + 2 + 2 .
a b c
a
b
c
(Vasile Cîrtoaje, 2002)
Solution. Using the Cauchy-Schwarz inequality, we have
v
X Š X 1
X 1‹
€X Š X 1 ‹ t€X
a
+
2
=
a2 + 2
bc
a
a2
bc
v
v

‹

t€X Š X 1
t€X Š X 1 ‹
≥
a2
+
2
bc
a2
bc
v
v
‹


t€X Š X 1
t€X Š X 1 ‹
2
=
a
+2
a
,
a2
a
Symmetric Nonrational Inequalities
321
hence
v
v
€X Š X 1 ‹
t€X Š X 1 ‹
t€X Š X 1 ‹
−2
a
+1=1+
a2
a
,
a
a
a2
™2
–v
v
t€X Š X 1 ‹
t€X Š X 1 ‹
a
a2
−1 ≥1+
,
a
a2
v
v
v
u
t€X Š X 1 ‹
t€X Š X 1 ‹
t
.
−1≥ 1+
a
a2
a
a2
The equality holds if and only if
€X
a2
Š X 1 ‹ X 1 ‹ €X Š
=
bc ,
bc
a2
which is equivalent to
(a2 − bc)(b2 − ca)(c 2 − a b) = 0.
Consequently, the equality occurs for a2 = bc, or b2 = ca, or c 2 = a b.
P 2.43. If a, b, c are positive real numbers, then
v

‹

‹
t
1
1
1
1 1 1
2
2
2
5 + 2(a + b + c ) 2 + 2 + 2 − 2 ≥ (a + b + c)
+ +
.
a
b
c
a b c
(Vasile Cîrtoaje, 2004)
Solution. Let us denote
x=
We have
and
a b c
+ + ,
b c a
y=
b c a
+ + .
a b c
‹
1 1 1
(a + b + c)
+ +
= x + y +3
a b c

‹
1
1
1
2(a + b + c ) 2 + 2 + 2 − 2 =
a
b
c
2
2
2
2
a
b
c
b
c 2 a2
=2
+
+
+2
+
+
+4
b2 c 2 a2
a2 b2 c 2
2
2
2

= 2(x 2 − 2 y) + 2( y 2 − 2x) + 4 = (x + y − 2)2 + (x − y)2 ≥ (x + y − 2)2 .
322
Vasile Cîrtoaje
Therefore,
v
t
‹
1
1
1
+
+
+
−2 ≥ x + y −2
a2 b2 c 2
‹

1 1 1
+ +
− 5.
= (a + b + c)
a b c
The equality occurs for a = b, or b = c, or c = a.
2(a2
b2
+ c2)

P 2.44. If a, b, c are real numbers, then
Æ
2(1 + a bc) + 2(1 + a2 )(1 + b2 )(1 + c 2 ) ≥ (1 + a)(1 + b)(1 + c).
(Wolfgang Berndt, 2006)
First Solution. Denoting
p = a + b + c, q = a b + bc + ca,
r = a bc,
the inequality becomes
Æ
2(p2 + q2 + r 2 − 2pr − 2q + 1) ≥ p + q − r − 1.
It suffices to show that
2(p2 + q2 + r 2 − 2pr − 2q + 1) ≥ (p + q − r − 1)2 ,
which is equivalent to
p2 + q2 + r 2 − 2pq + 2qr − 2pr + 2p − 2q − 2r + 1 ≥ 0,
(p − q − r + 1)2 ≥ 0.
The equality holds for p + 1 = q + r and q ≥ 1. The last condition follows from p + q −
r − 1 ≥ 0.
Second Solution. Since
2(1 + a2 ) = (1 + a)2 + (1 − a)2
and
(1 + b2 )(1 + c 2 ) = (b + c)2 + (bc − 1)2 ,
by the Cauchy-Schwarz inequality, we get
Æ
2(1 + a2 )(1 + b2 )(1 + c 2 ) ≥ (1 + a)(b + c) + (1 − a)(bc − 1)
= (1 + a)(1 + b)(1 + c) − 2(1 + a bc).
Symmetric Nonrational Inequalities
323
P 2.45. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that
v
v
v
t a2 + bc t b2 + ca t c 2 + a b
1
+
+
≥2+ p .
2
2
2
2
2
2
b +c
c +a
a +b
2
(Vo Quoc Ba Can, 2006)
Solution. We may assume that a ≥ b ≥ c. Then, it suffices to show that
v
v
v
t a2 + c 2 t b2 + c 2 t a b
1
+
+
≥2+ p .
2
2
2
2
2
2
b +c
c +a
a +b
2
Let us denote
x=
v
t a2 + c 2
b2
Since
x2 − y2 =
+ c2
,
y=
s
a
.
b
(a − b)(a b − c 2 )
≥ 0,
b(b2 + c 2 )
it follows that
x ≥ y ≥ 1.
From

‹
(x − y)(x y − 1)
1
1
x+ − y+
=
≥ 0,
x
y
xy
we have
v
t a2 + c 2
v
t b2 + c 2
s
v
a tb
+
.
b
a
+
≥
b2 + c 2
c 2 + a2
Therefore, it is enough to show that
v
v
s
a t b t ab
1
+
+
≥2+ p .
2
2
b
a
a +b
2
s
a
, the inequality becomes
Putting t =
b
s
1
1
t
t + −2≥ p −
.
2
t
t +1
2
We have
(t − 1)2
(t − 1)2
s

‹≤ 2
t +1
1
t
2(t 2 + 1) p +
2+1
t
2
2
(t − 1)
1
≤
= t + − 2.
t
t
The equality holds for a = b and c = 0 (or any cyclic permutation).
1
p −
2
s
t
=
2
t +1
324
Vasile Cîrtoaje
P 2.46. If a, b, c are nonnegative real numbers, then
Æ
a(2a + b + c) +
Æ
b(2b + c + a) +
Æ
c(2c + a + b) ≥
Æ
12(a b + bc + ca).
(Vasile Cîrtoaje, 2012)
Solution. By squaring, the inequality becomes
a2 + b2 + c 2 +
XÆ
bc(2b + c + a)(2c + a + b) ≥ 5(a b + bc + ca).
Using the Cauchy-Schwarz inequality yields
XÆ
bc(2b + c + a)(2c + a + b) =
≥
XÆ
(2b2 + bc + a b)(2c 2 + bc + ac)
X
X p
p
(2bc + bc + a bc) = 3(a b + bc + ca) +
a bc.
Therefore, it suffices to show that
X p
a bc ≥ 2(a b + bc + ca).
a2 + b2 + c 2 +
We can get this inequality by summing Schur’s inequality
a2 + b2 + c 2 +
X p
Xp
a bc ≥
a b(a + b)
and
Xp
a b (a + b) ≥ 2(a b + bc + ca).
The last inequality is equivalent to the obvious inequality
Xp
p
p
a b ( a − b )2 ≥ 0.
The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).
P 2.47. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that
a
Æ
(4a + 5b)(4a + 5c) + b
Æ
(4b + 5c)(4b + 5a) + c
Æ
(4c + 5a)(4c + 5b) ≥ 27.
(Vasile Cîrtoaje, 2010)
Symmetric Nonrational Inequalities
325
Solution. Assume that a ≥ b ≥ c, consider the non-trivial case b > 0, and write the
inequality in the following equivalent homogeneous forms:
X Æ
a (4a + 5b)(4a + 5c) ≥ 3(a + b + c)2 ,
X
X
X €p
Š2
p
2(
a2 −
a b) =
a
4a + 5b − 4a + 5c ,
X
X
25a(b − c)2
(b − c)2 ≥
,
p
p
( 4a + 5b + 4a + 5c)2
X
(b − c)2 Sa ≥ 0,
where
25a
.
Sa = 1 − p
p
( 4a + 5b + 4a + 5c)2
Since
and
25b
25b
Sb = 1 − p
≥1− p
=0
p
p
( 4b + 5c + 4b + 5a)2
( 4b + 9b)2
25c
25
25c
≥1− p
=1−
Sc = 1 − p
> 0,
p
p
36
( 9c + 9c)2
( 4c + 5a + 4c + 5b)2
we have
X
a2
(b − c)2 Sa ≥ (b − c)2 Sa + (a − c)2 S b ≥ (b − c)2 Sa + 2 (b − c)2 S b
b

‹
a
b
a
= (b − c)2
Sa + S b .
b
a
b
Thus, it suffices to prove that
a
b
Sa + S b ≥ 0.
a
b
We have
p
p
25a
a( 4a + 5b − 4a)2
Sa ≥ 1 − p
=1−
,
p
b2
( 4a + 5b + 4a)2
p
p
25b
b( 4b + 5a − 4b)2
Sb ≥ 1 − p
=1−
,
p
a2
( 4b + 4b + 5a)2
and hence
p
p
p
p
b
a
b ( 4a + 5b − 4a)2 a ( 4b + 5a − 4b)2
Sa + S b ≥ −
+ −
a
b
a
b
b
a
v
‚v
Œ

‹
t 4a2 5a t 4b2 5b
a b
=4
+
+
+
−
7
+
− 10
b2
b
a2
a
b a
q
p
= 4 4x 2 + 5x − 8 + 2 20x + 41 − 7x − 10,
326
Vasile Cîrtoaje
where
x=
a b
+ ≥ 2.
b a
To end the proof, we only need to show that x ≥ 2 yields
4
q
p
4x 2 + 5x − 8 + 2 20x + 41 ≥ 7x + 10.
By squaring, this inequality becomes
15x 2 − 60x − 228 + 32
p
20x + 41 ≥ 0.
Indeed,
15x 2 − 60x − 228 + 32
p
p
20x + 41 ≥ 15x 2 − 60x − 228 + 32 81 = 15(x − 2)2 ≥ 0.
The equality holds for a = b = c = 1, and also for c = 0 and a = b =
permutation).
3
(or any cyclic
2
P 2.48. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. Prove that
a
Æ
(a + 3b)(a + 3c) + b
Æ
(b + 3c)(b + 3a) + c
Æ
(c + 3a)(c + 3b) ≥ 12.
(Vasile Cîrtoaje, 2010)
Solution. Assume that a ≥ b ≥ c (b > 0), and write the inequality as
X Æ
a (a + 3b)(a + 3c) ≥ 4(a b + bc + ca),
X
X
X €p
Š2
p
2(
a2 −
a b) =
a
a + 3b − a + 3c ,
X
X
9a(b − c)2
(b − c)2 ≥
,
p
p
( a + 3b + a + 3c)2
X
(b − c)2 Sa ≥ 0,
where
Since
9a
Sa = 1 − p
.
p
( a + 3b + a + 3c)2
9b
9b
≥1− p
=0
Sb = 1 − p
p
p
2
( b + 3c + b + 3a)
( b + 4b)2
Symmetric Nonrational Inequalities
and
327
9
9c
9c
=1−
≥1− p
> 0,
Sc = 1 − p
p
p
2
2
16
( 4c + 4c)
( c + 3a + c + 3b)
we have
X
a2
(b − c)2 Sa ≥ (b − c)2 Sa + (a − c)2 S b ≥ (b − c)2 Sa + 2 (b − c)2 S b
b

‹
a
b
a
= (b − c)2
Sa + S b .
b
a
b
Thus, it suffices to prove that
b
a
Sa + S b ≥ 0.
a
b
We have
p
p
a( a + 3b − a)2
=1−
,
Sa ≥ 1 − p
p
b2
( a + 3b + a)2
p
p
b( b + 3a − b)2
9b
=1−
Sb ≥ 1 − p
,
p
a2
( b + b + 3a)2
9a
and hence
p
p
p
p
b
a
b ( a + 3b − a)2 a ( b + 3a − b)2
Sa + S b ≥ −
+ −
a
b
a
b
b
a
v
v
‚
Œ 
‹
t a2 3a t b2 3b
a b
=2
+
+
+
−
+
−6
b2
b
a2
a
b a
Æ
p
= 2 x 2 + 3x − 2 + 2 3x + 10 − x − 6,
where
a b
+ ≥ 2.
b a
To end the proof, it remains to show that
Æ
p
2 x 2 + 35x − 2 + 2 3x + 10 ≥ x + 6
x=
for x ≥ 2. By squaring, this inequality becomes
p
3x 2 − 44 + 8 3x + 10 ≥ 0.
Indeed,
p
3x 2 − 44 + 8 3x + 10 ≥ 12 − 44 + 32 = 0.
The equality holds for a = b = c = 1, and also for c = 0 and a = b =
permutation).
p
3 (or any cyclic
328
Vasile Cîrtoaje
P 2.49. Let a, b, c be nonnegative real numbers such that a2 + b2 + c 2 = 3. Prove that
Æ
p
p
p
2 + 7a b + 2 + 7bc + 2 + 7ca ≥ 3 3(a b + bc + ca).
(Vasile Cîrtoaje, 2010)
Solution. Consider a ≥ b ≥ c. Since the inequality is trivial for b = c = 0, we may
assume that b > 0. By squaring, the desired inequality becomes
XÆ
a2 + b2 + c 2 +
(2 + 7a b)(2 + 7ac) ≥ 10(a b + bc + ca),
X €p
Š2
p
6(a2 + b2 + c 2 − a b − bc − ca) =
2 + 7a b − 2 + 7ac ,
3
where
X
X
49a2 (b − c)2
(b − c)2 ≥
,
p
p
( 2 + 7a b + 2 + 7ac )2
X
(b − c)2 Sa ≥ 0,
49a2
Sa = 1 − p
,
p
( 6 + 21a b + 6 + 21ac)2
49b2
,
Sb = 1 − p
p
( 6 + 21a b + 6 + 21bc)2
49c 2
.
Sc = 1 − p
p
( 6 + 21ac + 6 + 21bc)2
Since 6 ≥ 2(a2 + b2 ) ≥ 4a b, we have
49a2
49a2
a
≥1− p
=1− ,
Sa ≥ 1 − p
p
p
b
( 4a b + 21a b + 4a b + 21ac)2
(5 a b + 2 a b)2
49b2
b
49b2
≥1− p
=1− ,
Sb ≥ 1 − p
p
p
a
( 4a b + 21a b + 4a b + 21bc)2
(5 a b + 2 a b)2
49c 2
49c 2
49
Sc ≥ 1 − p
≥1−
=1−
> 0.
p
2
2
(5c
+
5c)
100
( 4a b + 21ac + 4a b + 21bc)
Therefore,
X
(b − c)2 Sa ≥ (b − c)2 Sa + (c − a)2 S b
‹

a
b
2
2
≥ (b − c) 1 −
+ (c − a) 1 −
b
a
(a − b)2 (a b − c 2 )
=
≥ 0.
ab
p
The equality holds for a = b = c = 1, and also for c = 0 and a = b = 3 (or any cyclic
permutation).
Symmetric Nonrational Inequalities
329
P 2.50. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. Prove that
Pp
(a)
P
(b)
P
(c)
a(b + c)(a2 + bc) ≥ 6;
p
p
a(b + c) a2 + 2bc ≥ 6 3;
p
a(b + c) (a + 2b)(a + 2c) ≥ 18.
(Vasile Cîrtoaje, 2010)
Solution. Assume that a ≥ b ≥ c (b > 0).
(a) Write the inequality in the homogeneous form
XÆ
a(b + c)(a2 + bc) ≥ 2(a b + bc + ca).
First Solution. Write the homogeneous inequality as
XÆ
”p
—
Æ
a(b + c)
a2 + bc − a(b + c) ≥ 0,
p
X (a − b)(a − c) a(b + c)
≥ 0.
p
p
a2 + bc + a(b + c)
Since (c − a)(c − b) ≥ 0, it suffices to show that
p
p
(a − b)(a − c) a(b + c) (b − c)(b − a) b(c + a)
+ p
≥ 0.
p
p
p
a2 + bc + a(b + c)
b2 + ca + b(c + a)
This is true if
p
p
(b − c) b(c + a)
(a − c) a(b + c)
≥p
.
p
p
p
a2 + bc + a(b + c)
b2 + ca + b(c + a)
Since
Æ
a(b + c) ≥
Æ
b(c + a),
it suffices to show that
p
a−c
b−c
≥p
.
p
p
a2 + bc + a(b + c)
b2 + ca + b(c + a)
Moreover, since
p
a2 + bc ≥
Æ
a(b + c),
p
b2 + ca ≤
it is enough to show that
p
a−c
a2 + bc
≥p
b−c
b2 + ca
.
Æ
b(c + a),
330
Vasile Cîrtoaje
Indeed, we have
(a − c)2 (b2 + ca) − (b − c)2 (a2 + bc) = (a − b)(a2 + b2 + c 2 + 3a b − 3bc − 3ca) ≥ 0,
because
a2 + b2 + c 2 + 3a b − 3bc − 3ca = (a2 − bc) + (b − c)2 + 3a(b − c) ≥ 0.
The equality holds for a = b = c = 1, and for c = 0 and a = b =
permutation).
p
3 (or any cyclic
Second Solution. By squaring, the homogeneous inequality becomes
X
XÆ
a(b + c)(a2 + bc) + 2
bc(a + b)(a + c)(b2 + ca)(c 2 + a b) ≥ 4(a b + bc + ca)2 .
Since
(b2 + ca)(c 2 + a b) − bc(a + b)(a + c) = a(b + c)(b − c)2 ≥ 0,
it suffices to show that
X
X
a(b + c)(a2 + bc) + 2
bc(a + b)(a + c) ≥ 4(a b + bc + ca)2 ,
which is equivalent to
X
bc(b − c)2 ≥ 0.
(b) Write the inequality as
X
p
p
a(b + c) a2 + 2bc ≥ 2(a b + bc + ca) a b + bc + ca,
X
a(b + c)
X
p
”p
a2 + 2bc −
p
—
a b + bc + ca ≥ 0,
a(b + c)(a − b)(a − c)
≥ 0.
p
a2 + 2bc + a b + bc + ca
Since (c − a)(c − b) ≥ 0, it suffices to show that
p
a(b + c)(a − b)(a − c)
b(c + a)(b − c)(b − a)
+p
≥ 0.
p
p
2
a + 2bc + a b + bc + ca
b2 + 2ca + a b + bc + ca
This is true if
p
a(b + c)(a − c)
b(c + a)(b − c)
≥p
.
p
p
a2 + 2bc + a b + bc + ca
b2 + 2ca + a b + bc + ca
Since
(b + c)(a − c) ≥ (c + a)(b − c),
Symmetric Nonrational Inequalities
331
it suffices to show that
p
a2 + 2bc +
a
p
a b + bc + ca
Moreover, since
p
p
a2 + 2bc ≥ a b + bc + ca,
≥p
b2 + 2ca +
p
b
p
b2 + 2ca ≤
a b + bc + ca
p
.
a b + bc + ca,
it is enough to show that
p
a
a2 + 2bc
≥p
b
b2 + 2ca
.
Indeed, we have
a2 (b2 + 2ca) − b2 (a2 + 2bc) = 2c(a3 − b3 ) ≥ 0.
p
3 (or any cyclic
The equality holds for a = b = c = 1, and for c = 0 and a = b =
permutation).
(c) Write the inequality as follows:
X
Æ
Æ
a(b + c) (a + 2b)(a + 2c) ≥ 2(a b + bc + ca) 3(a b + bc + ca),
X
a(b + c)
X
p
Ӯ
(a + 2b)(a + 2c) −
Æ
—
3(a b + bc + ca) ≥ 0,
a(b + c)(a − b)(a − c)
≥ 0.
p
(a + 2b)(a + 2c) + 3(a b + bc + ca)
Since (c − a)(c − b) ≥ 0, it suffices to show that
p
b(c + a)(b − c)(b − a)
a(b + c)(a − b)(a − c)
+p
≥ 0.
p
p
(a + 2b)(a + 2c) + 3(a b + bc + ca)
(b + 2c)(b + 2a) + 3(a b + bc + ca)
This is true if
p
a(b + c)(a − c)
b(c + a)(b − c)
≥p
.
p
p
(a + 2b)(a + 2c) + 3(a b + bc + ca)
(b + 2c)(b + 2a) + 3(a b + bc + ca)
Since
(b + c)(a − c) ≥ (c + a)(b − c),
it suffices to show that
a
p
(a + 2b)(a + 2c) +
p
3(a b + bc + ca)
Moreover, since
Æ
Æ
(a + 2b)(a + 2c) ≥ 3(a b + bc + ca),
≥p
b
(b + 2c)(b + 2a) +
Æ
(b + 2c)(b + 2a) ≤
p
3(a b + bc + ca)
Æ
.
3(a b + bc + ca),
332
Vasile Cîrtoaje
it is enough to show that
a
p
(a + 2b)(a + 2c)
This is true if
p
p
≥p
b
(b + 2c)(b + 2a)
p
a
(a + 2b)(a + 2c)
≥p
b
(b + 2c)(b + 2a)
.
.
Indeed, we have
a(b + 2c)(b + 2a) − b(a + 2b)(a + 2c) = (a − b)(a b + 4bc + 4ca) ≥ 0.
The equality holds for a = b = c = 1, and for c = 0 and a = b =
permutation).
p
3 (or any cyclic
P 2.51. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. Prove that
a
p
p
p
bc + 3 + b ca + 3 + c a b + 3 ≥ 6.
(Vasile Cîrtoaje, 2010)
First Solution. Write the inequality in the homogeneous form
X p
a a b + 2bc + ca ≥ 2(a b + bc + ca).
If a = 0, then the inequality turns into
p
p
p
bc( b − c)2 ≥ 0.
Consider further a, b, c > 0. By squaring, the inequality becomes
X
X Æ
X
X
a2 b2 + 6a bc
a.
a b(a2 + b2 ) + 2
bc (bc + 2ca + a b)(ca + 2a b + bc) ≥ 4
Using the Cauchy-Schwarz inequality, we have
Æ
(bc + 2ca + a b)(ca + 2a b + bc) =
Æ
(a b + bc + ca + ca)(a b + bc + ca + a b)
≥ a b + bc + ca + a
p
bc,
and hence
X Æ
Xp
bc (bc + 2ca + a b)(ca + 2a b + bc) ≥ (a b + bc + ca)2 + a bc
bc.
Symmetric Nonrational Inequalities
333
Thus, it suffices to show that
X
Xp
X
X
a b(a2 + b2 ) + 2(a b + bc + ca)2 + 2a bc
bc ≥ 4
a2 b2 + 6a bc
a,
which is equivalent to
X
X
Xp
X
bc ≥ 2
a2 b2 + 2a bc
a,
a b(a2 + b2 ) + 2a bc
X
X
Xp
a b(a − b)2 ≥ 2a bc(
a−
bc),
X (a − b)2
Using the substitution x =
p
≥
Xp
p
( a − b)2 .
c
p
p
a, y = b, z = c, the last inequality becomes
X (x 2 − y 2 )2 X
≥
(x − y)2 ,
z2
( y − z)2 A + (z − x)2 B + (x − y)2 C ≥ 0,
where
A=
y +z− x
,
x2
B=
z+x− y
,
y2
C=
x + y −z
.
z2
Without loss of generality, assume that x ≥ y ≥ z > 0. Since B > 0 and C > 0, we have
( y − z)2 A + (z − x)2 B + (x − y)2 C ≥ ( y − z)2 A + (z − x)2 B ≥ ( y − z)2 (A + B).
Thus, we only need to show that A + B ≥ 0. Indeed,
A+ B =
y +z− x z+ x − y
y +z− x z+ x − y
2z
+
≥
+
= 2 > 0.
x2
y2
x2
x2
x
The equality holds for a = b = c = 1, and for a = 0 and b = c =
permutation).
p
3 (or any cyclic
Second Solution. Assume that a ≥ b ≥ c, and write the inequality as
X p
a a b + 2bc + ca ≥ 2(a b + bc + ca),
X €p
Š
a
a b + 2bc + ca − b − c ≥ 0,
X a(a b + ac − b2 − c 2 )
≥ 0,
p
a b + 2bc + ca + b + c
y
x
z
+
+
≥ 0,
b+c+A c+a+B a+ b+C
(*)
334
Vasile Cîrtoaje
where
A=
p
a b + 2bc + ca,
x = a2 (b + c) − a(b2 + c 2 ),
B=
p
bc + 2ca + a b,
C=
p
ca + 2a b + bc,
y = b2 (c + a) − b(c 2 + a2 ), z = c 2 (a + b) − c(a2 + b2 ).
Since x + y + z = 0, we can write the inequality as

‹

‹
1
1
1
1
x
−
+z
−
≥ 0.
b+c+A c+a+B
a+b+C c+a+B
We have
x = a b(a − b) + ac(a − c) ≥ 0, z = ac(c − a) + bc(c − b) ≤ 0.
Therefore, it suffices to show that
1
1
−
≥ 0,
b+c+A c+a+B
1
1
−
≤ 0,
a+b+C c+a+B
that is,
a − b + B − A ≥ 0,
b − c + C − B ≥ 0.
It is enough to prove that A ≤ B ≤ C. Indeed,
B 2 − A2 = c(a − b) ≥ 0,
C 2 − B 2 = a(b − c) ≥ 0.
Remark. We can also prove the inequality (*) as follows
X a b(a − b) + ac(a − c)
≥ 0,
b+c+A
X a b(a − b) X ba(b − a)
+
≥ 0,
b+c+A
c+a+B
‹

X
1
1
a b(a − b)
−
≥ 0,
b+c+A c+a+B
X
a b(a + b + C)(a − b)(a − b + B − A) ≥ 0,
X
c 2
a b(a + b + C)(a − b) 1 +
≥ 0.
A+ B
P 2.52. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that
P
p
(a)
(b + c) b2 + c 2 + 7bc ≥ 18;
p
P
p
(b)
(b + c) b2 + c 2 + 10bc ≤ 12 3.
(Vasile Cîrtoaje, 2010)
Symmetric Nonrational Inequalities
335
Solution. (a) Write the inequality in the equivalent homogeneous forms
X
p
(b + c) b2 + c 2 + 7bc ≥ 2(a + b + c)2 ,
X”
—
p
(b + c) b2 + c 2 + 7bc − b2 − c 2 − 4bc ≥ 0,
X (b + c)2 (b2 + c 2 + 7bc) − (b2 + c 2 + 4bc)2
≥ 0,
p
(b + c) b2 + c 2 + 7bc + b2 + c 2 + 4bc
X
bc(b − c)2
≥ 0.
p
(b + c) b2 + c 2 + 7bc + b2 + c 2 + 4bc
3
(or any cyclic permuta2
tion), and for a = 3 and b = c = 0 (or any cyclic permutation).
The equality holds for a = b = c = 1, for a = 0 and b = c =
(b) Write the inequality as
X
Æ
(b + c) 3(b2 + c 2 + 10bc) ≤ 4(a + b + c)2 ,
X”
—
Æ
2b2 + 2c 2 + 8bc − (b + c) 3(b2 + c 2 + 10bc) ≥ 0,
X 4(b2 + c 2 + 4bc)2 − 3(b + c)2 (b2 + c 2 + 10bc)
≥ 0,
p
2b2 + 2c 2 + 8bc + (b + c) 3(b2 + c 2 + 10bc)
X
(b − c)4
≥ 0.
p
2b2 + 2c 2 + 8bc + (b + c) 3(b2 + c 2 + 10bc)
The equality holds for a = b = c = 1.
P 2.53. Let a, b, c be nonnegative real numbers such then a + b + c = 2. Prove that
p
p
p
p
a + 4bc + b + 4ca + c + 4a b ≥ 4 a b + bc + ca.
(Vasile Cîrtoaje, 2012)
Solution. Without loss of generality, assume that c = min{a, b, c}. Using Minkowski’s
inequality gives
Ç
p
p
p p
p
p
p
p
p
a + 4bc + b + 4ca ≥ ( a + b)2 + 4c( a + b)2 = ( a + b) 1 + 4c.
Therefore, it suffices to show that
p p
p
p
p
( a + b) 1 + 4c ≥ 4 a b + bc + ca − c + 4a b.
336
Vasile Cîrtoaje
By squaring, this inequality becomes
p
Æ
(a + b + 2 a b)(1 + 4c) + 8 (a b + bc + ca)(c + 4a b) ≥ 16(a b + bc + ca) + c + 4a b.
According to Lemma below, it suffices to show that
p
(a + b + 2 a b)(1 + 4c) + 8(2a b + bc + ca) ≥ 16(a b + bc + ca) + c + 4a b,
which is equivalent to
a+ b−c+2
p
a b + 8c
p
a b ≥ 4(a b + bc + ca).
Write this inequality in the homogeneous form
p
p
(a + b + c)(a + b − c + 2 a b) + 16c a b ≥ 8(a b + bc + ca).
p
1
Due to homogeneity, we may assume that a + b = 1. Let us denote d = a b, 0 ≤ d ≤ .
2
We need to show that f (c) ≥ 0 for 0 ≤ c ≤ d, where
f (c) = (1 + c)(1 − c + 2d) + 16cd − 8d 2 − 8c
= (1 − 2d)(1 + 4d) + 2(9d − 4)c − c 2 .
Since f (c) is concave, it suffices to show that f (0) ≥ 0 and f (d) ≥ 0. Indeed,
f (0) = (1 − 2d)(1 + 4d) ≥ 0,
f (d) = (3d − 1)2 ≥ 0.
Thus, the proof is completed. The equality holds for a = b = 1 and c = 0 (or any cyclic
permutation).
Lemma (by Nguyen Van Quy). Let a, b, c be nonnegative real numbers such then
c = min{a, b, c},
a + b + c = 2.
Then,
Æ
(a b + bc + ca)(c + 4a b) ≥ 2a b + bc + ca.
Proof. By squaring, the inequality becomes
c[a b + bc + ca − c(a + b)2 ] ≥ 0.
We need to show that
(a + b + c)(a b + bc + ca) − 2c(a + b)2 ≥ 0.
We have
(a + b + c)(a b + bc + ca) − 2c(a + b)2 ≥ (a + b)(b + c)(c + a) − 2c(a + b)2
= (a + b)(a − c)(b − c) ≥ 0.
Symmetric Nonrational Inequalities
337
P 2.54. If a, b, c are nonnegative real numbers, then
p
p
p
p
a2 + b2 + 7a b + b2 + c 2 + 7bc + c 2 + a2 + 7ca ≥ 5 a b + bc + ca.
(Vasile Cîrtoaje, 2012)
Solution (by Nguyen Van Quy). Assume that c = min{a, b, c}. Using Minkowski’s inequality yields
Ç
p
p
p
p
2
2
2
2
b + c + 7bc + a + c + 7ca ≥ (a + b)2 + 4c 2 + 7c( a + b)2 .
Therefore, it suffices to show that
Ç
p
p
p
p
(a + b)2 + 4c 2 + 7c( a + b)2 ≥ 5 a b + bc + ca − a2 + b2 + 7a b.
By squaring, this inequality becomes
p
Æ
2c 2 + 7c a b + 5 (a2 + b2 + 7a b)(a b + bc + ca) ≥ 15a b + 9c(a + b).
Due to homogeneity, we may assume that a + b = 1. Let us denote x = a b. We need to
1
show that f (x) ≥ 0 for c 2 ≤ x ≤ , where
4
Æ
p
f (x) = 2c 2 + 7c x + 5 (1 + 5x)(c + x) − 15x − 9c.
Since
−7c
5(5c − 1)2
<0
f 00 (x) = p − p
4 x 3 4 [5x 2 + (5c + 1)x + c]3
 ‹
1
f (c) is concave. Thus, it suffices to show that f (c 2 ) ≥ 0 and f
≥ 0.
4
Write the inequality f (c 2 ) ≥ 0 as
Æ
5 (1 + 5c 2 )(c + c 2 ) ≥ 6c 2 + 9c.
By squaring, this inequality turns into
c(89c 3 + 17c 2 − 56c + 25) ≥ 0,
which is true since
89c 3 + 17c 2 − 56c + 25 ≥ 12c 2 − 56c + 25 = (1 − 2c)(25 − 6c) ≥ 0.
 ‹
1
Write the inequality f
≥ 0 as
4
p
8c 2 − 22c + 15( 4c + 1 − 1) ≥ 0.
338
Vasile Cîrtoaje
Making the substitution t =
p
4c + 1, t ≥ 1, the inequality becomes
(t − 1)(t 3 + t 2 − 12t + 18) ≥ 0.
This is true since
t 3 + t 2 − 12t + 18 ≥ 2t 2 − 12t + 18 = 2(t − 3)2 ≥ 0.
Thus, the proof is completed. The equality holds for a = b and c = 0 (or any cyclic
permutation).
P 2.55. If a, b, c are nonnegative real numbers, then
p
a2 + b2 + 5a b +
p
b2 + c 2 + 5bc +
p
c 2 + a2 + 5ca ≥
Æ
21(a b + bc + ca).
(Nguyen Van Quy, 2012)
Solution. Without loss of generality, assume that c = min{a, b, c}. Using Minkowski’s
inequality, we have
Æ
(a + c)2 + 3ac +
Æ
(b + c)2 + 3bc ≥
Ç
p
p
(a + b + 2c)2 + 3c( a + b)2 .
Therefore, it suffices to show that
Ç
p
p
Æ
p
(a + b + 2c)2 + 3c( a + b)2 ≥ 21(a b + bc + ca) − a2 + b2 + 5a b.
By squaring, this inequality becomes
2c 2 + 3c
p
ab +
Æ
21(a2 + b2 + 5a b)(a b + bc + ca) ≥ 12a b + 7c(a + b).
Due to homogeneity, we may assume that a + b = 1. Let us denote x = a b. We need to
1
show that f (x) ≥ 0 for c 2 ≤ x ≤ , where
4
Æ
p
f (x) = 2c 2 + 3c x + 21(1 + 3x)(c + x) − 12x − 7c.
Since
p
−3c
21(3c − 1)2
<0
f (x) = p − p
4 x 3 4 [3x 2 + (3c + 1)x + c]3
 ‹
1
f (c) is concave. Thus, it suffices to show that f (c 2 ) ≥ 0 and f
≥ 0.
4
00
Symmetric Nonrational Inequalities
339
Write the inequality f (c 2 ) ≥ 0 as
Æ
21(1 + 3c 2 )(c + c 2 ) ≥ 7(c + c 2 ).
By squaring, this inequality turns into
c(c + 1)(1 − 2c)(3 − c) ≥ 0,
which is clearly true.  ‹
1
≥ 0 as
Write the inequality f
4
Æ
8c 2 − 22c + 7 3(4c + 1) − 12 ≥ 0.
1
Using the substitution 3t 2 = 4c + 1, t ≥ p , the inequality becomes
3
(t − 1)2 (3t 2 + 6t − 4) ≥ 0.
This is true since
p
3t 2 + 6t − 4 ≥ 1 + 2 3 − 4 > 0.
Thus, the proof is completed. The equality holds for a = b = c.
P 2.56. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. Prove that
a
p
a2 + 5 + b
p
b2 + 5 + c
p
c2 + 5 ≥
v
t2
3
(a + b + c)2 .
(Vasile Cîrtoaje, 2010)
Solution. Write the inequality in the homogeneous form
X Æ
p
a 3a2 + 5(a b + bc + ca) ≥ 2 (a + b + c)2 .
Due to homogeneity, we may assume that
a b + bc + ca = 1.
By squaring, the inequality becomes
X
X Æ
X
X
X
a4 + 2
bc (3b2 + 5)(3c 2 + 5) ≥ 12
a2 b2 + 19a bc
a+3
a b(a2 + b2 ).
340
Vasile Cîrtoaje
Applying Lemma below for x = 3b2 , y = 3c 2 and d = 5, we have
2
Æ
(3b2 + 5)(3c 2 + 5) ≥ 3(b2 + c 2 ) + 10 −
9 2
(b − c 2 )2 ,
20
hence
9
bc(b2 − c 2 )2 ,
20
X Æ
X
X
9 X
bc(b2 − c 2 )2
2
bc (3b2 + 5)(3c 2 + 5) ≥ 3
bc(b2 + c 2 ) + 10(
bc)2 −
20
X
X
X
9 X
= 10
a2 b2 + 20a bc
a+3
a b(a2 + b2 ) −
bc(b2 − c 2 )2 .
20
Therefore, it suffices to show that
X
X
X
X
9 X
bc(b2 − c 2 )2 ≥
a4 + 10
a2 b2 + 20a bc
a+3
a b(a2 + b2 ) −
20
X
X
X
≥ 12
a2 b2 + 19a bc
a+3
a b(a2 + b2 ),
2bc
Æ
(3b2 + 5)(3c 2 + 5) ≥ 3bc(b2 + c 2 ) + 10bc −
which is equivalent to
X
a4 − 2
X
a2 b2 + a bc
X
a−
9 X
bc(b2 − c 2 )2 ≥ 0.
20
Since
X
X Š
X
X Š
€X
€X
Š
€X
2
a4 − 2
a2 b2 + a bc
a =2
a4 −
a2 b2 − 2
a2 b2 − a bc
a
=
X
X
(b2 − c 2 )2 −
a2 (b − c)2 ,
we can write the inequality as
X
(b − c)2 Sa ≥ 0,
where
Sa = (b + c)2 − a2 −
9
bc(b + c)2 .
10
In addition, since
Sa ≥ (b + c)2 − a2 − bc(b + c)2 = (b + c)2 − a2 −
=
a(b + c)3 − a2 (a b + bc + ca)
,
a b + bc + ca
it is enough to show that
X
(b − c)2 Ea ≥ 0,
bc(b + c)2
,
a b + bc + ca
Symmetric Nonrational Inequalities
341
where
Ea = a(b + c)3 − a2 (a b + bc + ca).
Assume that a ≥ b ≥ c, b > 0. Since
E b = b(c + a)3 − b2 (a b + bc + ca) ≥ b(c + a)3 − b2 (c + a)(c + b)
≥ b(c + a)3 − b2 (c + a)2 = b(c + a)2 (c + a − b) ≥ 0,
Ec = c(a + b)3 − c 2 (a b + bc + ca) ≥ c(a + b)3 − c 2 (a + b)(b + c)
≥ c(a + b)3 − c 2 (a + b)2 = c(a + b)2 (a + b − c) ≥ 0
and
Ea E b
(b + c)3 (c + a)3
+
=
+
− 2(a b + bc + ca)
a2 b2
a
b
b3 + 2b2 c a3 + 2a2 c
+
− 2(a b + bc + ca)
≥
a
b
(a2 − b2 )2 + 2c(a + b)(a − b)2
=
≥ 0,
ab
we get

‹
X
Eb
2
2
2
2
2 Ea
(b − c) Ea ≥ (b − c) Ea + (a − c) E b ≥ a (b − c)
+
≥ 0.
a2 b2
The equality holds for a = b = c = 1, and also for a = b =
permutation).
p
3 and c = 0 (or any cyclic
Lemma. If x ≥ 0, y ≥ 0 and d > 0, then
2
Æ
(x + d)( y + d) ≥ x + y + 2d −
1
(x − y)2 .
4d
Proof. We have
2
Æ
2(d x + d y + x y)
2(d x + d y + x y)
≥ (x+d)+( y+d)
(x + d)( y + d) − 2d = p
(x + d)( y + d) + d
+d
2
=
4(d x + d y + x y)
(x − y)2
(x − y)2
=x+y−
≥x+y−
.
x + y + 4d
x + y + 4d
4d
342
Vasile Cîrtoaje
P 2.57. Let a, b, c be nonnegative real numbers such that a2 + b2 + c 2 = 1. Prove that
p
p
p
a 2 + 3bc + b 2 + 3ca + c 2 + 3a b ≥ (a + b + c)2 .
(Vasile Cîrtoaje, 2010)
Solution. Let q = a b + bc + ca. Write the inequality as
X p
a 2 + 3bc ≥ 1 + 2q.
By squaring, the inequality becomes
X
X Æ
1 + 3a bc
a+2
bc (2 + 3a b)(2 + 3ac) ≥ 4q + 4q2 .
Applying Lemma from the preceding P 2.56 for x = 3a b, y = 3ac2 and d = 2, we have
9
(2 + 3a b)(2 + 3ac) ≥ 3a(b + c) + 4 − a2 (b − c)2 ,
8
X Æ
X
X
X
9
2
bc (2 + 3a b)(2 + 3ac) ≥ 3a bc
a(b − c)2
(b + c) + 4
bc − a bc
8
X
X
9
= 6a bc
a + 4q − a bc
a(b − c)2 .
8
Therefore, it suffices to show that
2
Æ
1 + 3a bc
X
a + 4q + 6a bc
X
X
9
a − a bc
a(b − c)2 ≥ 4q + 4q2 ,
8
which is equivalent to
1 + 9a bc
X
a − 4q2 ≥
X
9
a bc
a(b − c)2 .
8
Since
a4 + b4 + c 4 = 1 − 2(a2 b2 + b2 c 2 + c 2 a2 ) = 1 − 2q2 + 4a bc
X
a,
from Schur’s inequality of fourth degree
a4 + b4 + c 4 + 2a bc
X
a≥
€X
a2
Š €X
Š
ab ,
we get
1 ≥ 2q2 + q − 6a bc
X
a.
Thus, it is enough to prove that
(2q2 + q − 6a bc
X
a) + 9a bc
X
a − 4q2 ≥
X
9
a bc
a(b − c)2 ;
8
Symmetric Nonrational Inequalities
343
that is,
8(q − 2q2 + 3a bc
Since
X
a) ≥ 9a bc
X
a(b − c)2 .
X
X
€X Š €X Š
€X Š2
q − 2q2 + 3a bc
a=
a2
ab − 2
a b + 3a bc
a
X
X
X
=
bc(b2 + c 2 ) − 2
b2 c 2 =
bc(b − c)2 ,
we need to show that
X
bc(8 − 9a2 )(b − c)2 ≥ 0.
Also, since
8 − 9a2 = 8(b2 + c 2 ) − a2 ≥ b2 + c 2 − a2 ,
it suffices to prove that
X
bc(b2 + c 2 − a2 )(b − c)2 ≥ 0.
Assume that a ≥ b ≥ c. It is enough to show that
bc(b2 + c 2 − a2 )(b − c)2 + ca(c 2 + a2 − b2 )(c − a)2 ≥ 0.
This is true if
a(c 2 + a2 − b2 )(a − c)2 ≥ b(a2 − b2 − c 2 )(b − c)2 .
For the non-trivial case a2 − b2 − c 2 ≥ 0, this inequality follows from
a ≥ b,
c 2 + a2 − b2 ≥ a2 − b2 − c 2 , (a − c)2 ≥ (b − c)2 .
1
1
The equality holds for a = b = c = p , and for a = 0 and b = c = p (or any cyclic
3
2
permutation).
P 2.58. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that
(a)
(b)
a
a
v
t 2a + bc
3
v
t a(1 + b + c)
3
+b
+b
v
t 2b + ca
3
+c
v
t b(1 + c + a)
3
v
t 2c + a b
+c
3
≥ 3;
v
t c(1 + a + b)
3
≥ 3.
(Vasile Cîrtoaje, 2010)
344
Vasile Cîrtoaje
Solution. (a) If two of a, b, c are zero, then the inequality is trivial. Otherwise, by
Hölder’s inequality, we have
‚
v
Œ
P
X t 2a + bc 2
( a)3
9
≥P
=P
a
.
a
3a
3
2a + bc
2a + bc
Therefore, it suffices to show that
X
Since
a
≤ 1.
2a + bc
bc
2a
=1−
,
2a + bc
2a + bc
we can write this inequality as
X
bc
≥ 1.
2a + bc
By the Cauchy-Schwarz inequality, we have
P
P
X bc
( bc)2
( bc)2
P
P
=
≥P
= 1.
2a + bc
bc(2a + bc) 2a bc a + b2 c 2
The equality holds for a = b = c = 1, and for a = 0 and b = c =
permutation).
3
(or any cyclic
2
(b) Write the inequality in the homogeneous form
X Æ
a a(a + 4b + 4c) ≥ (a + b + c)2 .
By squaring, the inequality becomes
X Æ
X
X
bc bc(b + 4c + 4a)(c + 4a + 4b) ≥ 3
a2 b2 + 6a bc
a.
Applying the Cauchy-Schwarz inequality, we have
Æ
(b + 4c + 4a)(c + 4a + 4b) =
Æ
(4a + b + c + 3c)(4a + b + c + 3b)
p
≥ 4a + b + c + 3 bc,
hence
p
Æ
bc bc(b + 4c + 4a)(c + 4a + 4b) ≥ (4a + b + c)bc bc + 3b2 c 2 ,
X Æ
X
X
p
bc bc(b + 4c + 4a)(c + 4a + 4b) ≥
(4a + b + c)bc bc + 3
b2 c 2 .
Symmetric Nonrational Inequalities
345
Thus, it is enough to show that
X
X
p
a.
(4a + b + c)bc bc ≥ 6a bc
Replacing a, b, c by a2 , b2 , c 2 , respectively, this inequality becomes
X
X
(4a2 + b2 + c 2 )b3 c 3 ≥ 6a2 b2 c 2
a2 ,
X
X
€X Š €X
Š
a2
b3 c 3 + 3a2 b2 c 2
bc ≥ 6a2 b2 c 2
a2 ,
X Š
€X Š €X
Š
€X
a2
a3 b3 − 3a2 b2 c 2 ≥ 3a2 b2 c 2
a2 −
ab .
Since
X
a3 b3 − 3a2 b2 c 2 =
€X
ab
Š €X
a2 b2 − a bc
and
X
a2 −
X
ab =
X Š 1 €X Š X
ab
a2 (b − c)2 ,
a =
2
1X
(b − c)2 ,
2
we can write the inequality as
X
(b − c)2 Sa ≥ 0,
where
Sa = a 2
€X
a2
Š €X
Š
a b − 3a2 b2 c 2 .
Assume that a ≥ b ≥ c. Since Sa ≥ S b ≥ 0 and
€X Š €X Š
S b + Sc = (b2 + c 2 )
a2
a b − 6a2 b2 c 2
€X Š €X Š
≥ 2bc
a2
a b − 6a2 b2 c 2
€X Š
≥ 2bca2
a b − 6a2 b2 c 2 = 2a2 bc(a b + ac − 2bc) ≥ 0,
we get
X
(b − c)2 Sa ≥ (c − a)2 S b + (a − b)2 Sc ≥ (a − b)2 (S b + Sc ) ≥ 0.
The equality holds for a = b = c = 1, and for a = 0 and b = c =
permutation).
3
(or any cyclic
2
P 2.59. If a, b, c are nonnegative real numbers such that a + b + c = 3, then
Æ
Æ
Æ
8(a2 + bc) + 9 + 8(b2 + ca) + 9 + 8(c 2 + a b) + 9 ≥ 15.
(Vasile Cîrtoaje, 2013)
346
Vasile Cîrtoaje
Solution. Let q = a b + bc + ca and
A = (3a − b − c)2 + 8q,
B = (3b − c − a)2 + 8q,
C = (3c − a − b)2 + 8q.
Since
8(a2 + bc) + 9 = 8(a2 + q) + 9 − 8a(b + c) = 8(a2 + q) + 9 − 8a(3 − a)
= (4a − 3)2 + 8q = (3a − b − c)2 + 8q = A,
we can rewrite the inequality as follows
Xp
A ≥ 15,
X p
[ A − (3a + b + c)] ≥ 0,
X 2bc − ca − a b
≥ 0,
p
A + 3a + b + c
˜
X•
b(c − a)
c(b − a)
+
≥ 0,
p
p
A + 3a + b + c
A + 3a + b + c
X
X
c(a − b)
c(b − a)
+
≥ 0,
p
p
B + 3b + c + a
A + 3a + b + c
X
p
p
p
c(a − b)( C + 3c + a + b)[ A − B + 2(a − b)] ≥ 0,
•
˜
X
p
4(a + b − c)
c(a − b)2 ( C + 3c + a + b) p
+ 1 ≥ 0.
p
A+ B
Without loss of generality, assume that a ≥ b ≥ c. Since a + b − c > 0, it suffices to show
that
•
˜
p
4(c + a − b)
2
b(a − c) ( B + 3b + c + a) p
+1 ≥
p
A+ C
•
˜
p
4(a − b − c)
−
1
.
a(b − c)2 ( A + 3a + b + c) p
p
B+ C
This inequality follows from the inequalities
b2 (a − c)2 ≥ a2 (b − c)2 ,
p
p
a( B + 3b + c + a) ≥ b( A + 3a + b + c),
4(a − b − c)
4(c + a − b)
+1≥ p
− 1.
p
p
p
A+ C
B+ C
Write the second inequality as
a2 B − b2 A
p
p + (a − b)(a + b + c) ≥ 0.
a B+b A
Symmetric Nonrational Inequalities
347
Since
a2 B − b2 A = (a − b)(a + b + c)(a2 + b2 − 6a b + bc + ca) + 8q(a2 − b2 )
≥ (a − b)(a + b + c)(a2 + b2 − 6a b) ≥ −4a b(a − b)(a + b + c),
it suffices to show that
−4a b
p
p + 1 ≥ 0.
a B+b A
p
p
p
p
p
p
Indeed, since A > 8q ≥ 2 a b and B ≥ 8q ≥ 2 a b, we have
p
p
p
p
p
a B + b A − 4a b > 2(a + b) a b − 4a b = 2 a b(a + b − 2 a b) ≥ 0.
The third inequality holds if
2(a − b − c)
1≥ p
p .
B+ C
p
p
Clearly, it suffices to show that B ≥ a and C ≥ a. We have
B − a2 = 8q − 2a(3b − c) + (3b − c)2 ≥ 8a b − 2a(3b − c) = 2a(b + c) ≥ 0
and
C − a2 = 8q − 2a(3c − b) + (3c − b)2 ≥ 8a b − 2a(3c − b) = 2a(5b − 3c) ≥ 0.
The equality holds for a = b = c = 1, and also for a = 3 and b = c = 0 (or any cyclic
permutation).
P 2.60. Let a, b, c be nonnegative real numbers such that a + b + c = 3. If k ≥
p
a2 + bc + k +
p
b2 + ca + k +
p
9
, then
8
p
c 2 + a b + k ≥ 3 2 + k.
Solution. We will show that
XÆ
XÆ
Æ
8(a2 + bc + k) ≥
(3a + b + c)2 + 8k − 9 ≥ 6 2(k + 2).
The right inequality is equivalent to
XÆ
Æ
(2a + 3)2 + 8k − 9 ≥ 6 2(k + 2),
and follows immediately from Jensen’s inequality applied to the convex function f :
[0, ∞) → R defined by
Æ
f (x) = (2x + 3)2 + 8k − 9.
348
Vasile Cîrtoaje
Using the substitution
A1 = 8(a2 + bc + k),
B1 = 8(b2 + ca + k),
C1 = 8(c 2 + a b + k),
A2 = (3a + b + c)2 + 8k − 9, B2 = (3b + c + a)2 + 8k − 9, C2 = (3c + a + b)2 + 8k − 9,
we can write the left inequality as follows:
p
B1 − B2
C1 − C2
A1 − A2
p +p
p +p
p ≥ 0,
A1 + A2
B1 + B2
C1 + C2
2bc − ca − a b 2ca − a b − bc 2a b − bc − ca
+ p
+ p
≥ 0,
p
p
p
p
A1 + A2
B1 + B2
C1 + C2
X b(c − a)
c(b − a)
≥ 0,
p +p
p
p
A1 + A2
A1 + A2
X c(a − b)
c(b − a)
p +p
p ≥ 0,
p
B1 + B2
A1 + A2
X
p
p
p
p
p
p
c(a − b)( C1 + C2 )[( A1 − B1 ) + ( A2 − B2 )] ≥ 0,
X
p
p
2(a + b − c)
2a + 2b + c
2
c(a − b) ( C1 + C2 ) p
≥ 0.
p +p
p
A1 + B1
A2 + B2
Without loss of generality, assume that a ≥ b ≥ c. Clearly, the desired inequality is true
for b + c ≥ a. Consider further the case b + c < a. Since a + b − c > 0, it suffices to show
that
p
p
2(b + c − a)
2b + 2c + a
2
+
a(b − c) ( A1 + A2 ) p
p +p
p
B1 + C1
B2 + C2
p
p
2(c + a − b)
2c + 2a + b
2
+b(a − c) ( B1 + B2 ) p
≥ 0.
p
p +p
C1 + A1
C2 + AC2
Since
b2 (a − c)2 ≥ a2 (b − c)2 ,
it suffices to show that
p
p
2(b + c − a)
2b + 2c + a
+
b( A1 + A2 ) p
p +p
p
B1 + C1
B2 + C2
p
p
2(c + a − b)
2c + 2a + b
+a( B1 + B2 ) p
≥ 0.
p +p
p
C1 + A1
C2 + A2
From
a2 B1 − b2 A1 = 8c(a3 − b3 ) + 8k(a2 − b2 ) ≥ 0
Symmetric Nonrational Inequalities
349
and
a2 B2 − b2 A2 = (a − b)(a + b + c)(a2 + b2 + 6a b + bc + ca) + (8k − 9)(a2 − b2 ) ≥ 0,
we get a
p
B1 ≥ b
p
A1 and a
p
B2 ≥ b
p
A2 , hence
p
p
p
p
a( B1 + B2 ) ≥ b( A1 + A2 ).
Therefore, it is enough to show that
2b + 2c + a
2c + 2a + b
2(b + c − a)
2(c + a − b)
p +p
p +p
p
p +p
p ≥ 0.
B1 + C1
C1 + A1
B2 + C2
C2 + A2
This is true if
p
and
p
−2b
2b
p +p
p ≥0
B1 + C1
C1 + A1
−2a
2a
2a
p +p
p ≥ 0.
p +p
B1 + C1
C1 + A1
C2 + A2
The first inequality is true because A1 − B1 = 8(a − b)(a + b − c) ≥ 0. The second
inequality can be written as
1
p
Since
C1 +
1
p
C1 +
p
A1
p
A1
+p
+p
1
C2 +
1
C2 +
p
A2
p
A2
≥p
≥p
1
B1 +
.
p
C1
p
C2 +
4
C1 +
p
A1 +
p
A2
,
it suffices to show that
4
p
p
p
p
p
B1 + 3 C1 ≥ A1 + A2 + C2 .
Taking into account of
C1 − C2 = 4(2a b − bc − ca) ≥ 0,
C1 − B1 = 8(b − c)(a − b − c) ≥ 0,
A2 − A1 = 4(a b − 2bc + ca) ≥ 0,
we have
4
p
B1 + 3
p
C1 −
p
A1 −
p
A2 −
p
p
A1 −
p
A2 −
p
p
B1 + 2 B1 −
p
p
= 2(3 B1 − A2 ).
p
≥4
p
B1 + 2
C1 −
C2 ≥ 4
p
p
A2
A2
350
Vasile Cîrtoaje
In addition,
9B1 − A2 = 64k − 8a2 + 72b2 − 4a b + 68ac
≥ 72 − 8a2 + 72b2 − 4a b + 68ac
= 8(a + b + c)2 − 8a2 + 72b2 − 4a b + 68ac
= 4(20b2 + 2c 2 + 3a b + 4bc + 21ac) ≥ 0.
Thus, the proof is completed. The equality holds for a = b = c = 1. If k = 9/8, then the
equality holds also for a = 3 and b = c = 0 (or any cyclic permutation).
P 2.61. If a, b, c are nonnegative real numbers such that a + b + c = 3, then
p
p
p
p
a3 + 2bc + b3 + 2ca + c 3 + 2a b ≥ 3 3.
(Nguyen Van Quy, 2013)
Solution. Since
(a3 + 2bc)(a + 2bc) ≥ (a2 + 2bc)2 ,
it suffices to prove that
X a2 + 2bc
p
≥ 3 3.
p
a + 2bc
By Hölder’s inequality, we have
X 2
2 X
”X
—3
a + 2bc
(a2 + 2bc)(a + 2bc) ≥
(a2 + 2bc) = (a + b + c)6 .
p
a + 2bc
Therefore, it suffices to show that
(a + b + c)4 ≥ 3
X
(a2 + 2bc)(a + 2bc).
which is equivalent to
(a + b + c)4 ≥
X
(a2 + 2bc)(a2 + 6bc + ca + a b).
Indeed,
(a + b + c)4 −
X
X
(a2 + 2bc)(a2 + 6bc + ca + a b) = 3
a b(a − b)2 ≥ 0.
The equality holds for a = b = c = 1, and also for a = 3 and b = c = 0 (or any cyclic
permutation).
Symmetric Nonrational Inequalities
351
P 2.62. If a, b, c are positive real numbers, then
p
a2 + bc
+
b+c
p
b2 + ca
+
c+a
p
p
3 2
c2 + a b
≥
.
a+b
2
(Vasile Cîrtoaje, 2006)
Solution. According to the well-known inequality
(x + y + z)2 ≥ 3(x y + yz + z x),
it suffices to show that
X
p
x, y, z ≥ 0,
(b2 + ca)(c 2 + a b
3
≥ .
(c + a)(a + b)
2
Replacing a, b, c by a2 , b2 , c 2 , respectively, the inequality becomes
X
Æ
2
(b2 + c 2 ) (b4 + c 2 a2 )(c 4 + a2 b2 ) ≥ 3(a2 + b2 )(b2 + c 2 )(c 2 + a2 ).
Multiplying the Cauchy-Schwarz inequalities
Æ
(b2 + c 2 )(b4 + c 2 a2 ) ≥ b3 + ac 2 ,
Æ
(c 2 + b2 )(c 4 + a2 b2 ) ≥ c 3 + a b2 ,
we get
Æ
(b2 + c 2 ) (b4 + c 2 a2 )(c 4 + a2 b2 ) ≥ (b3 + ac 2 )(c 3 + a b2 )
= b3 c 3 + a(b5 + c 5 ) + a2 b2 c 2 .
Therefore, it suffices to show that
X
X
2
b3 c 3 + 2
a(b5 + c 5 ) + 6a2 b2 c 2 ≥ 3(a2 + b2 )(b2 + c 2 )(c 2 + a2 ).
This inequality is equivalent to
X
X
X
2
b3 c 3 + 2
bc(b4 + c 4 ) ≥ 3
b2 c 2 (b2 + c 2 ),
X
bc[2b2 c 2 + 2(b4 + c 4 ) − 3bc(b2 + c 2 )] ≥ 0,
X
bc(b − c)2 (2b2 + bc + 2c 2 ) ≥ 0.
The equality holds for a = b = c.
352
Vasile Cîrtoaje
P 2.63. If a, b, c are nonnegative real numbers, no two of which are zero,then
p
p
p
bc + 4a(b + c)
ca + 4b(c + a)
a b + 4c(a + b) 9
+
+
≥ .
b+c
c+a
a+b
2
(Vasile Cîrtoaje, 2006)
Solution. Let us denote
A = bc + 4a(b + c),
B = ca + 4b(c + a),
C = a b + 4c(a + b).
By squaring, the inequality becomes
X
p
X
A
BC
81
+2
≥
.
2
(b + c)
(c + a)(a + b)
4
Further, we use the following identity due to Sung-Yoon Kim:
(b + c)2 BC − 4[a(b2 + c 2 ) + 2bc(b + c) + 3a bc]2 = a bc(b − c)2 (a + 4b + 4c),
which yields
2a(b2 + c 2 ) + 4bc(b + c) + 6a bc
,
b+c
P
P
p
X
4 a(b2 + c 2 ) + 8 bc(b + c) + 36a bc
BC
2
≥
,
(c + a)(a + b)
(a + b)(b + c)c + a)
P
p
X
12 bc(b + c) + 36a bc
BC
2
≥
.
(c + a)(a + b)
(a + b)(b + c)c + a)
On the other hand, according to the known inequality Iran-1996,
X a b + bc + ca
9
≥ ,
2
(b + c)
4
p
BC ≥
(see Remark from the proof of P 1.71), we have
X a b + bc + ca
X a
X
X a
9
A
=
+
3
≥
+
3
.
(b + c)2
(b + c)2
b+c
4
b+c
Thus, it suffices to show that
3
X
P
12 bc(b + c) + 36a bc
a
+
≥ 18.
b+c
(a + b)(b + c)c + a)
This is equivalent to Schur’s inequality of degree three
X
X
a3 + 3a bc ≥
bc(b + c).
The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).
Symmetric Nonrational Inequalities
353
P 2.64. If a, b, c are nonnegative real numbers, no two of which are zero,then
p
p
p
a a2 + 3bc b b2 + 3ca c c 2 + 3a b
+
+
≥ a + b + c.
b+c
c+a
a+b
(Cezar Lupu, 2006)
Solution. Using the AM-GM inequality, we have
p
2a(a2 + 3bc)
a a2 + 3bc
2a(a2 + 3bc)
2a3 + 6a bc
≥
= p
=
,
b+c
S + 5bc
2 (b + c)2 (a2 + 3bc) (b + c)2 + (a2 + 3bc)
where S = a2 + b2 + c 2 . Thus, it suffices to show that
X 2a3 + 6a bc
S + 5bc
≥ a + b + c.
Write this inequality as
X 2a2 + 6bc
a
− 1 ≥ 0,
S + 5bc
or, equivalently,
AX + BY + X Z ≥ 0,
where
A=
1
,
S + 5bc
X = a3 + a bc − a(b2 + c 2 ),
B=
1
,
S + 5ca
C=
1
,
S + 5a b
Y = b3 + a bc − b(c 2 + a2 ),
Z = c 3 + a bc − c(a2 + b2 ).
Without loss of generality, assume that a ≥ b ≥ c. We have
A ≥ B ≥ C,
X = a(a2 − b2 ) + ac(b − c) ≥ 0,
Z = c(c 2 − b2 ) + ac(b − a) ≤ 0
and, according to Schur’s inequality of third degree,
X
X
X +Y +Z =
a3 + 3a bc −
a(b2 + c 2 ) ≥ 0.
Therefore,
AX + BY + C Z ≥ BX + BY + BZ = B(X + Y + Z) ≥ 0.
The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).
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Vasile Cîrtoaje
P 2.65. If a, b, c are nonnegative real numbers, no two of which are zero,then
v
v
v
t
t
t
2a(b + c)
2b(c + a)
2c(a + b)
+
+
≥ 2.
(2b + c)(b + 2c)
(2c + a)(c + 2a)
(2a + b)(a + 2b)
(Vasile Cîrtoaje, 2006)
p
p
p
Solution. Making the substitution x = a, y = b, z = c, the inequality becomes
v
X t
2( y 2 + z 2 )
x
≥ 2.
(2 y 2 + z 2 )( y 2 + 2z 2 )
We claim that
v
t
(2 y 2
2( y 2 + z 2 )
y +z
≥ 2
.
2
2
2
+ z )( y + 2z )
y + yz + z 2
Indeed, be squaring and direct calculation, this inequality reduces to y 2 z 2 ( y − z)2 ≥ 0.
Thus, it suffices to show that
X
x( y + z)
≥ 2,
+ yz + z 2
y2
which is just the inequality in P 1.68. The equality holds for a = b = c, and also for
a = 0 and b = c (or any cyclic permutation).
P 2.66. If a, b, c are nonnegative real numbers such that a b + bc + ca = 3, then
v
v
v
v
s
s
t bc
t ab
t bc
t ab
ca
ca
+
+
≤
1
≤
+
+
.
3a2 + 6
3b2 + 6
3c 2 + 6
6a2 + 3
6b2 + 3
6c 2 + 3
(Vasile Cîrtoaje, 2011)
Solution. By the Cauchy-Schwarz inequality, we have
‚
v
Xt
bc
3a2 + 6
Œ2
≤
X
‹ X
a b + bc + ca
3a2 + 6
hence
‚
v
Xt
bc
2
3a + 6
Œ2
≤
X
a2
‹
bc
,
a b + bc + ca
1
.
+2
Therefore, to prove the original left inequality, it suffices to show that
X
a2
1
≤ 1.
+2
Symmetric Nonrational Inequalities
355
This inequality is equivalent to
a2
≥ 1.
a2 + 2
Indeed, by the Cauchy-Schwarz inequality, we get
X
X
a2
(a + b + c)2
(a + b + c)2
P
P
≥
=
= 1.
a2 + 2
(a2 + 2)
a2 + 6
The equality occurs for a = b = c = 1. By Hölder’s inequality, we have
v
Œ2
‚
X t bc
”X
— €X Š3
2 2
2
b
c
(6a
+
3)
≥
bc .
6a2 + 3
To prove the original right inequality, it suffices to show that
X
(a b + bc + ca)3 ≥
b2 c 2 (6a2 + a b + bc + ca),
which is equivalent to
(a b + bc + ca)[(a b + bc + ca)2 −
X
b2 c 2 ] ≥ 18a2 b2 c 2 ,
2a bc(a b + bc + ca)(a + b + c) ≥ 18a2 b2 c 2 ,
X
2a bc
a(b − c)2 ≥ 0.
The equality occurs for a = b = c = 1, and for a = 0 and bc = 3 (or any cyclic
permutation).
P 2.67. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. If k > 1, than
a k (b + c) + b k (c + a) + c k (a + b) ≥ 6.
Solution. Let
E = a k (b + c) + b k (c + a) + c k (a + b).
We consider two cases.
Case 1: k ≥ 2. Applying Jensen’s inequality to the convex function f (x) = x k−1 , x ≥ 0,
we get
E = (a b + ac)a k−1 + (bc + ba)b k−1 + (ca + c b)c k−1
˜
•
(a b + ac)a + (bc + ba)b + (ca + c b)c k−1
≥ 2(a b + bc + ca)
2(a b + bc + ca)
2
k−1
a (b + c) + b2 (c + a) + c 2 (a + b)
=6
.
6
356
Vasile Cîrtoaje
Thus, it suffices to show that
a2 (b + c) + b2 (c + a) + c 2 (a + b) ≥ 6.
Write this inequality as
(a b + bc + ca)(a + b + c) − 3a bc ≥ 6,
a + b + c ≥ 2 + a bc.
It is true since
Æ
a+b+c ≥
and
3(a b + bc + ca) = 3
a+b+c
a bc ≤
3

‹3
= 1.
Case 2: 1 < k ≤ 2. We have
E = a k−1 (3 − bc) + b k−1 (3 − ca) + c k−1 (3 − a b)
= 3(a k−1 + b k−1 + c k−1 ) − a k−1 b k−1 c k−1 (a b)2−k + (bc)2−k + (ca)2−k .
Since 0 ≤ 2 − k < 1, f (x) = x 2−k is concave for x ≥ 0. Thus, by Jensen’s inequality, we
have
‹

a b + bc + ca 2−k
(a b)2−k + (bc)2−k + (ca)2−k ≤ 3
= 3,
3
and hence
E ≥ 3(a k−1 + b k−1 + c k−1 ) − 3a k−1 b k−1 c k−1 .
Consequently, it suffices to show that
a k−1 + b k−1 + c k−1 ≥ a k−1 b k−1 c k−1 + 2.
Due to symmetry, we may assume that a ≥ b ≥ c. In addition, write the inequality as
a k−1 + b k−1 − 2 ≥ (a k−1 b k−1 − 1)c k−1 ,
a
k−1
+b
k−1
− 2 ≥ (a
Let
x=
p
k−1 k−1
b
3 − ab
− 1)
a+b
a b, 1 ≤ x ≤

p
3.
By the AM-GM inequality, we have
a + b ≥ 2x,
a k−1 + b k−1 ≥ 2x k−1 .
‹k−1
.
Symmetric Nonrational Inequalities
357
Thus, it suffices to show that
2(x
k−1
− 1) ≥ (x
2k−2
3 − x2
− 1)
2x
k−1
.
Since x ≥ 1, this is true if
2 ≥ (x
k−1
3 − x2
+ 1)
2x
k−1
,
which can be written as
2≥
3 − x2
2
Since
1≥
k−1
3 − x2
+
2x
k−1
.
3 − x2
3 − x2
≥
,
2
2x
the conclusion follows. Thus, the proof is completed. The equality holds for a = b =
c = 1.
P 2.68. Let a, b, c be nonnegative real numbers such that a + b + c = 2. If 2 ≤ k ≤ 3, than
a k (b + c) + b k (c + a) + c k (a + b) ≤ 2.
Solution. Denote by Ek (a, b, c) the left hand side of the inequality, assume that a ≤ b ≤
c and show that
Ek (a, b, c) ≤ Ek (0, a + b, c) ≤ 2.
The left inequality is equivalent to
a b k−1
(a
+ b k−1 ) ≤ (a + b)k − a k − b k .
c
Clearly, it suffices to consider c = b, when the inequality becomes
2a k + b k−1 (a + b) ≤ (a + b)k .
Since 2a k ≤ a k−1 (a + b), it remains to show that
a k−1 + b k−1 ≤ (a + b)k−1 ,
358
Vasile Cîrtoaje
which is true since

‹k−1
b
a
b
a k−1 + b k−1 a k−1
=
+
≤
+
= 1.
k−1
(a + b)
a+b
a+b
a+b a+b
The right inequality, Ek (0, a + b, c) ≤ 2, is equivalent to
cd(c k−1 + d k−1 ) ≤ 2,
where d = a + b, and hence c + d = 2. By the Power-Mean inequality (or Jensen’s
k−1
inequality applied to the concave function t 2 ), we have
c k−1 + d k−1
2
c
k−1
+d
1/(k−1)
k−1
≤
c2 + d 2
2
c2 + d 2
≤2
2
1/2
,
(k−1)/2
.
Thus, it suffices to show that
cd
which is equivalent to
c2 + d 2
2
(k−1)/2
≤ 1,
cd(2 − cd)(k−1)/2 ≤ 1.
Since 2 − cd ≥ 1, we have
cd(2 − cd)(k−1)/2 ≤ cd(2 − cd) = 1 − (1 − cd)2 ≤ 1.
The equality holds for a = 0 and b = c = 1 (or any cyclic permutation).
P 2.69. Let a, b, c be nonnegative real numbers, no two of which are zero. If m > n ≥ 0,
than
bm + c m
c m + am
am + bm
(b
+
c
−
2a)
+
(c
+
a
−
2b)
+
(a + b − 2c) ≥ 0.
bn + c n
c n + an
an + bn
(Vasile Cîrtoaje, 2006)
Solution. Write the inequality as
AX + BY + C Z ≥ 0,
Symmetric Nonrational Inequalities
where
359
c m + am
am + bm
bm + c m
,
B
=
,
C
=
,
bn + c n
c n + an
an + bn
X = b + c − 2a, Y = c + a − 2b, Z = a + b − 2c, X + Y + Z = 0.
A=
Without loss of generality, assume that a ≤ b ≤ c, which involves X ≥ Y ≥ Z and X ≥ 0.
Since
2(AX + BY + C Z) = (2A − B − C)X + (B + C)X + 2(BY + C Z)
= (2A − B − C)X − (B + C)(Y + Z) + 2(BY + C Z)
= (2A − B − C)X + (B − C)(Y − Z),
it suffices to show that B ≥ C and 2A − B − C ≥ 0. The inequality B ≥ C can be written
as
b n c n (c m−n − b m−n ) + a n (c m − b m ) − a m (c n − b n ) ≥ 0,
b n c n (c m−n − b m−n ) + a n [c m − b m − a m−n (c n − b n )] ≥ 0.
This is true since c m−n ≥ b m−n and
c m − b m − a m−n (c n − b n ) ≥ c m − b m − b m−n (c n − b n ) = c n (c m−n − b m−n ) ≥ 0.
The inequality 2A − B − C ≥ 0 follows from
2A ≥ b m−n + c m−n ,
b m−n ≥ C,
c m−n ≥ B.
Indeed, we have
2A − b m−n − c m−n =
b m−n − C =
(b n − c n )(b m−n − c m−n )
≥ 0,
bn + c n
a n (b m−n − a m−n )
≥ 0,
an + bn
a n (c m−n − a m−n )
≥ 0.
c n + an
The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).
c m−n − B =
P 2.70. Let a, b, c be positive real numbers such that a bc = 1. Prove that
p
p
p
a2 − a + 1 + b2 − b + 1 + c 2 − c + 1 ≥ a + b + c.
(Vasile Cîrtoaje, 2012)
360
Vasile Cîrtoaje
First Solution. Among a − 1, b − 1 and c − 1 there are two with the same sign. Let
(b − 1)(c − 1) ≥ 0, that is,
1
t ≤ , t = b + c − 1.
a
By Minkowsky’s inequality, we have
v
v
u
u
‹2
‹
p
p
t
3 t
1
1 2 3
2
2
+ +
+
b − b+1+ c −c+1=
b−
c−
2
4
2
4
p
≥ t 2 + 3.
Thus, it suffices to show that
p
p
a2 − a + 1 + t 2 + 3 ≥ a + b + c,
which is equivalent to
p
a2 − a + 1 + f (t) ≥ a + 1,
where
f (t) =
p
t 2 + 3 − t.
Clearly, f (t) is decreasing for t ≤ 0. Since
f (t) = p
3
,
t2 + 3 + t
 ‹
1
, and it suffices to show that
f (t) is also decreasing for t ≥ 0. Then, f (t) ≥ f
a
 ‹
p
1
a2 − a + 1 + f
≥ a + 1,
a
which is equivalent to
p
v
t1
1
a2 − a + 1 +
+ 3 ≥ a + + 1.
a2
a
By squaring, this inequality becomes
v

‹
t
1
2
2
2 (a − a + 1) 2 + 3 ≥ 3a + − 1.
a
a
Indeed, by the Cauchy-Schwarz inequality, we have
v

‹ v

‹
t
t
1
1
2
2 (a − a + 1) 2 + 3 = [(2 − a)2 + 3a2 ] 2 + 3
a
a
2−a
2
≥
+ 3a = 3a + − 1.
a
a
Symmetric Nonrational Inequalities
361
The equality holds for a = b = c.
Second Solution. If the inequality
p
x2
1
− x +1− x ≥
2

3
−1
2
x + x +1
‹
holds for all x > 0, then it suffices to prove that
1
1
1
+
+
≥ 1,
a2 + a + 1 b2 + b + 1 c 2 + c + 1
which is just the known inequality in P 1.44. Indeed, the above inequality is equivalent
to
1− x
(1 − x)(2 + x)
,
≥
p
2
2(x 2 + x + 1)
x − x +1+ x
p
(x − 1)[(x + 2) x 2 − x + 1 − x 2 − 2] ≥ 0,
3x 2 (x − 1)2
≥ 0.
p
(x + 2) x 2 − x + 1 + x 2 + 2
P 2.71. Let a, b, c be positive real numbers such that a bc = 1. Prove that
p
p
p
16a2 + 9 + 16b2 + 9 + 16b2 + 9 ≥ 4(a + b + c) + 3.
(MEMO, 2012)
First Solution (by Vo Quoc Ba Can). Since
p
16a2 + 9 − 4a = p
9
16a2
+ 9 + 4a
,
the inequality is equivalent to
1
X
p
16a2 + 9 + 4a
≥
1
.
3
By the AM-GM inequality, we have
2
p
16a2 + 9 ≤
16a2 + 9
+ 2a + 3,
2a + 3
p
16a2 + 9
18(2a2 + 2a + 1)
2( 16a2 + 9 + 4a) ≤
+ 10a + 3 =
.
2a + 3
2a + 3
362
Vasile Cîrtoaje
Thus, it suffices to show that
X
2a + 3
≥ 3.
2a2 + 2a + 1
If the inequality
2a + 3
3
≥ 8/5
2a2 + 2a + 1
a + a4/5 + 1
holds for all a > 0, then it suffices to show that
X
1
≥ 1,
8/5
a + a4/5 + 1
which follows immediately from the inequality in P 1.44. Therefore, using the substitution x = a1/5 , x > 0, we need to show that
3
2x 5 + 3
≥ 8
,
10
5
2x + 2x + 1
x + x4 + 1
which is equivalent to
2x 4 (x 5 − 3x 2 + x + 1) + x 4 − 4x + 3 ≥ 0.
This is true since, by the AM-GM inequality, we have
p
3
x 5 + x + 1 ≥ 3 x 5 · x · 1 = 3x 2
and
x4 + 3 = x4 + 1 + 1 + 1 ≥ 4
p
4
x 4 · 1 · 1 · 1 = 4x.
The equality holds for a = b = c = 1.
Second Solution. Making the substitution
p
p
p
x = 16a2 + 9 − 4a, y = 16b2 + 9 − 4b, z = 16c 2 + 9 − 4c,
which involves
a=
9 − x2
,
8x
b=
9 − y2
,
8y
c=
9 − z2
,
8z
we need to show that
(9 − x 2 )(9 − y 2 )(9 − z 2 ) = 512x yz
yields
x + y + z ≥ 3.
Use the contradiction method. Assume that x + y + z < 3 and show that
(9 − x 2 )(9 − y 2 )(9 − z 2 ) > 512x yz.
x, y, z > 0,
Symmetric Nonrational Inequalities
363
According to the AM-GM inequality, we get
p
p
p
4
4
3 + x = 1 + 1 + 1 + x ≥ 4 x, 3 + y ≥ 4 4 y, 3 + z ≥ 4 z,
hence
p
(3 + x)(3 + y)(3 + z) ≥ 64 4 x yz.
Therefore, it suffices to prove that
(3 − x)(3 − y)(3 − z) > 8(x yz)3/4 .
By the AM-GM inequality,
1>
x + y + z 3
≥ x yz,
3
and hence
(3 − x)(3 − y)(3 − z) = 9(3 − x − y − z) + 3(x y + yz + z x) − x yz
> 3(x y + yz + z x) − x yz ≥ 9(x yz)2/3 − x yz
> 8(x yz)2/3 > 8(x yz)3/4 .
P 2.72. Let a, b, c be positive real numbers such that a bc = 1. Prove that
p
p
p
25a2 + 144 + 25b2 + 144 + 25c 2 + 144 ≤ 5(a + b + c) + 24.
(Vasile Cîrtoaje, 2012)
First Solution. Since
p
144
25a2 + 144 − 5a = p
,
25a2 + 144 + 5a
the inequality is equivalent to
X
1
1
≤ .
p
6
25a2 + 144 + 5a
If the inequality
1
1
≤ p
p
25a2 + 144 + 5a
6 5a18/13 + 4
holds for all a > 0, then it suffices to show that
1
X
p
5a18/13 + 4
≤ 1,
364
Vasile Cîrtoaje
which follows immediately from P 2.31. Therefore, using the substitution x = a1/13 ,
x > 0, we only need to show that
p
p
25x 26 + 144 + 5x 13 ≥ 6 5x 18 + 4.
By squaring, the inequality becomes
p
10x 13 ( 25x 26 + 144 + 5x 13 − 18x 5 ) ≥ 0.
This is true if
25x 26 + 144 ≥ (18x 5 − 5x 13 )2 ,
which is equivalent to
5x 18 + 4 ≥ 9x 10 .
By the AM-GM inequality, we have
5x 18 + 4 = x 18 + x 18 + x 18 + x 18 + x 18 + 1 + 1 + 1 + 1
p
9
≥ 9 x 18 · x 18 · x 18 · x 18 · x 18 · 1 · 1 · 1 · 1 = 9x 10 .
The equality holds for a = b = c = 1.
Second Solution. Making the substitution
p
p
p
8x = 25a2 + 144 − 5a, 8 y = 25b2 + 144 − 5b, 8z = 25c 2 + 144 − 5c,
which involves
9 − 4x 2
a=
,
5x
9 − 4 y2
b=
,
5y
9 − 4z 2
c=
,
5z

‹
3
x, y, z ∈ 0,
,
2
we need to show that
(9 − 4x 2 )(9 − 4 y 2 )(9 − 4z 2 ) = 125x yz
involves
x + y + z ≤ 3.
Use the contradiction method. Assume that x + y + z > 3 and show that
(9 − 4x 2 )(9 − 4 y 2 )(9 − 4z 2 ) < 125x yz.
Since
9 − 4x 2 < 3(x + y + z) −
3( y + z − x)( y + z + 3x)
12x 2
=
,
x + y +z
x + y +z
it suffices to show that
27AB ≤ 125x yz(x + y + z)3 ,
Symmetric Nonrational Inequalities
365
where
A = ( y + z − x)(z + x − y)(x + y − z),
B = ( y + z + 3x)(z + x + 3 y)(x + y + 3z).
Consider the nontrivial case A ≥ 0. By the AM-GM inequality, we have
B≤
125
(x + y + z)3 .
27
Therefore, it suffices to show that
A ≤ x yz,
which is a well known inequality (equivalent to Schur’s inequality of degree three).
P 2.73. If a, b are positive real numbers such that a b + bc + ca = 3, then
p
(a)
p
(b)
a2 + 3 +
a+b+
p
p
b2 + 3 +
b+c+
p
p
b2 + 3 ≥ a + b + c + 3;
c+a≥
p
4(a + b + c) + 6.
(Lee Sang Hoon, 2007)
Solution. (a) First Solution (by Pham Thanh Hung). By squaring, the inequality becomes
XÆ
(b2 + 3)(c 2 + 3) ≥ 3(1 + a + b + c).
Since
(b2 + 3)(c 2 + 3) = (b + c)(b + a)(c + a)(c + b) = (b + c)2 (a2 + 3) ≥
1
(b + c)2 (a + 3)2 ,
4
we have
XÆ
X Š
1X
1€ X
(b2 + 3)(c 2 + 3) ≥
(b + c)(a + 3) =
2
bc + 6
a = 3(1 + a + b + c).
2
2
The equality holds for a = b = c = 1.
Second Solution. Write the inequality as follows:
Æ
Æ
Æ
(a + b)(a + c) + (b + c)(b + a) + (c + a)(c + b) ≥ a + b + c + 3,
”
— X €p
Š2
Æ
p
2 a + b + c − 3(a b + bc + ca) ≥
a+b− a+c ,
1
a+b+c+
p
3(a b + bc + ca)
X
X
(b − c)2 ≥
p
(b − c)2
2 ,
p
a+b+ a+c
366
Vasile Cîrtoaje
X
where
€p
Sa =
a+b+
p
Sa (b − c)2
2 ≥ 0,
p
p
a+b+ a+c
a+c
Š2
−a−b−c−
Æ
(a + b)(a + c) −
Æ
3(a b + bc + ca).
The last inequality is true since
Sa = 3(a + b + c) + 2
>2
Æ
Æ
a2 + (a b + bc + ca) −
Æ
3(a b + bc + ca)
3(a b + bc + ca) > 0.
Third Solution. Use the substitution
p
p
p
x = a2 + 3 − a, y = b2 + 3 − b, z = c 2 + 3 − c,
x, y, z > 0.
We need to show that
x + y + z ≥ 3.
We have
X
X Æ
Æ
[ (b + a)(b + c) − b][ (c + a)(c + b) − c]
X
X Æ
X Æ
X
Æ
=
(b + c) (a + b)(a + c) −
b (c + a)(c + b) −
c (b + a)(b + c) +
bc
X
=
bc = 3.
yz =
Thus, we get
x + y +z ≥
Æ
3(x y + yz + z x) = 3.
(b) By squaring, we get the inequality in (a). Otherwise, using the substitution
x=
p
b + c,
y=
p
c + a, z =
p
a + b,
the inequality becomes
x + y +z ≥
Ç
2(x 2 + y 2 + z 2 ) +
Æ
3(2x 2 y 2 + 2 y 2 z 2 + 2z 2 x 2 − x 4 − y 4 − z 4 ).
By squaring two times, we get
Æ
2(x y + yz + z x) − x 2 − y 2 − z 2 ≥ 3(2x 2 y 2 + 2 y 2 z 2 + 2z 2 x 2 − x 4 − y 4 − z 4 ),
X
(x − y)2 (x + y − z)2 ≥ 0.
Symmetric Nonrational Inequalities
367
P 2.74. If a, b, c are nonnegative real numbers such that a + b + c = 3, then
Æ
(5a2 + 3)(5b2 + 3) +
Æ
(5b2 + 3)(5c 2 + 3) +
Æ
(5c 2 + 3)(5a2 + 3) ≥ 24.
(Nguyen Van Quy, 2012)
Solution. Assume that a ≥ b ≥ c, which involves 1 ≤ a ≤ 3 and b + c ≤ 2. Denote
A = 5a2 + 3,
B = 5b2 + 3,
C = 5c 2 + 3,
and write the inequality as follows:
p p
p
p
A ( B + C) + BC ≥ 24,
p q
p
p
A · A(B + C + 2 BC ) ≥ 24 − BC.
This is true if
p
p
A(B + C + 2 BC) ≥ (24 − BC)2 ,
which is equivalent to
A(A + B + C + 48) ≥ (A + 24 −
p
BC)2 .
Applying Lemma below for k = 5/3 and m = 4/15 yields
p
5 BC ≥ 4(b − c)2 + 25bc + 15.
Therefore, it suffices to show that
25A(A + B + C + 48) ≥ [5A + 120 − 4(b − c)2 − 25bc − 15]2 ,
which is equivalent to
25(5a2 + 3)[5(a2 + b2 + c 2 ) + 57] ≥ [25a2 + 120 − 4(b − c)2 − 25bc]2 .
Since
5(a2 + b2 + c 2 ) + 57 = 5a2 + 5(b + c)2 − 10bc + 57 = 2(5a2 − 15a + 51 − 5bc)
and
25a2 + 120 − 4(b − c)2 − 25bc = 25a2 + 120 − 4(b + c)2 − 9bc = 3(7a2 + 8a + 28 − 3bc),
we need to show that
50(5a2 + 3)(5a2 − 15a + 51 − 5bc) ≥ 9(7a2 + 8a + 28 − 3bc)2 .
368
Vasile Cîrtoaje
From bc ≤ (b + c)2 /4 and (a − b)(a − c) ≥ 0, we get
bc ≤ (3 − a)2 /4,
bc ≥ a(b + c) − a2 = 3a − 2a2 .
Consider a fixed, a ≥ 1, and denote x = bc. So, we only need to prove that f (x) ≥ 0
for
a2 − 6a + 9
,
3a − 2a2 ≤ x ≤
4
where
f (x) = 50(5a2 + 3)(5a2 − 15a + 51 − 5x) − 9(7a2 + 8a + 28 − 3x)2 .
2
a − 6a + 9
2
Since f is concave, it suffices to show that f (3a − 2a ) ≥ 0 and f
≥ 0.
4
Indeed,
f (3a − 2a2 ) = 3(743a4 − 2422a3 + 2813a2 − 1332a + 198)
= 3(a − 1)2 [(a − 1)(743a − 193) + 5] ≥ 0,
f
a2 − 6a + 9
4
375
(25a4 − 140a3 + 286a2 − 252a + 81)
16
375
=
(a − 1)2 (5a − 9)2 ≥ 0.
16
=
Thus, the proof is completed. The equality holds for a = b = c = 1, and also for a = 9/5
and b = c = 3/5 (or any cyclic permutation).
Lemma. Let b, c ≥ 0 such that b + c ≤ 2. If k > 0 and 0 ≤ m ≤
Æ
k
, then
2k + 2
(k b2 + 1)(kc 2 + 1) ≥ m(b − c)2 + k bc + 1.
Proof. By squaring, the inequality becomes
(b − c)2 [k − 2m − 2kmbc − m2 (b − c)2 ] ≥ 0.
This is true since
k − 2m − 2kmbc − m2 (b − c)2 = k − 2m − 2m(k − 2m)bc − m2 (b + c)2
m(k − 2m)
(b + c)2 − m2 (b + c)2
2
km
= k − 2m −
(b + c)2 ≥ k − 2m − 2km ≥ 0.
2
≥ k − 2m −
Symmetric Nonrational Inequalities
369
P 2.75. If a, b, c are nonnegative real numbers such that a + b + c = 3, then
p
a2 + 1 +
p
b2 + 1 +
p
c2 + 1 ≥
v
t 4(a2 + b2 + c 2 ) + 42
3
.
(Vasile Cîrtoaje, 2014)
Solution. Assume that a ≥ b ≥ c, which involves a ≥ 1 and b + c ≤ 2. By squaring, the
inequality becomes
p p
p
p
a2 + b2 + c 2 + 33
A ( B + C) + BC ≥
,
6
q
p
p
a2 + b2 + c 2 + 33
A(B + C + 2 BC ) + BC ≥
,
6
where
A = a2 + 1,
B = b2 + 1,
C = c 2 + 1.
Applying Lemma from the preceding problem P 2.74 for k = 1 and m = 1/4 gives
p
BC ≥
1
(b − c)2 + bc + 1.
4
Therefore, it suffices to show that
v
t
1
1
a2 + b2 + c 2 + 33
A[B + C + (b − c)2 + 2bc + 2] + (b − c)2 + bc + 1 ≥
,
2
4
6
which is equivalent to
6
Æ
6
2(a2 + 1)[3(b + c)2 + 8 − 4bc] ≥ 2a2 − (b + c)2 + 54 − 4bc,
Æ
2(a2 + 1)(3a2 − 18a + 35 − 4bc) ≥ a2 + 6a + 45 − 4bc.
From bc ≤ (b + c)2 /4 and (a − b)(a − c) ≥ 0, we get
bc ≤ (3 − a)2 /4,
bc ≥ a(b + c) − a2 = 3a − 2a2 .
Consider a fixed, a ≥ 1, and denote x = bc. So, we only need to prove that f (x) ≥ 0
for
a2 − 6a + 9
3a − 2a2 ≤ x ≤
,
4
where
f (x) = 72(a2 + 1)(3a2 − 18a + 35 − 4x) − (a2 + 6a + 45 − 4x)2 .
370
Vasile Cîrtoaje
Since f is concave, it suffices to show that f (3a − 2a ) ≥ 0 and f
2
a2 − 6a + 9
4
≥ 0.
Indeed,
f (3a − 2a2 ) = 9(79a4 − 228a3 + 274a2 − 180a + 55)
= 9(a − 1)2 (79a2 − 70a + 55 ≥ 0,
f
a2 − 6a + 9
4
= 144(a4 − 6a3 + 13a2 − 12a + 4)
= 144(a − 1)2 (a − 2)2 ≥ 0.
The equality holds for a = b = c = 1, and also for a = 2 and b = c = 1/2 (or any cyclic
permutation).
P 2.76. If a, b, c are nonnegative real numbers such that a + b + c = 3, then
p
p
p
p
(a)
a2 + 3 + b2 + 3 + c 2 + 3 ≥ 2(a2 + b2 + c 2 ) + 30;
p
p
p
p
(b)
3a2 + 1 + 3b2 + 1 + 3c 2 + 1 ≥ 2(a2 + b2 + c 2 ) + 30.
(Vasile Cîrtoaje, 2014)
Solution. Assume that a ≥ b ≥ c, which involves a ≥ 1 and b + c ≤ 2.
(a) By squaring, the inequality becomes
p p
p
p
a2 + b2 + c 2 + 21
A ( B + C) + BC ≥
,
2
q
p
p
a2 + b2 + c 2 + 21
A(B + C + 2 BC ) + BC ≥
,
2
where
A = a2 + 3,
B = b2 + 3,
C = c 2 + 3.
Applying Lemma from problem P 2.74 for k = 1/3 and m = 1/9 gives
p
BC ≥
1
(b − c)2 + bc + 3.
3
Therefore, it suffices to show that
v
t
2
1
a2 + b2 + c 2 + 21
A[B + C + (b − c)2 + 2bc + 6] + (b − c)2 + bc + 3 ≥
,
3
3
2
Symmetric Nonrational Inequalities
371
which is equivalent to
2
Æ
3(a2 + 3)[5(b + c)2 + 36 − 8bc] ≥ 3a2 + (b + c)2 + 45 − 4bc,
Æ
3(a2 + 3)(5a2 − 30a + 81 − 8bc) ≥ 2a2 − 3a + 27 − 2bc.
From bc ≤ (b + c)2 /4 and (a − b)(a − c) ≥ 0, we get
bc ≤ (3 − a)2 /4,
bc ≥ a(b + c) − a2 = 3a − 2a2 .
Consider a fixed, a ≥ 1, and denote x = bc. So, we only need to prove that f (x) ≥ 0
for
a2 − 6a + 9
3a − 2a2 ≤ x ≤
,
4
where
f (x) = 3(a2 + 3)(5a2 − 30a + 81 − 8x) − (2a2 − 3a + 27 − 2x)2 .
2
a − 6a + 9
2
Since f is concave, it suffices to show that f (3a − 2a ) ≥ 0 and f
≥ 0.
4
Indeed,
f (3a − 2a2 ) = 27a2 (a − 1)2 ≥ 0,
f
a2 − 6a + 9
4
27 4
(a − 8a3 + 22a2 − 24a + 9)
4
27
=
(a − 1)2 (a − 3)2 ≥ 0.
4
=
The equality holds for a = b = c = 1, and also for a = 3 and b = c = 0 (or any cyclic
permutation).
(b) By squaring, the inequality becomes
p p
p
p
27 − a2 − b2 − c 2
A ( B + C) + BC ≥
,
2
q
p
p
27 − a2 − b2 − c 2
A(B + C + 2 BC ) + BC ≥
,
2
where
A = 3a2 + 1,
B = 3b2 + 1,
C = 3c 2 + 1.
Applying Lemma from problem P 2.74 for k = 3 and m = 1/3 gives
p
BC ≥
1
(b − c)2 + 3bc + 1.
3
372
Vasile Cîrtoaje
Therefore, it suffices to show that
v
t
1
27 − a2 − b2 − c 2
2
,
A[B + C + (b − c)2 + 6bc + 2] + (b − c)2 + 3bc + 1 ≥
3
3
2
which is equivalent to
Æ
2 3(3a2 + 1)[11(b + c)2 + 12 − 8bc] ≥ 75 − 3a2 − 5(b + c)2 − 4bc,
Æ
3(3a2 + 1)(11a2 − 66a + 111 − 8bc) ≥ 15 + 15a − 4a2 − 2bc.
From bc ≤ (b + c)2 /4 and (a − b)(a − c) ≥ 0, we get
bc ≤ (3 − a)2 /4,
bc ≥ a(b + c) − a2 = 3a − 2a2 .
Consider a fixed, a ≥ 1, and denote x = bc. So, we only need to prove that f (x) ≥ 0
for
a2 − 6a + 9
,
3a − 2a2 ≤ x ≤
4
where
f (x) = 3(3a2 + 1)(11a2 − 66a + 111 − 8x) − (15 + 15a − 4a2 − 2x)2 .
2
a − 6a + 9
2
≥ 0.
Since f is concave, it suffices to show that f (3a − 2a ) ≥ 0 and f
4
Indeed,
f (3a − 2a2 ) = 27(a − 1)2 (3a − 2)2 ≥ 0,
f
a2 − 6a + 9
4
27
(9a4 − 48a3 + 94a2 − 80a + 25)
4
27
=
(a − 1)2 (3a − 5)2 ≥ 0.
4
=
The equality holds for a = b = c = 1, and also for a = 5/3 and b = c = 2/3 (or any
cyclic permutation).
Remark. Similarly, we can prove the following generalization.
• Let a, b, c be nonnegative real numbers such that a + b + c = 3. If k > 0, then
v
t 8k(a2 + b2 + c 2 ) + 3(9k2 + 10k + 9)
p
p
p
ka2 + 1 + k b2 + 1 + kc 2 + 1 ≥
,
3(k + 1)
with equality for a = b = c = 1, and also for a =
cyclic permutation).
3k − 1
3k + 1
and b = c =
(or any
2k
4k
Symmetric Nonrational Inequalities
373
P 2.77. If a, b, c are nonnegative real numbers such that a + b + c = 3, then
Æ
Æ
Æ
(32a2 + 3)(32b2 + 3) + (32b2 + 3)(32c 2 + 3) + (32c 2 + 3)(32a2 + 3) ≤ 105.
(Vasile Cîrtoaje, 2014)
Solution. Assume that a ≤ b ≤ c, which involves a ≤ 1 and b + c ≥ 2. Denote
A = 32a2 + 3,
B = 32b2 + 3,
C = 32c 2 + 3,
and write the inequality as follows:
p p
p
p
A ( B + C) + BC ≤ 105,
p Æ
p
p
A · B + C + 2 BC ≤ 105 − BC.
By Lemma below, we have
p
BC ≤ 5(b + c)2 + 12bc + 3 ≤ 8(b + c)2 + 3 ≤ 8(a + b + c)2 + 3 = 75 < 105.
Therefore, we can write the desired inequality as
p
p
A(B + C + 2 BC) ≤ (105 − BC)2 ,
which is equivalent to
A(A + B + C + 210) ≤ (A + 105 −
p
BC)2 .
According to Lemma below, it suffices to show that
A(A + B + C + 210) ≤ [A + 105 − 5(b2 + c 2 ) − 22bc − 3]2 ,
which is equivalent to
[32a2 + 105 − 5(b2 + c 2 ) − 22bc]2 ≥ (32a2 + 3)[32(a2 + b2 + c 2 ) + 219].
Since
32(a2 + b2 + c 2 ) + 219 = 32a2 + 32(b + c)2 − 64bc + 219 = 64a2 − 192a + 507 − 64bc
and
32a2 +105−5(b2 +c 2 )−22bc = 32a2 +105−5(b+c)2 −12bc = 3(9a2 +10a+20−4bc),
we need to show that
9(9a2 + 10a + 20 − 4bc)2 ≥ (32a2 + 3)(64a2 − 192a + 507 − 64bc).
374
Vasile Cîrtoaje
From bc ≤ (b + c)2 /4, we get
bc ≤ (3 − a)2 /4.
Consider a fixed, 0 ≤ a ≤ 1, and denote x = bc. So, we only need to prove that f (x) ≥ 0
for
a2 − 6a + 9
x≤
,
4
where
f (x) = 9(9a2 + 10a + 20 − 4x)2 − (32a2 + 3)(64a2 − 192a + 507 − 64x).
Since
f 0 (x) = 72(4x − 9a2 − 10a − 20) + 64(32a2 + 3)
≤ 72[(a2 − 6a + 9) − 9a2 − 10a − 20) + 64(32a2 + 3)
= 8[184a(a − 1) + (44a − 75)] < 0,
2
a − 6a + 9
f is decreasing, hence f (x) ≥ f
. Therefore, it suffices to show that
4
2
a − 6a + 9
f
≥ 0. We have
4
2
a − 6a + 9
f
=9[9a2 + 10a + 20 − (a2 − 6a + 9)]2
4
− (32a2 + 3)[64a2 − 192a + 507 − 16(a2 − 6a + 9)]
=9(8a2 + 16a + 11)2 − (32a2 + 3)(48a2 − 96a + 363)
=192a(a − 1)2 (18 − 5a) ≥ 0.
Thus, the proof is completed. The equality holds for a = b = c = 1, and also for a = 0
and b = c = 3/2 (or any cyclic permutation).
Lemma. If b, c ≥ 0 such that b + c ≥ 2, then
Æ
(32b2 + 3)(32c 2 + 3) ≤ 5(b2 + c 2 ) + 22bc + 3.
Proof. By squaring, the inequality becomes
(5b2 + 5c 2 + 22bc)2 − 322 b2 c 2 ≥ 96(b2 + c 2 ) − 6(5b2 + 5c 2 + 22bc),
5(b − c)2 (5b2 + 5c 2 + 54bc) ≥ 66(b − c)2 .
It suffices to show that
5(5b2 + 5c 2 + 10bc) ≥ 100,
which is equivalent to the obvious inequality (b + c)2 ≥ 4.
Symmetric Nonrational Inequalities
375
P 2.78. If a, b, c are positive real numbers, then
c + a
b+c
a
+
b
a − 3 + b − 3 + c − 3 ≥ 2.
(Vasile Cîrtoaje, 2012)
Solution. Without loss of generality, assume that a ≥ b ≥ c.
Case 1: a > b + c. We have
b+c
a + b
c + a
b+c
b+c
a − 3 + c − 3 + b − 3 ≥ a − 3 = 3 − a > 2.
Case 2: a ≤ b + c. We have
b+c
a + b
c + a
c + a
b+c
+
+
+
−
3
−
3
−
3
≥
−
3
−
3
a
c
b
a
b

‹ b+c
c + a
2b b + a
(a − b)(2b − a)
= 3−
+ 3−
≥6−
−
=2+
≥ 2.
a
b
a
b
ab
a
Thus, the proof is completed. The equality holds for = b = c (or any cyclic permuta2
tion).
P 2.79. If a, b, c are real numbers such that a bc 6= 0, then
b + c c + a a + b +
≥ 2.
+
a b
c First Solution. Let
|a| = max{|a|, |b|, |c|}.
We have
b + c c + a a + b b + c c + a a + b +
≥
+
+
+
a b
c a a
a ≥
|(−b − c) + (c + a) + (a + b)|
= 2.
|a|
The equality holds for a = 1, b = −1 and |c| ≤ 1 (or any permutation).
Second Solution. Since the inequality remains unchanged by replacing a, b, c with
−a, −b, −c, it suffices to consider two cases: a, b, c > 0, and a < 0 and b, c > 0.
376
Vasile Cîrtoaje
Case 1: a, b, c > 0. We have
b + c c + a a + b  a b ‹  b c ‹ c a +
=
+
+
+
+
+
≥ 6.
+
a b
c b a
c
b
a c
Case 2: a < 0 and b, c > 0. Replacing a by −a, we need to show that
b + c |a − c| |a − b|
+
+
≥2
a
b
c
for all a, b, c > 0. Without loss of generality, assume that b ≥ c.
For b ≥ c ≥ a, we have
b + c |a − c| |a − b|
b+c
+
+
≥
≥ 2.
a
b
c
a
For b ≥ a ≥ c, we have
b + c |a − c| |a − b|
b+c a−c
(a − b)2 + c(b − a)
+
+
−2≥
+
−2=
≥ 0.
a
b
c
a
b
ab
For a ≥ b ≥ c, we have
b+c a−c a−b
b + c |a − c| |a − b|
+
+
=
+
+
a
b
c
a
b
c

‹

‹
a b
a−b
1 1
(a − b)2 (a − b)(a b − c 2 )
+ −2 +
+c
−
=
+
≥ 0.
=
b a
c
a b
ab
a bc
Third Solution. Using the substitution
x=
b+c
,
a
y=
c+a
a+b
, z=
,
b
c
we need to show that
x + y + z + 2 = x yz,
x, y, z ∈ R
involves
|x| + | y| + |z| ≥ 2.
If x yz ≤ 0, then
−x − y − z = 2 − x yz ≥ 2,
hence
|x| + | y| + |z| ≥ |x + y + z| = | − x − y − z| ≥ −x − y − z ≥ 2.
If x yz > 0, then either x, y, z > 0 or only one of x, y, z is positive (for instance, x > 0
and y, z < 0).
Symmetric Nonrational Inequalities
377
Case 1: x, y, z > 0. We need to show that x + y + z ≥ 2. We have
x yz = x + y + z + 2 > 2
and, by the AM-GM inequality, we get
p
p
3
x + y + z ≥ 3 3 x yz > 3 2 > 2,
Case 2: x > 0 and y, z < 0. Replacing y, z by − y, −z, we need to prove that
x − y − z + 2 = x yz
involves
x + y +z ≥2
for all x, y, z > 0. Since
x + y + z − 2 = x + y + z − (x yz − x + y + z) = x(2 − yz),
we need to show that yz ≤ 2. Indeed, we have
p
x + 2 = y + z + x yz ≥ 2 yz + x yz,
p
x(1 − yz) + 2(1 − yz ) ≥ 0,
p
p
(1 − yz )[x(1 + yz ) + 2] ≥ 0,
hence
yz ≤ 1 < 2.
P 2.80. Let a, b, c be nonnegative real numbers, no two of which are zero, and let
x=
2a
,
b+c
y=
2b
2c
, z=
.
c+a
a+b
Prove that
(a)
(b)
x + y +z+
p
x+
p
p
xy+
y+
p
yz +
p
z x ≥ 6;
p
p
z ≥ 8 + x yz.
378
Vasile Cîrtoaje
Solution. (a) First Solution. Since
p
p
p
2 bc(a + b)(c + a) 2 bc (a + bc)
p
yz =
≥
(a + b)(c + a)
(a + b)(c + a)
p
2a(b + c) bc + 2bc(b + c)
4a bc + 2bc(b + c)
≥
,
=
(a + b)(b + c)(c + a)
(a + b)(b + c)(c + a)
we have
Xp
yz ≥
12a bc + 2
P
bc(b + c)
(a + b)(b + c)(c + a)
8a bc
+ 2 = x yz + 2.
=
(a + b)(b + c)(c + a)
Therefore, it suffices to show that
x + y + z + x yz ≥ 4,
which is equivalent to Schur’s inequality of degree three
X
a3 + b3 + c 3 + 3a bc ≥
a b(a + b).
The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).
Second Solution. Write the inequality as
X
Xp
p
4
(x − 1) ≥
( y − z)2 .
Since
X
X (a − b) + (a − c) X a − b X b − a X (a − b)2
=
+
=
(x − 1) =
b+c
b+c
c+a
(b + c)(c + a)
=
and
p
p
( y − z)2 =
X
(b − c)2
,
(a + b)(a + c)
2(b − c)2 (a + b + c)2
2 ,
p
p
(a + b)(a + c) b2 + a b + c 2 + ac
we can write the inequality as
X
(b − c)2 Ea ≥ 0,
where

Ea = (b + c) 2 − p

(a + b + c)
2  .
p
2
2
b + a b + c + ac
2
Symmetric Nonrational Inequalities
379
By Minkowski’s inequality, we have
€p
b2 + a b +
p
c 2 + ac
Therefore,
Š2
p p
≥ (b+c)2 +a( b+ c)2 ≥ (b+c)2 +a(b+c) = (b+c)(a+b+c).
‹

a+b+c
= b + c − a,
Ea ≥ (b + c) 2 −
b+c
and hence
X
X
(b − c)2 Ea ≥
(b − c)2 (b + c − a) ≥ 0.
The right inequality is just Schur’s inequality of third degree.
Third Solution. Using the Cauchy-Schwarz inequality, we have
a
b
c
(a + b + c)2
(a + b + c)2
+
+
≥
=
.
b+c c+a a+b
a(b + c) + b(c + a) + c(a + b) 2(a b + bc + ca)
Using Hölder’s inequality, we have
‚s
v
Œ2
s
t b
a
c
(a + b + c)3
+
+
≥ 2
.
b+c
c+a
a+b
a (b + c) + b2 (c + a) + c 2 (a + b)
Thus, it suffices to prove that
(a + b + c)2
2(a + b + c)3
+ 2
≥ 12.
a b + bc + ca a (b + c) + b2 (c + a) + c 2 (a + b)
Due to homogeneity, we may assume that a + b + c = 1. Substituting q = a b + bc + ca,
3q ≤ 1, the inequality becomes
1
2
+
≥ 12.
q q − 3a bc
From the fourth degree Schur’s inequality
6a bc p ≥ (p2 − q)(4q − p2 ),
p = a + b + c,
we obtain
6a bc ≥ (4q − 1)(1 − q).
Therefore,
1
2
1
+
− 12 ≥ +
q q − 3a bc
q q−
=
2
(4q−1)(1−q)
2
− 12
(1 − 3q)(1 − 4q)2
4
1
+ 2
− 12 =
≥ 0.
q 4q − 3q + 1
q(4q2 − 3q + 1)
380
Vasile Cîrtoaje
(b) By squaring, the inequality becomes
p
p
p
x + y + z + 2 x y + 2 yz + 2 z x ≥ 8 + x yz.
As we have shown in the first solution of the inequality in (a),
p
p
p
x y + yz + z x ≥ x yz + 2.
Thus, it suffices to show that
x + y + z + 2(x yz + 2) ≥ 8 + x yz,
which is equivalent to
x + y + z + x yz ≥ 4.
This inequality reduces to Schur’s inequality of third degree. The equality holds for
a = b = c, and also for a = 0 and b = c (or any cyclic permutation).
P 2.81. Let a, b, c be nonnegative real numbers, no two of which are zero, and let
x=
2a
,
b+c
y=
2b
2c
, z=
.
c+a
a+b
Prove that
p
1 + 24x +
p
1 + 24 y +
p
1 + 24z ≥ 15.
(Vasile Cîrtoaje, 2005)
Solution (by Vo Quoc Ba Can). Assume that c = min{a, b, c}, hence z = min{x, y, z},
z ≤ 1. By Hölder’s inequality
‚s
v
Œ2
t b
2
a
+
a (b + c) + b2 (c + a) ≥ (a + b)3 ,
b+c
c+a
we get
p
x+
p 2
y ≥
2(a + b)3
2(a + b)3
≥
c(a + b)2 + a b(a + b − 2c)
c(a + b)2 + 14 (a + b)2 (a + b − 2c)
8(a + b)
8
=
.
a + b + 2c
1+z
Using this result and Minkowski’s inequality, we have
=
p
1 + 24x +
p
1 + 24 y ≥
q
(1 + 1)2
v
t
p
48
p 2
+ 24( x + y) ≥ 2 1 +
.
1+z
Symmetric Nonrational Inequalities
381
Therefore, it suffices to show that
2
v
t
1+
p
48
+ 1 + 24z ≥ 15.
1+z
Using the substitution
p
1 + 24z = 5t,
1
≤ t ≤ 1,
5
the inequality turns into
v
t t 2 + 47
2
≥ 3 − t.
25t 2 + 23
By squaring, this inequality becomes
25t 4 − 150t 3 + 244t 2 − 138t + 19 ≤ 0,
which is equivalent to the obvious inequality
(t − 1)2 (5t − 1)(5t − 19) ≤ 0.
The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).
P 2.82. If a, b, c are positive real numbers, then
v
t
v
v
t
t
7a
7b
7c
+
+
≤ 3.
a + 3b + 3c
b + 3c + 3a
c + 3a + 3b
(Vasile Cîrtoaje, 2005)
First Solution. Making the substitution
x=
we have
v
t
7a
,
a + 3b + 3c
y=
v
t
v
t
7b
7c
, z=
,
b + 3c + 3a
c + 3a + 3b

(x 2 − 7)a + 3x 2 b + 3x 2 c = 0





3 y 2 a + ( y 2 − 7)b + 3 y 2 c = 0 ,


2
2
2


 3z a + 3z b + (z − 7)c = 0
382
Vasile Cîrtoaje
which involves
x2 − 7
3 y2
3z 2
3x 2
3x 2
y2 − 7 3 y2
3z 2
z2 − 7
=0 ;
that is,
F (x, y, z) = 0,
where
F (x, y, z) = 4x 2 y 2 z 2 + 8
X
x2 y2 + 7
X
x 2 − 49.
We need to show that F (x, y, z) = 0 involves x + y + z ≤ 3, where x, y, z > 0. To
do this, we use the contradiction method. Assume that x + y + z > 3 and show that
F (x, y, z) > 0. Since F (x, y, z) is strictly increasing in each of its arguments, it is enough
to prove that x + y + z = 3 involves F (x, y, z) ≥ 0. Assume that x = max{x, y, z} and
denote
y +z
t=
, 0 < t ≤ 1 ≤ x.
2
We will show that
F (x, y, z) ≥ F (x, t, t) ≥ 0.
We have
F (x, y, z) − F (x, t, t) = (8x 2 + 7)( y 2 + z 2 − 2t 2 ) − 4(x 2 + 2)(t 4 − y 2 z 2 )
1
= (8x 2 + 7)( y − z)2 − (x 2 + 2)(t 2 + yz)( y − z)2
2
1
≥ (8x 2 + 7)( y − z)2 − 2(x 2 + 2)( y − z)2
2
1
= (4x 2 − 1)( y − z)2 ≥ 0
2
and
3− x 3− x
F (x, t, t) = F x,
,
2
2

‹
=
1
(x − 1)2 (x − 2)2 (x 2 − 6x + 23) ≥ 0.
4
a
= b = c (or any cyclic permutation).
8
Second Solution. Due to homogeneity, we may assume that a + b + c = 3, when the
inequality becomes
v
X t 7a
≤ 3.
9 − 2a
The equality holds for a = b = c, and also for
Using the substitution
v
t 7a
x=
,
9 − 2a
v
v
t 7b
t 7c
y=
, z=
,
9 − 2b
9 − 2c
Symmetric Nonrational Inequalities
383
we need to show that if x, y, z are positive real number such that
X
x2
1
= ,
2
2x + 7 3
then
x + y + z ≤ 3.
For the sake of contradiction, assume that x + y + z > 3 and show that F (x, y, z) > 0,
where
X x2
1
F (x, y, z) =
− .
2
2x + 7 3
Since F (x, y, z) is strictly increasing in each of its arguments, it is enough to prove that
x + y + z = 3 involves F (x, y, z) ≥ 0. This is just the inequality in P 1.30. We give here
another proof. By the Cauchy-Schwarz inequality, we have
P
P
X x2
( x 3/2 )2
( x 3/2 )2
= P
≥P
.
2x 2 + 7
x(2x 2 + 7) 2 x 3 + 21
Therefore, it suffices to show that
X
X
3(
x 3/2 )2 ≥ 2
x 3 + 21,
which is equivalent to the homogeneous inequality
X
x3 + 6
X
7
(x y)3/2 ≥ (x + y + z)3 .
9
Replacing x, y, z by x 2 , y 2 , z 2 , we need to prove that G(x, y, z) ≥ 0, where
G(x, y, z) =
X
x6 + 6
X
7
x 3 y 3 − (x 2 + y 2 + z 2 )3 .
9
Assume that x = max{x, y, z} and denote
v
t y 2 + z2
t=
, 0 < t ≤ x.
2
We will show that
G(x, y, z) ≥ G(x, t, t) ≥ 0.
We have
G(x, y, z) − G(x, t, t) = y 6 + z 6 + 6 y 3 z 3 − 8t 4 + 6x 3 ( y 3 + z 3 − 2t 3 ).
Since y 3 + z 3 − 2t 3 ≥ 0, we get
6x 3 ( y 3 + z 3 − 2t 3 ) ≥ 4x 3 ( y 3 + z 3 − 2t 3 ) ≥ ( y 3 + z 3 + 2t 3 )( y 3 + z 3 − 2t 3 ).
384
Vasile Cîrtoaje
Thus,
G(x, y, z) − G(x, t, t) ≥ y 6 + z 6 + 6 y 3 z 3 − 8t 4 + [( y 3 + z 3 )2 − 4t 6 ]
= 2( y 6 + z 6 + 4 y 3 z 3 − 6t 6 ) = 2[( y 2 + z 2 )3 − 3 y 2 z 2 ( y 2 + z 2 ) + 4 y 3 z 3 − 6t 6 ]
= 4(t 6 − 3t 2 y 2 z 2 + 2 y 3 z 3 ) = 4(t 2 − yz)2 (t 2 + 2 yz) ≥ 0.
Also,
7
G(x, t, t) = x 6 + 2t 6 + 6(t 6 + 2x 3 t 3 ) − (x 2 + 2t 2 )3
9
=
2
(x − t)2 (x − 2t)2 (x 2 + 6x t + 2t 2 ) ≥ 0.
9
P 2.83. If a, b, c are positive real numbers such that a + b + c = 3, then
Æ
3
a2 (b2 + c 2 ) +
Æ
3
b2 (c 2 + a2 ) +
Æ
3
p
3
c 2 (a2 + b2 ) ≤ 3 2.
(Michael Rozenberg, 2013)
Solution. By Hölder’s inequality, we have
”X Æ
3
—3 ”X
—2 X b 2 + c 2
.
a2 (b2 + c 2 ) ≤
a(b + c) ·
(b + c)2
Therefore, it suffices to show that
X b2 + c 2
27
≤
,
2
(b + c)
2(a b + bc + ca)2
which is equivalent to the homogeneous inequalities
X b2 + c 2
p4
−
1
≤
− 3,
(b + c)2
6q2
X
p4
2bc
+
≥ 3,
(b + c)2 6q2
where
p = a + b + c,
q = a b + bc + ca.
According to P 1.61, the following inequality holds
X
p2
9
2bc
+
≥ .
2
(b + c)
q
2
Symmetric Nonrational Inequalities
385
Thus, it is enough to show that
p4
9 p2
−
+ 2 ≥ 3,
2
q
6q
which is equivalent to
p2
−3
q
2
≥ 0.
The equality holds for a = b = c = 1.
P 2.84. If a, b, c are nonnegative real numbers, no two of which are zero, then
1
1
1
2
1
+
+
≥
+p
.
a+b b+c c+a
a+b+c
a b + bc + ca
(Vasile Cîrtoaje, 2005)
Solution. Using the notation
p = a + b + c,
q = a b + bc + ca,
r = a bc,
we can write the inequality as
p2 + q
1
2
≥ +p .
pq − r
p
q
According to P 3.57-(a) in Volume 1, for fixed p and q, the product r = a bc is minimal
when two of a, b, c are equal or one of a, b, c is zero. Therefore, it suffices to prove the
inequality for b = c = 1 and for a = 0. For a = 0, the inequality reduces to
1 1
2
+ ≥p ,
b c
bc
which is obvious. For b = c = 1, the inequality becomes as follows:
1
2
1
2
,
+
≥
+p
2 a+1
a+2
2a + 1
1
1
2
2
−
≥p
−
,
2 a+2
2a + 1 a + 1
p
a
2(a + 1 − 2a + 1)
≥
,
p
2(a + 2)
(a + 1) 2a + 1
386
Vasile Cîrtoaje
2a2
a
.
≥
p
p
2(a + 2) (a + 1) 2a + 1(a + 1 + 2a + 1)
So, we need to show that
2a
1
≥
.
p
p
2(a + 2) (a + 1) 2a + 1(a + 1 + 2a + 1)
Consider two cases: 0 ≤ a ≤ 1 and a > 1.
Case 1: 0 ≤ a ≤ 1. Since
p
p
p
p
p
2a + 1(a + 1 + 2a + 1) ≥ 2a + 1( 2a + 1 + 2a + 1) = 2(2a + 1),
it suffices to prove that
1
a
≥
,
2(a + 2) (a + 1)(2a + 1)
which is equivalent to 1 − a ≥ 0.
Case 2: a > 1. We will show first that
p
(a + 1) 2a + 1 > 3a.
Indeed, by squaring, we get the obvious inequality
a3 + a(a − 2)2 + 1 > 0.
Therefore, it suffices to show that
1
2a
≥
,
2(a + 2) (a + 1)(3a + 2a + 1)
which is equivalent to (a−1)2 ≥ 0. The proof is completed. The equality holds for a = 0
and b = c.
P 2.85. If a, b ≥ 1, then
1
1
1
1
+ ≥p
+p
.
p
2
3a + 1
3a b + 1
3b + 1
Symmetric Nonrational Inequalities
387
Solution. Using the substitution
2
y=p
,
3b + 1
2
,
x=p
3a + 1
x, y ∈ (0, 1],
the desired inequality can be written as
v
t
3
≥ x + y − 1.
xy
2
2
x y − x2 − y2 + 4
Consider the non-trivial case x + y − 1 ≥ 0, and denote
t = x + y − 1,
p = x y,
1 ≥ p ≥ t ≥ 0.
Since
x 2 + y 2 = (x + y)2 − 2x y = (t + 1)2 − 2p,
we need to prove that
p
v
t
3
≥ t.
p2 + 2p − t 2 − 2t + 3
By squaring, we get the inequality
(p − t)[(3 − t 2 )p + t(1 − t)(3 + t)] ≥ 0,
which is clearly true. The equality holds for a = b = 1.
P 2.86. Let a, b, c be positive real numbers such that a ≥ 1 ≥ b ≥ c and a bc = 1. Prove
that
1
1
1
3
+p
+p
≥ .
p
3a + 1
3c + 1 2
3b + 1
(Vasile Cîrtoaje, 2007)
Solution. Let b1 = 1/b, b1 ≥ 1. We claim that
1
1
1
+p
≥ .
p
3b + 1
3b1 + 1 2
This inequality is equivalent to
1
+
p
3b + 1
v
t
b
1
≥ .
b+3 2
Making the substitution
1
= t,
p
3b + 1
1
≤ t < 1,
2
388
Vasile Cîrtoaje
the inequality becomes
v
t 1 − t2
≥ 1 − t.
1 + 8t 2
By squaring, we get t(1 − t)(1 − 2t)2 ≥ 0, which is clearly true. Similarly, we have
1
1
1
≥ ,
+p
p
3c + 1
3c1 + 1 2
where c1 = 1/c, c1 ≥ 1. Using these inequality, it suffices to show that
1
1
1
1
+ ≥p
+p
,
p
2
3a + 1
3c1 + 1
3b1 + 1
which is equivalent to
1
p
3b1 c1 + 1
+
1
1
1
≥p
.
+p
2
3c1 + 1
3b1 + 1
From the inequality in the preceding P 2.85, the conclusion follows. The equality holds
for a = b = c = 1.
1
P 2.87. Let a, b, c be positive real numbers such that a + b + c = 3. If k ≥ p , then
2
(a bc)k (a2 + b2 + c 2 ) ≤ 3.
(Vasile Cîrtoaje, 2006)
Solution. Since
a+b+c
abc ≤
3

‹3
= 1,
p
it suffices to prove the desired inequality for k = 1/ 2. Write the inequality in the
homogeneous form
a+b+c
(a bc) (a + b + c ) ≤ 3
3
k
2
2
2

‹3k+2
.
According to P 3.57-(a) in Volume 1, for fixed a+b+c and a b+bc+ca, the product a bc is
maximal when two of a, b, c are equal. Therefore, it suffices to prove the homogeneous
inequality for b = c = 1; that is, f (a) ≥ 0, where
f (a) = (3k + 2) ln(a + 2) − (3k + 1) ln 3 − k ln a − ln(a2 + 2).
Symmetric Nonrational Inequalities
389
From
f 0 (a) =
2a
3k + 2 k
− − 2
a+2
a a +2
p
p
2(a − 1)(ka2 − 2a + 2k)
2(a − 1)(a − 2)2
=
=
,
a(a + 2)(a2 + 2)
a(a + 2)(a2 + 2)
it follows that f is decreasing on (0, 1] and increasing on [1, ∞); therefore, f (a) ≥
f (1) = 0. This completes the proof. The equality holds for a = b = c.
P 2.88. Let p and q be nonnegative real numbers such that p2 ≥ 3q, and let
v
v
t 2p + w
t 2p − 2w
+2
,
g(p, q) =
3
3
s
 s
2p + 2w
2p − w

+2
,

3
3
h(p, q) =
p
p
p

 p + p + q,
where w =
p
p2 ≤ 4q
p2 ≥ 4q
,
p2 − 3q. If a, b, c are nonnegative real numbers such that
a + b + c = p,
a b + bc + ca = q,
then
p
(a)
with equality for a =
(b)
a+b+
p
b+c+
p
c + a ≥ g(p, q),
p + 2w
p−w
and b = c =
(or any cyclic permutation);
3
3
p
p
p
a + b + b + c + c + a ≤ h(p, q),
p+w
p − 2w
and b = c =
(or any cyclic permutation) - when
3
3
p2 ≤ 4q, and for a = 0, b + c = p and bc = q (or any cyclic permutation) - when p2 ≥ 4q.
with equality for a =
(Vasile Cîrtoaje, 2013)
Solution. Consider the non-trivial case p > 0. Since
b + c = p − a,
390
Vasile Cîrtoaje
(a + b)(a + c) = a2 + q
and
p
a+b+
p
a+c =
Ç
a+p+2
we get
p
a+b+
p
b+c+
where
f (a) =
From
p
p−a+
Ç
p
Æ
a2 + q,
c + a = f (a),
a+p+2
Æ
a2 + q.
p
a2 + q + 2a
−1
f (a) = p
+ p
,
q
2 p − a 2 a 2 + q · a + p + 2p a 2 + q
0
it follows that f 0 (a) has the same sign as F (a), where
p
p
( a2 + q + 2a)2
−2(3a2 − 2pa + q)(a + a2 + q)
−1
F (a) =
+
=
.
p
p
p − a (a2 + q)(a + p + 2 a2 + q) (p − a)(a2 + q)(a + p + 2 a2 + q)
Therefore, f (a) is increasing on [a1 , a2 ] and decreasing on [0, a1 ] and [a2 , p], where
p
p
p − p2 − 3q
p + p2 − 3q
a1 =
, a2 =
.
3
3
(a) From
0 ≤ (b + c)2 − 4bc = (p − a)2 − 4(a2 − pa + q) = −(3a2 − 2pa + 4q − p2 ),
we get a ∈ [0, a4 ], where
p+2
p
p2 − 3q
.
3
Since a2 ≤ a4 , f (a) is increasing on [a1 , a2 ] and decreasing on [0, a1 ] ∪ [a2 , a4 ]; therefore,
f (a) ≥ min{ f (a1 ), f (a4 )}.
a4 =
We need to show that
min{ f (a1 ), f (a4 )} = g(p, q).
Indeed, from 2a1 + a4 = p and
q
Æ
a12 + q = 2a1 (p − a1 ),
q
2 a42 + q = a4 + p,
we get
f (a1 ) = f (a4 ) = g(p, q).
Symmetric Nonrational Inequalities
391
(b) Consider the cases 3q ≤ p2 ≤ 4q and p2 ≥ 4q.
Case 1: 3q ≤ p2 ≤ 4q. From
(b + c)2 − 4bc = (p − a)2 − 4(a2 − pa + q) = −(3a2 − 2pa + 4q − p2 ) ≥ 0,
we get a ∈ [a3 , a4 ], where
a3 =
p−2
p
p2 − 3q
≥ 0,
3
a4 =
p+2
p
p2 − 3q
.
3
Since a3 ≤ a1 ≤ a2 ≤ a4 , it follows that f (a) is increasing on [a1 , a2 ] and decreasing on
[a3 , a1 ] ∪ [a2 , a4 ]. Thus,
f (a) ≤ max{ f (a2 ), f (a3 )}.
To complete the proof, we need to show that
max{ f (a2 ), f (a3 )} = h(p, q).
Indeed, from 2a2 + a3 = p and
q
Æ
a22 + q = 2a2 (p − a2 ),
q
2 a32 + q = a3 + p,
we get
f (a2 ) = f (a3 ) = h(p, q).
Case 2: p2 ≥ 4q. Assume that a = min{a, b, c}, a ≤ p/3. From
0 ≤ bc = q − a(b + c) = q − a(p − a) = a2 − pa + q,
we get a ∈ [0, a5 ] ∪ [a6 , p], where
p
p − p2 − 4q
a5 =
,
2
a6 =
p+
p
p2 − 4q
.
2
Since a6 > p/3, it follows that a ∈ [0, a5 ]. Since a1 ≤ a5 ≤ a2 , f (a) is decreasing on
[0, a1 ] and increasing on [a1 , a5 ]; thus,
f (a) ≤ max{ f (0), f (a5 )}.
It remains to show that max{ f (0), f (a5 )} = h(p, q). Indeed, from a52 + q = pa5 and
p
a5 +
p
p − a5 =
r
p+2
q
pa5 − a52 =
we get
f (0) = f (a5 ) = h(p, q).
Æ
p
p + 2 q,
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Vasile Cîrtoaje
Remark. Note the following particular cases:
(a) If a, b, c are nonnegative real numbers such that
a + b + c = a b + bc + ca = 4,
then
p
p
p
p
2(1 + 10) p
≤ a + b + b + c + c + a ≤ 2(1 + 2),
p
3
with left equality for a = 8/3 and b = c = 2/3 (or any cyclic permutation), and right
equality for a = 0 and b = c = 2 (or any cyclic permutation).
(b) If a, b, c are nonnegative real numbers such that
a + b + c = 4,
a b + bc + ca = 5,
then
p
p
p
p
p
2+2 3≤ a+ b+ b+c+ c+a ≤
p
p
10 + 2 7
,
p
3
with left equality for a = 2 and b = c = 1 (or any cyclic permutation), and right equality
for a = 2/3 and b = c = 5/3 (or any cyclic permutation).
(c) If a, b, c are nonnegative real numbers such that
a + b + c = 11,
then
3
p
6≤
p
a+b+
p
b+c+
a b + bc + ca = 7,
p
c+a≤
Æ
p
p
11 + 11 + 7,
with left equality for a = 31/3 and b = c = 1/3 (or any cyclic permutation), and right
equality for a = 0, b + c = 11 and bc = 7 (or any cyclic permutation).
P 2.89. Let a, b, c, d be positive real numbers such that a2 + b2 + c 2 + d 2 = 1. Prove that
p
p
p
p
p
p
p
p
1 − a + 1 − b + 1 − c + 1 − d ≥ a + b + c + d.
(Vasile Cîrtoaje, 2007)
Symmetric Nonrational Inequalities
393
First Solution. We can obtain the desired inequality by summing the inequalities
p
p
p
p
1 − a + 1 − b ≥ c + d,
p
p
p
p
1 − c + 1 − d ≥ a + b.
Since
p
Æ
p
4
1 − a + 1 − b ≥ 2 (1 − a)(1 − b)
and
p
c+
p
d ≤2
v
tc + d
≤2
2
v
t
2
2
4 c +d
2
,
the former inequality above holds if
(1 − a)(1 − b) ≥
c2 + d 2
.
2
Indeed,
2(1 − a)(1 − b) − c 2 − d 2 = 2(1 − a)(1 − b) + a2 + b2 − 1 = (a + b − 1)2 ≥ 0.
Similarly, we can prove the second inequality. The equality holds for a = b = c = d =
Second Solution. We can obtain the desired inequality by summing the inequalities
p
p
1
1 − a − a ≥ p (1 − 4a2 ),
2 2
p
p
p
1
1 − c − c ≥ p (1 − 4c 2 ),
2 2
p
1− b−
p
1−d −
p
1
b ≥ p (1 − 4b2 ),
2 2
1
d ≥ p (1 − 4d 2 ).
2 2
To prove the first inequality, we write it as
1 − 2a
1
p
p ≥ p (1 − 2a)(1 + 2a).
1−a+ a
2 2
1
. We need to show that
2
p
p
p
2 2 ≥ (1 + 2a)( 1 − a + a).
p
p
p
p
Since 1 − a + a ≤ 2[(1 − a) + a] = 2, we have
p
p
p
p
2 2 − (1 + 2a)( 1 − a + a) ≥ 2(1 − 2a) ≥ 0.
Case 1: 0 < a ≤
1
.
2
394
Vasile Cîrtoaje
1
≤ a < 1. We need to show that
2
p
p
p
2 2 ≤ (1 + 2a)( 1 − a + a).
p
Since 1 + 2a ≥ 2 2a, it suffices to prove that
Æ
1 ≤ a(1 − a) + a.
Case 2:
Indeed,
1−a−
Æ
p
p
p
a(1 − a) = 1 − a ( 1 − a − a) =
p
1 − a (1 − 2a)
≤ 0.
p
p
1−a+ a
P 2.90. Let a, b, c, d be positive real numbers. Prove that
p
A + 2 ≥ B + 4,
where
‹
1 1 1 1
A = (a + b + c + d)
+ + +
− 16,
a b c d
‹

1
1
1
1
2
2
2
2
B = (a + b + c + d ) 2 + 2 + 2 + 2 − 16.
a
b
c
d

(Vasile Cîrtoaje, 2004)
Solution. By squaring, the inequality becomes
A2 + 4A ≥ B.
Let us denote
f (x, y, z) =
y z
x
+ + − 3,
y
z
x
F (x, y, z) =
y 2 z2
x2
+
+ 2 − 3,
y 2 z2
x
where x, y, z > 0. By the AM-GM inequality, it follows that f (x, y, z) ≥ 0 and F (x, y, z) ≥
0. We can check that
A = f (a, b, c) + f (b, a, d) + f (c, d, a) + f (d, c, b)
= f (c, b, a) + f (d, a, b) + f (a, d, c) + f (b, c, d)
and
B = F (a, b, c) + F (b, a, d) + F (c, d, a) + F (d, c, b).
Symmetric Nonrational Inequalities
395
Since
F (x, y, z) = [ f (x, y, z) + 3]2 − 2[ f (z, y, x) + 3] − 3
= f 2 (x, y, z) + 6 f (x, y, z) − 2 f (z, y, x),
we get
B = f 2 (a, b, c) + f 2 (b, a, d) + f 2 (c, d, a) + f 2 (d, c, b) + 4A.
Therefore,
A2 + 4A − B = [ f (a, b, c) + f (b, a, d) + f (c, d, a) + f (d, c, b)]2
− f 2 (a, b, c) − f 2 (b, a, d) − f 2 (c, d, a) − f 2 (d, c, b) ≥ 0.
The equality holds for a = b = c = d.
P 2.91. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = 1.
Prove that
p
p
p
3a1 + 1 + 3a2 + 1 + · · · + 3an + 1 ≥ n + 1.
First Solution. Without loss of generality, assume that a1 = max{a1 , a2 , · · · , an }. Write
the inequality as follows:
p
p
p
( 3a1 + 1 − 2) + ( 3a2 + 1 − 1) + · · · + ( 3an + 1 − 1) ≥ 0,
a1 − 1
p
3a1 + 1 + 2
+p
a2
3a2 + 1 + 1
+ ··· + p
an
3an + 1 + 1
≥ 0,
a2 + · · · + an
≥p
,
3an + 1 + 1
3a1 + 1 + 2
3a2 + 1 + 1
1
1
1
1
a2 p
+ · · · + an p
≥ 0.
−p
−p
3an + 1 + 1
3a1 + 1 + 2
3a1 + 1 + 2
3a2 + 1 + 1
a2
p
+ ··· + p
an
The last inequality is clearly true. The equality holds for a1 = 1 and a2 = · · · = an = 0
(or any cyclic permutation).
Second Solution. We use the induction method. For n = 1, the inequality is an equality.
We claim that
Æ
p
p
3a1 + 1 + 3an + 1 ≥ 3(a1 + an ) + 1 + 1.
By squaring, this becomes
Æ
(3a1 + 1)(an + 1) ≥
Æ
3(a1 + an ) + 1,
396
Vasile Cîrtoaje
which is equivalent to a1 an ≥ 0. Thus, to prove the original inequality, it suffices to
show that
Æ
p
p
3(a1 + an ) + 1 + 3a2 + 1 + · · · + 3an−1 + 1 ≥ n.
Using the substitution b1 = a1 + an and b2 = a2 , · · · , bn−1 = an−1 , this inequality becomes
Æ
Æ
Æ
3b1 + 1 + 3b2 + 1 + · · · + 3bn−1 + 1 ≥ n
for b1 + b2 + · · · + bn−1 = 1. Clearly, this is true by the induction hypothesis.
P 2.92. Let 0 ≤ a < b and a1 , a2 , . . . , an ∈ [a, b]. Prove that
€p
p
p Š2
a1 + a2 · · · + an − n n a1 a2 · · · an ≤ (n − 1)
b− a .
(Vasile Cîrtoaje, 2005)
Solution. Based on Lemma below, it suffices to consider that
a1 = · · · = ak = a,
ak+1 = · · · = an = b,
where k ∈ {1, 2, · · · , n − 1}. The original inequality becomes
k
(n − k − 1)a + (k − 1)b + na n b
n−k
n
p
≥ (2n − 2) a b.
Clearly, this inequality follows by the weighted AM-GM inequality. For n ≥ 3, the equality holds when a = 0, one of ai is also 0 and the other ai are equal to b.
Lemma. Let 0 ≤ a < b and a1 , a2 , . . . , an ∈ [a, b]. Then, the expression
p
a1 + a2 · · · + an − n n a1 a2 · · · an
is maximal when a1 , a2 , . . . , an ∈ {a, b}.
Proof. We use the contradiction method. Consider that a2 , . . . , an are fixed and define
the function
p
f (a1 ) = a1 + a2 + · · · + an − n n a1 a2 · · · an .
For the sake of contradiction, assume that there exits a1 ∈ (a, b) such that
f (a1 ) > f (a)
p
p
p
n
and f (a1 ) > f (b). Let us denote x i = n ai for all i, c = n a and d = b (c < x 1 < d).
From
f (a1 ) − f (a) = x 1n − c n − n(x 1 − c)x 2 · · · x n
= (x 1 − c)(x 1n−1 + x 1n−2 c + · · · + c n−1 − nx 2 · · · x n ),
Symmetric Nonrational Inequalities
397
we get
x 1n−1 + x 1n−2 c + · · · + c n−1 > nx 2 · · · x n .
(*)
Analogously, from
f (a1 ) − f (b) = x 1n − d n − n(x 1 − d)x 2 · · · x n
= (x 1 − d)(x 1n−1 + x 1n−2 d + · · · + d n−1 − nx 2 · · · x n ),
we get
nx 2 · · · x n > x 1n−1 + x 1n−2 d + · · · + d n−1 .
(**)
Summing up (*) and (**) yields
x 1n−1 + x 1n−2 c + · · · + c n−1 > x 1n−1 + x 1n−2 d + · · · + d n−1 ,
which is clearly false.
P 2.93. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an = 1. Prove that
1
p
1 + (n2
− 1)a1
+p
1
1 + (n2 − 1)a2
+ ··· + p
1
1 + (n2 − 1)an
≥ 1.
Solution. For the sake of contradiction, assume that
1
p
1 + (n2
− 1)a1
+p
1
1 + (n2
− 1)a2
+ ··· + p
1
1 + (n2 − 1)an
It suffices to show that a1 a2 · · · an > 1. Let
xi = p
Since ai =
1 − x i2
(n2 − 1)x i2
1
1 + (n2 − 1)ai
, 0 < x i < 1,
i = 1, 2, · · · , n.
for all i, we need to show that
x1 + x2 + · · · + x n < 1
implies
(1 − x 12 )(1 − x 22 ) · · · (1 − x n2 ) > (n2 − 1)n x 12 x 22 · · · x n2 .
< 1.
398
Vasile Cîrtoaje
Using the AM-GM inequality gives
i Y
Y
Y h€X Š2
(1 − x 12 ) >
x 1 − x 12 =
(x 2 + · · · + x n )(2x 1 + x 2 + · · · + x n )
≥ (n2 − 1)n
Y€ p
n−1
x2 · · · x n ·
q
n+1
Š
x 12 x 2 · · · x n = (n2 − 1)n x 12 x 22 · · · x n2 .
The equality holds for a1 = a2 = · · · = an = 1.
P 2.94. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an = 1. Prove that
n
X
i=1
1
1+
p
1 + 4n(n − 1)ai
First Solution. Write the inequality as follows
n p
X
1 + 4n(n − 1)ai − 1
a1
i=1
≥
1
.
2
≥ 2n(n − 1),
v
n u
X
X 1
4n(n − 1)
t1
+
≥
2n(n
−
1)
+
.
a1
a1
a12
i=1
By squaring, the inequality becomes
v
™
u
–
X u 1
X
4n(n
−
1)
4n(n
−
1)
1
1
2
2
t
+
+
≥
2n
(n
−
1)
+
.
2
2
a
a
a
a
a
a
i
j
i
j
i
j
1≤i< j≤n
1≤i< j≤n
The Cauchy-Schwarz inequality gives
v
™
u
–
u 1
4n(n
−
1)
1
4n(n
−
1)
1
4n(n − 1)
t
≥
+ p
.
+
+
2
2
ai
aj
ai a j
ai a j
ai
aj
Thus, it suffices to show that
X
1≤i< j≤n
p
1
n(n − 1)
≥
,
ai a j
2
which follows immediately from the AM-GM inequality. The equality holds for a1 =
a2 = · · · = an = 1.
Symmetric Nonrational Inequalities
399
Second Solution. For the sake of contradiction, assume that
n
X
i=1
1
1+
p
1 + 4n(n − 1)ai
<
1
.
2
It suffices to show that a1 a2 · · · an > 1. Using the substitution
xi
1
,
=
p
2n 1 + 1 + 4n(n − 1)ai
which yields
ai =
n − xi
(n − 1)x i2
i = 1, 2, · · · , n,
, 0 < x i < n,
i = 1, 2, · · · , n,
we need to show that
x1 + x2 + · · · + x n < n
implies
(n − x 1 )(n − x 2 ) · · · (n − x n ) > (n − 1)n x 12 x 22 · · · x n2 .
By the AM-GM inequality, we have
x1 x2 · · · x n ≤
x + x + · · · + x n
1
2
n
<1
n
and
n − x i > (x 1 + x 2 + · · · + x n ) − x i ≥ (n − 1)
v
t x1 x2 · · · x n
n−1
xi
,
i = 1, 2, · · · , n.
Therefore, we get
(n − x 1 )(n − x 2 ) · · · (n − x n ) > (n − 1)n x 1 x 2 · · · x n > (n − 1)n x 12 x 22 · · · x n2 .
P 2.95. If f is a convex function on a real interval I and a1 , a2 , . . . , an ∈ I, then
a + a + ··· + a 1
2
n
f (a1 ) + f (a2 ) + · · · + f (an ) + n(n − 2) f
≥
n
≥ (n − 1)[ f (b1 ) + f (b2 ) + · · · + f (bn )],
where
bi =
1 X
aj,
n − 1 j6=i
i = 1, 2, · · · , n.
(Tiberiu Popoviciu, 1965)
400
Vasile Cîrtoaje
Solution. Without loss of generality, we may assume that n ≥ 3 and a1 ≤ a2 ≤ · · · ≤ an .
There is an integer m, 1 ≤ m ≤ n − 1, such that
a1 ≤ · · · ≤ am ≤ a ≤ am+1 ≤ · · · ≤ an ,
b1 ≥ · · · ≥ bm ≥ a ≥ bm+1 ≥ · · · ≥ bn .
We can get the desired inequality by summing the following two inequalities:
f (a1 ) + f (a2 ) + · · · + f (am ) + n(n − m − 1) f (a) ≥
≥ (n − 1)[ f (bm+1 ) + f (bm+2 ) + · · · + f (bn )],
(*)
f (am+1 ) + f (am+2 ) + · · · + f (an ) + n(m − 1) f (a) ≥
≥ (n − 1)[ f (b1 ) + f (b2 ) + · · · + f (bm )].
(**)
In order to prove (*), we apply Jensen’s inequality to get
f (a1 ) + f (a2 ) + · · · + f (am ) + (n − m − 1) f (a) ≥ (n − 1) f (b),
where
b=
a1 + a2 + · · · + am + (n − m − 1)a
.
n−1
Thus, it suffices to show that
(n − m − 1) f (a) + f (b) ≥ f (bm+1 ) + f (bm+2 ) + · · · + f (bn ).
Since
a ≥ bm+1 ≥ bm+2 ≥ · · · ≥ bn , (n − m − 1)a + b = bm+1 + bm+2 + · · · + bn ,
−
→
−
→
we see that A n−m = (a, · · · , a, b) majorizes B n−m = (bm+1 , bm+2 , · · · , bn ). Therefore,
the inequality is a consequence of Karamata’s inequality.
Similarly, we can prove the inequality (**), by summing Jensen’s inequality
f (am+1 ) + f (am+2 ) + · · · + f (an ) + (m − 1) f (a)
≥ f (c)
n−1
and the inequality
f (c) + (m − 1) f (a) ≥ f (b1 ) + f (b2 ) + · · · + f (bm ),
where
am+1 + am+2 + · · · + an + (m − 1)a
.
n−1
The last inequality follows by Karamata’s inequality, because
c=
c + (m − 1)a = b1 + b2 + · · · + bm ,
−
→
−
→
therefore C m = (c, a, · · · , a) majorizes D m = (b1 , b2 , · · · , bm ).
b1 ≥ b2 ≥ · · · ≥ bm ≥ a,
Symmetric Nonrational Inequalities
401
P 2.96. Let a1 , a2 , . . . , an (n ≥ 3) be positive real numbers such that a1 a2 · · · an = 1. Prove
that
n
X
1
1
≤ .
p
2
(n − 1)2 + 4nai
i=1 n − 1 +
Solution. Use the contradiction method. Assume that
n
X
i=1
1
n−1+
p
(n − 1)2
+ 4nai
>
1
,
2
and show that a1 a2 · · · an < 1. Using the substitution
1
n−1+
p
which involves
ai =
(n − 1)2 + 4nai
(n − 1)2 ci
,
(n − ci )2
=
n − ci
,
2n(n − 1)
0 < ci < n,
i = 1, 2, · · · , n,
i = 1, 2, · · · , n,
we need to show that
c1 + c2 + · · · + cn < n
implies
n − c 2 n − c 2
1
2
n−1
n−1
Clearly, it suffices to show that
···
n − c 2
n
n−1
> c1 c2 · · · cn .
c1 + c2 + · · · + cn = n
implies
n − c 2 n − c 2
1
2
n − c 2
n
≥ c1 c2 · · · cn .
n−1
n−1
n−1
Popoviciu’s inequality (see the preceding P 2.95) applied to the convex function f (x) =
− ln x, x > 0 gives
n − c n−1 n − c n−1
1
2
n−1
n−1
···
···
n − c n−1
n
≥ c1 c2 · · · cn
c + c + · · · + c n(n−2)
1
2
n
,
n
n−1
n − c n−1 n − c n−1 n − c n−1
n
1
2
···
≥ c1 c2 · · · cn ,
n−1
n−1
n−1
n − c 2 n − c 2 n − c 2
2
n
1
2
···
≥ (c1 c2 · · · cn ) n−1 .
n−1
n−1
n−1
Thus, it suffices to show that
2
(c1 c2 · · · cn ) n−1 ≥ c1 c2 · · · cn .
402
Vasile Cîrtoaje
For n = 3, this inequality is an equality, while for n ≥ 4, it is equivalent to c1 c2 · · · cn ≤ 1.
Indeed, by the AM-GM inequality, we have
c1 c2 · · · cn ≤
c + c + · · · + c n
1
2
n
= 1.
n
The equality holds for a1 = a2 = · · · = an = 1.
P 2.97. If a1 , a2 , . . . , an are positive real numbers such that a1 a2 · · · an = 1, then
a1 + a2 + · · · + an ≥ n − 1 +
v
u 2
t a1 + a22 + · · · + an2
n
.
Solution. Let us denote
a=
a1 + a2 + · · · + an
,
n
b=
v P
u
t 2 1≤i< j≤n ai a j
n(n − 1)
,
where b ≥ 1 (by the AM-GM inequality). We need to show that
na − n + 1 ≥
v
t n2 a2 − n(n − 1)b2
n
.
By squaring, this inequality becomes
(n − 1)[n(a − 1)2 + b2 − 1] ≥ 0,
which is clearly true. The equality holds for a1 = a2 = · · · = an = 1.
P 2.98. If a1 , a2 , . . . , an are positive real numbers such that a1 a2 · · · an = 1, then
q
(n − 1)(a12 + a22 + · · · + an2 ) + n −
Æ
n(n − 1) ≥ a1 + a2 + · · · + an .
(Vasile Cîrtoaje, 2006)
Symmetric Nonrational Inequalities
403
Solution. We use the induction method. For n = 2, the inequality is equivalent to the
obvious inequality
1
≥ 2.
a1 +
a1
Assume now that the inequality holds for n − 1 numbers, n ≥ 3, and prove that it holds
also for n numbers. Let a1 = min{a1 , a2 , . . . , an }, and denote
x=
f (a1 , a2 , . . . , an ) =
q
a2 + a3 + · · · + an
,
n−1
y=
p
n−1
(n − 1)(a12 + a22 + · · · + an2 ) + n −
a2 a3 · · · an ,
Æ
n(n − 1) − (a1 + a2 + · · · + an ).
By the AM-GM inequality, we have x ≥ y. We will show that
f (a1 , a2 , . . . , an ) ≥ f (a1 , y, · · · , y) ≥ 0.
(*)
Write the left inequality as
q
q
p
a12 + a22 + · · · + an2 − a12 + (n − 1) y 2 ≥ n − 1 (x − y).
To prove this inequality, we use the induction hypothesis, written in the homogeneous
form
q
”
—
Æ
(n − 2)(a22 + a32 + · · · + an2 ) + n − 1 − (n − 1)(n − 2) y ≥ (n − 1)x,
which is equivalent to
a22 + · · · + an2 ≥ (n − 1)A2 ,
where
A = k x − (k − 1) y,
k=
v
tn − 1
n−2
.
So, we need to prove that
q
q
p
a12 + (n − 1)A2 − a12 + (n − 1) y 2 ≥ n − 1 (x − y).
Write this inequality as
q
A2 − y 2
x−y
≥p
.
q
2
2
n−1
a1 + (n − 1)A2 + a1 + (n − 1) y 2
Since x ≥ y and
A2 − y 2 = k(x − y)[k x − (k − 2) y],
we need to show that
q
a12
k[k x − (k − 2) y]
1
≥p
.
q
2
n−1
+ (n − 1)A2 + a1 + (n − 1) y 2
404
Vasile Cîrtoaje
In addition, since a1 ≤ y, it suffices to show that
p
We claim that
Æ
k[k x − (k − 2) y]
1
≥p
.
p
n−1
y 2 + (n − 1)A2 + n y
p
y 2 + (n − 1)A2 ≤ k n − 1 [k x − (k − 1) y] .
If this inequality is true, then it is enough to prove that
p
k[k x − (k − 2) y]
k n − 1[k x − (k − 1) y] +
p
ny
≥p
1
n−1
.
Rewrite this inequality as
k[k x − (k − 2) y]
≥ 1,
k[k x − (k − 1) y] + m y
m=
s
n
.
n−1
Since m < k, we have
k[k x − (k − 2) y]
k[k x − (k − 2) y]
>
= 1.
k[k x − (k − 1) y] + m y
k[k x − (k − 1) y] + k y
Using the relation a1 y n−1 = 1, we can write the right inequality in (*) as
Æ
(n − 1)[(n − 1) y 2n + 1] ≥ (n − 1) y n − t y n−1 + 1,
where
t = n−
Æ
n(n − 1).
This inequality is true if
(n − 1)[(n − 1) y 2n + 1] ≥ [(n − 1) y n − t y n−1 + 1]2 ,
that is,
2(n − 1)t y 2n−1 − t 2 y 2n−2 − 2(n − 1) y n + 2t k y n−1 + n − 2 ≥ 0.
Since
t 2 = n(2t − 1),
we can write this inequality in the form
2t y n−1 B + C ≥ 0,
where
B = (n − 1) y n − n y n−1 + 1,
C = n y 2n−2 − 2(n − 1) y n + n − 2.
Symmetric Nonrational Inequalities
405
This inequality is true if b ≥ 0 and C ≥ 0. Indeed, by the AM-GM inequality, we have
Æ
n
(n − 1) y n + 1 ≥ n y (n−1)n · 1 = n y n−1
and
n y 2n−2 + n − 2 ≥ (2n − 2)
Æ
2n−2
y n(2n−2) · 1n−2 = 2(n − 1) y n .
The proof is completed. The equality holds for a1 = a2 = · · · = an = 1.
P 2.99. Let a1 , a2 , . . . , an (n ≥ 3) be positive real numbers such that a1 a2 · · · an = 1. If
0<p≤
then
1
p
1 + pa1
+p
1
1 + pa2
2n − 1
,
(n − 1)2
+p
1
1 + pan
≤p
n
1+p
.
(Vasile Cîrtoaje and Gabriel Dospinescu, 2006)
Solution. We use the contradiction method. Assume that the reverse inequality holds,
namely
1
1
n
1
,
+p
>p
+p
p
1 + pan
1 + pa1
1 + pa2
1+p
and show that this implies a1 a2 · · · an < 1. Using the substitution
p
p
p
1+p
, 0 < x i < p + 1, i = 1, 2, · · · , n,
1 + pai =
xi
we need to show that x 1 + x 2 + · · · + x n > n yields
1+p
1+p
1+p
−1
− 1 ···
− 1 < pn.
x n2
x 12
x 22
Clearly, it suffices to prove that
x1 + x2 + · · · + x n = n
yields
1+p
x 12
−1
1+p
x 22
1+p
− 1 ···
− 1 ≤ pn.
x n2
Denoting
p
1 + p = q,
1<q≤
n
,
n−1
406
Vasile Cîrtoaje
we need to show that
(q2 − x 12 )(q2 − x 22 ) · · · (q2 − x n2 ) ≤ (q2 − 1)n (x 1 x 2 · · · x n )2
(*)
for all x i ∈ (0, q) satisfying x 1 + x 2 + · · · + x n = n. Applying Popoviciu’s inequality (see
P 2.95) to the convex function
n
f (x) = − ln
− x , 0 < x < 1,
n−1
we get
(x 1 x 2 · · · x n )n−1 ≥ [n − (n − 1)x 1 ][n − (n − 1)x 2 ] · · · [n − (n − 1)x n ].
(**)
On the other hand, Jensen’s inequality applied to the convex function
f (x) = ln
yields
n − (n − 1)x
q−x
[n − (n − 1)x 1 ][n − (n − 1)x 2 ] · · · [n − (n − 1)x n ]
1
≥
.
(q − x 1 )(q − x 2 ) · · · (q − x n )
(q − 1)n
Multiplying this inequality and (**) gives
(x 1 x 2 · · · x n )n−1 ≥
(q − x 1 )(q − x 2 ) · · · (q − x n )
.
(q − 1)n
Therefore, in order to prove (*), we still have to show that
(x 1 x 2 · · · x n )n−3 (q + x 1 )(q + x 2 ) · · · (q + x n ) ≤ (q + 1)n .
This is true because, by the AM-GM inequality, we have
x + x + · · · + x n
1
2
n
=1
x1 x2 · · · x n ≤
n
and
x 1 + x 2 + · · · + x n n
(q + x 1 )(q + x 2 ) · · · (q + x n ) ≤ q +
= (q + 1)n .
n
The equality occurs for a1 = a2 = · · · = an = 1.
Remark. For p =
2n − 1
, we get the following inequality
(n − 1)2
n
X
i=1
1
p
(n − 1)2 + (2n − 1)ai
≤ 1,
which holds for all positive numbers a1 , a2 , . . . , an (n ≥ 3) satisfying a1 a2 · · · an = 1.
Symmetric Nonrational Inequalities
407
P 2.100. If a1 , a2 , . . . , an (n ≥ 3) are positive real numbers such that a1 a2 · · · an = 1, then
n Ç
X
‚
(n − 1)2 ai4
+ 2n − 1 ≥
i=1
n
X
n
X
Œ2
;
ai
i=1
ai
q
‚
(n − 1)2 ai2
+ 2n − 1 ≥
n
X
Œ2
.
ai
i=1
i=1
(Vasile Cîrtoaje, 2007)
Solution. According to the preceding P 2.99, the following inequality holds for any real
m:
n
X
1
≤ 1.
Æ
(n − 1)2 + (2n − 1)aim
i=1
On the other hand, by the Cauchy-Schwarz inequality, we have
‚
n
X
i=1
Œ‚
1
Æ
(n − 1)2 + (2n − 1)aim
n
X
ai2
q
Œ
(n − 1)2 + (2n − 1)aim
i=1
ai2
q
≥
n
X
Œ2
ai
,
i=1
hence
n
X
‚
‚
(n − 1)2 + (2n − 1)aim
≥
i=1
n
X
Œ2
ai
.
i=1
Setting m = −4 and m = −2 gives the desired inequalities. The equality occurs for
a1 = a2 = · · · = an = 1.
P 2.101. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an ≥ 1. If k > 1,
then
X
a1k
≥ 1.
a1k + a2 + · · · + an
(Vasile Cîrtoaje, 2006)
p
First Solution. Let us denote r = n a1 a2 · · · an and bi = ai /r for i = 1, 2, · · · , n. Note
that r ≥ 1 and b1 b2 · · · bn = 1. The desired inequality becomes
X
b1k
b1k + (b2 + · · · + bn )/r k−1
≥ 1,
408
Vasile Cîrtoaje
and we see that it suffices to prove it for r = 1; that is, for a1 a2 · · · an = 1. On this
hypothesis, we will show that there exists a positive number p such that
p
a1k
a1k + a2 + · · · + an
≥
a1
p.
p
p
a1 + a2 + · · · + an
If this is true, by adding this inequality and the analogous inequalities for a2 , . . . , an , we
get the desired inequality. Write the claimed inequality as
p
a2 + · · · + anp ≥ (a2 · · · an )k−p (a2 + · · · + an ).
For
(n − 1)k + 1
, p > 1,
n
this inequality turns into the homogeneous inequality
p=
p−1
p
a2 + · · · + anp ≥ (a2 · · · an ) n−1 (a2 + · · · + an ).
Based on the AM-GM inequality
p−1
(a2 · · · an ) n−1 ≤
a + · · · + a p−1
2
n
,
n−1
we only need to show that
p
p
a2 + · · · + a n
n−1
≥
a + · · · + a p
2
n
,
n−1
which is just Jensen’s inequality applied to the convex function f (x) = x p . The equality
holds for a1 = a2 = · · · = an = 1.
Second Solution. By the Cauchy-Schwarz inequality, we have
‹2

P k+1
P k+1
P
2
k+1
a
X
1
a1 + 2 1≤i< j≤n (a1 a2 ) 2
a1k
.
≥P
= P k+1
P
a1k + a2 + · · · + an
a1 (a1k + a2 + · · · + an )
a1 + 2 1≤i< j≤n a1 a2
Thus, it suffices to show that
X
(a1 a2 )
k+1
2
≥
1≤i< j≤n
X
a1 a2 .
1≤i< j≤n
Jensen’s inequality applied to the convex function f (x) = x
X
(a1 a2 )
1≤i< j≤n
k+1
2
n(n − 1)
≥
2
k+1
2
yields
‚ P
Πk+1
2 1≤i< j≤n a1 a2 2
n(n − 1)
Symmetric Nonrational Inequalities
409
On the other hand, by the AM-GM inequality, we get
X
2
2
a1 a2 ≥ (a1 a2 · · · an ) n ≥ 1.
n(n − 1) 1≤i< j≤n
Therefore,
‚ P
Πk+1
2 1≤i< j≤n a1 a2 2
n(n − 1)
hence
X
(a1 a2 )
k+1
2
1≤i< j≤n
n(n − 1)
≥
2
≥
‚ P
Œ
2 1≤i< j≤n a1 a2
n(n − 1)
‚ P
Œ
2 1≤i< j≤n a1 a2
n(n − 1)
=
,
X
a1 a2 .
1≤i< j≤n
P 2.102. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an ≥ 1. If
−2
≤ k < 1,
n−2
then
a1k
X
≤ 1.
a1k + a2 + · · · + an
(Vasile Cîrtoaje, 2006)
p
Solution. Let us denote r = n a1 a2 · · · an and bi = ai /r for i = 1, 2, · · · , n. Notice that
r ≥ 1 and b1 b2 · · · bn = 1. The desired inequality becomes
b1k
X
b1k + (b2 + · · · + bn )r 1−k
≤ 1,
and we see that it suffices to prove it for r = 1; that is, for a1 a2 · · · an = 1. On this
hypothesis, we will prove the desired inequality by summing the inequality
a1k
a1k + a2 + · · · + an
p
≤
a1
p
p
p
a1 + a2 + · · · + an
and the analogous inequalities for a2 , . . . , an , where
p=
(n − 1)k + 1
,
n
−1
≤ p < 1.
n−2
Rewrite this inequality as
1−p
p
a2 + · · · + an ≥ (a2 · · · an ) n−1 (a2 + · · · + anp ).
(*)
410
Vasile Cîrtoaje
To prove it, we use the weighted AM-GM inequality
1−p
1 + (n − 2)p
n − 1 1+(n−2)p
a2 + a3 + · · · + an ≥
a2 n−1 (a3 · · · an ) n−1 ,
1−p
1−p
which is equivalent to
1−p
1 + (n − 2)p
n−1 p
a2 + a3 + · · · + an ≥
a2 (a2 a3 · · · an ) n−1 .
1−p
1−p
Adding this inequality and the analogous inequalities for a3 , · · · , an yields the inequality
(*). Thus, the proof is completed. The equality holds for a1 = a2 = · · · = an = 1.
P 2.103. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an ≥ 1. If k > 1,
then
X
a1
≤ 1.
k
a1 + a2 + · · · + an
(Vasile Cîrtoaje, 2006)
1
.
n−1
Case 1: 1 < k ≤ n + 1. By the AM-GM inequality, the hypothesis a1 a2 · · · an ≥ 1 yields
a1 + a2 + · · · + an ≥ n. Therefore, it suffices to prove that the desired inequality holds
for a1 + a2 + · · · + an ≥ n. Actually, we only need to consider that a1 + a2 + · · · + an = n.
Indeed, if we denote p = (a1 + a2 + · · · + an )/n and bi = ai /p for i = 1, 2, · · · , n, then
the desired inequality becomes
Solution. We consider two cases: 1 < k ≤ n + 1 and k ≥ n −
b1
X
p k−1 b1k
+ b2 + · · · + bn
≤ 1,
p ≥ 1.
Clearly, it suffices to consider the case p = 1; that is, a1 + a2 + · · · + an = n. On this
hypothesis, we can rewrite the desired inequality as
X
a1
≤ 1.
k
a1 − a1 + n
Since k > 1, by Bernoulli’s inequality, we have
a1k − a1 + n ≥ 1 + k(a1 − 1) − a1 + n = n − k + 1 + (k − 1)a1 > 0.
Thus, it is enough to show that
X
a1
≤ 1,
n − k + 1 + (k − 1)a1
Symmetric Nonrational Inequalities
411
which is equivalent to
X
1
≥ 1.
n − k + 1 + (k − 1)a1
This inequality follows immediately from the AM-HM inequality
€X Š X 1 ‹
x1
≥ n2 ,
x1
for x i = n − k + 1 + (k − 1)ai , i = 1, 2, · · · , n.
1
p
. Let us denote r = n a1 a2 · · · an and bi = ai /r for i = 1, 2, · · · , n.
n−1
Note that r ≥ 1 and b1 b2 · · · bn = 1. The desired inequality becomes
Case 2: k ≥ n −
b1
X
b1k r k−1
+ b2 + · · · + b n
≤ 1,
and we see that it suffices to prove it for r = 1; that is, for a1 a2 · · · an = 1. On this
hypothesis, it suffices to show that
p
(n − 1)a1
a1k + a2 + · · · + an
+
a1
p
p
p
a1 + a2 + · · · + an
≤1
for a suitable real p; then, adding this inequality and the analogous inequalities for
p
a2 , · · · an yields the desired inequality. Let us denote t = n−1 a2 · · · an . By the AM-GM
inequality, we have
a2 + · · · an ≥ (n − 1)t,
p
a2 + · · · + anp ≥ (n − 1)t p .
Thus, it suffices to show that
(n − 1)a1
a1k + (n − 1)t
p
+
a1
p
a1 + (n − 1)t p
≤ 1.
Since a1 = 1/t n−1 , this inequality is equivalent to
(n − 1)t n+q − (n − 1)t q − t q−np + 1 ≥ 0,
where
q = (n − 1)(k − 1).
Choosing
p=
(n − 1)(k − n − 1)
,
n
the inequality becomes as follows
(n − 1)t n+q − (n − 1)t q − t n(n−1) + 1 ≥ 0,
412
Vasile Cîrtoaje
2
(n − 1)t q (t n − 1) − (t n − 1)(t n
(t n − 1)[(t q − t n
2
−2n
−2n
2
) + (t q − t n
+ tn
−3n
2
−3n
+ · · · + 1) ≥ 0,
) + · · · + (t q − 1)] ≥ 0.
The last inequality is clearly true for q ≥ n2 − 2n; that is, for k ≥ n −
proof is completed. The equality holds for a1 = a2 = · · · = an = 1.
1
. Thus, the
n−1
P 2.104. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an ≥ 1. If
−1 −
then
2
≤ k < 1,
n−2
a1
X
a1k
+ a2 + · · · + an
≥ 1.
(Vasile Cîrtoaje, 2006)
p
Solution. Let us denote r = n a1 a2 · · · an and bi = ai /r for i = 1, 2, · · · , n. Note that
r ≥ 1 and b1 b2 · · · bn = 1. The desired inequality becomes
b1
X
b1k /r 1−k
+ b2 + · · · + bn
≥ 1,
and we see that it suffices to prove it for r = 1; that is, for a1 a2 · · · an = 1. On this
hypothesis, by the Cauchy-Schwarz inequality, we have
P
P
X
( a1 )2
( a1 )2
a1
= P
≥P
P
P .
a1k + a2 + · · · + an
a1 (a1k + a2 + · · · + an ) ( a1 )2 + a11+k − a12
Thus, we still have to show that
X
a12 ≥
X
a11+k .
Case 1: −1 ≤ k < 1. Using Chebyshev’s inequality and the AM-GM inequality yields
X
a12 ≥
X
X
1 €X 1−k Š €X 1+k Š
a1
a1
≥ (a1 a2 · · · an )(1−k)/n
a11+k =
a11+k .
n
2
≤ k < −1. It is convenient to replace the numbers a1 , a2 , · · · , an
n−1
(n−1)/2 (n−1)/2
by a1
, a2
, · · · , an(n−1)/2 , respectively. So, we need to show that a1 a2 · · · an = 1
involves
X
X
q
a1n−1 ≥
a1 ,
Case 2: −1 −
Symmetric Nonrational Inequalities
where
q=
413
(n − 1)(1 + k)
,
2
−1 ≤ q < 0.
By the AM-GM inequality, we get
X
a1n−1 =
X
X 1
1 X n−1
(a2 + · · · + ann−1 ) ≥
a2 · · · an =
.
n−1
a1
Thus, it suffice to show that
X
X 1
q
≥
a1 .
a1
By Chebyshev’s inequality and the AM-GM inequality, we have
X 1
€X Š X
1 €X −1−q Š €X q Š
q
q
a1 .
a1 =
a1 ≥ (a1 a2 · · · an )−(1+q)/n
≥
a1
a1
n
Thus, the proof is completed. The equality holds for a1 = a2 = · · · = an = 1.
P 2.105. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an = 1. If k ≥ 0,
then
X
1
≤ 1.
k
a1 + a2 + · · · + an
(Vasile Cîrtoaje, 2006)
Solution. Consider two cases: 0 ≤ k ≤ 1 and k ≥ 1.
Case 1: 0 ≤ k ≤ 1. By the Cauchy-Schwarz inequality and the AM-GM inequality, we
have
a11−k + 1 + · · · + 1
1
≤ p
p
p
( a1 + a2 + · · · + an )2
a1k + a2 + · · · + an
=P
a11−k + n − 1
a11−k + n − 1
P
≤P
,
p
a1 + 2 1≤i< j≤n ai a j
a1 + n(n − 1)
hence
X
a11−k + n(n − 1)
P
≤
.
a1 + n(n − 1)
a1k + a2 + · · · + an
P
1
Therefore, it suffices to show that
X
a11−k ≤
X
a1 .
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Vasile Cîrtoaje
Indeed, by Chebyshev’s inequality and the AM-GM inequality, we have
X
a1 =
X
a1k · a11−k ≥
€X
Š X
1 €X k Š €X 1−k Š
a1
a1
≥ (a1 a2 · · · an )k/n
a11−k =
a11−k .
n
Case 2: k ≥ 1. Write the inequality as
X
p
n−1
a1k + a2 + · · · + an
where
p=
+
a1
p
p
p
a1 + a2 + · · · + an
− 1 ≤ 0,
(n − 1)(k − 1)
≥ 0.
n
To complete the proof, it suffices to show that
p
n−1
a1k + a2 + · · · + an
≤1−
a1
p
p.
p
a1 + a2 + · · · + an
Let
x=
p
n−1
x > 0.
a1 ,
By the AM-GM inequality, we have
p
n−1
n−1
a2 + · · · + an ≥ (n − 1) n−1 a2 · · · an = n−1
p =
a1
x
and
p
a2 + · · · + anp ≥
Æ
n−1
(a2 · · · an ) p =
n−1
n−1
.
q p =
xp
a1
n−1
Thus, it suffices to show that
n−1
x (n−1)k +
n−1
x
≤1−
x (n−1)p
x (n−1)p +
n−1
xp
,
which is equivalent to
x
x (n−1)k+1
x
(n−1)k+1
+n−1
≤
1
,
x np + n − 1
− x np+1 − (n − 1)(x − 1) ≥ 0,
x q (x n−1 − 1) − (n − 1)(x − 1) ≥ 0,
where
q = (n − 1)(k − 1) + 1 ≥ 1.
Symmetric Nonrational Inequalities
415
The last inequality is true since
x q (x n−1 − 1) − (n − 1)(x − 1) = (x q − 1)(x n−1 − 1) + (x n−1 − 1) − (n − 1)(x − 1),
and
(x q − 1)(x n−1 − 1) ≥ 0,
(x n−1 − 1) − (n − 1)(x − 1) = (x − 1)[(x n−2 − 1) + (x n−3 − 1) + · · · + (x − 1)] ≥ 0.
This completes the proof. The equality holds for a1 = a2 = · · · = an = 1.
P 2.106. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an ≥ n.
If 1 < k ≤ n + 1, then
a1
a1k + a2 + · · · + an
+
a2
a1 + a2k + · · · + an
+ ··· +
an
≤ 1.
a1 + a2 + · · · + ank
(Vasile Cîrtoaje, 2006)
Solution. Using the substitutions
s=
and
x1 =
a1 + a2 + · · · + an
,
n
an
a1
a
, x2 = 2 , · · · , x n = ,
s
s
s
the desired inequality becomes
x1
s k−1 x 1k
+ ··· +
+ x2 + · · · + x n
xn
≤ 1,
x 1 + x 2 + · · · + s k−1 x nk
where s ≥ 1 and x 1 + x 2 + · · · + x n = n. Clearly, if this inequality holds for s = 1,
then it holds for any s ≥ 1. Therefore, we need only to consider the case s = 1, when
a1 + a2 + · · · + an = n, and the desired inequality is equivalent to
a1
a1k
− a1 + n
+
a2
a2k
− a2 + n
+ ··· +
an
≤ 1.
ank − an + n
By Bernoulli’s inequality, we have
a1k − a1 + n ≥ 1 + k(a1 − 1) − a1 + n = n − k + 1 + (k − 1)a1 ≥ 0.
Consequently, it suffices to prove that
n
X
i=1
ai
≤ 1.
n − k + 1 + (k − 1)ai
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Vasile Cîrtoaje
For k = n + 1, this inequality is an equality. Otherwise, for 1 < k < n + 1, we rewrite the
inequality as
n
X
1
≥ 1,
n
−
k
+
1
+
(k − 1)ai
i=1
which follows from the AM-HM inequality as follows:
n
X
i=1
n2
1
≥ Pn
= 1.
n − k + 1 + (k − 1)ai
i=1 [n − k + 1 + (k − 1)ai ]
The equality holds for a1 = a2 = · · · = an = 1.
P 2.107. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an ≤ n.
If 0 ≤ k < 1, then
1
a1k
+ a2 + · · · + an
+
1
a1 + a2k
+ · · · + an
+ ··· +
1
≥ 1.
a1 + a2 + · · · + ank
Solution. By the AM-HM inequality
X
n2
n2
P
P
P
=
≥
a1k + a2 + · · · + an
(a1k + a2 + · · · + an )
a1k + (n − 1) a1
1
and Jensen’s inequality
X
a1k
 X ‹k
1
≤n
a1 ,
n
we get
1
X
a1k + a2 + · · · + an
n2
≥
n
P
1
n
a1
k
+ (n − 1)
P
≥ 1.
a1
The equality holds for a1 = a2 = · · · = an = 1.
P 2.108. Let a1 , a2 , . . . , an be positive real numbers. If k > 1, then
X ak + ak + · · · + ak
2
3
n
a2 + a3 + · · · + an
≤
n(a1k + a2k + · · · + ank )
a1 + a2 + · · · + an
.
(Wolfgang Berndt, Vasile Cîrtoaje, 2006)
Symmetric Nonrational Inequalities
417
Solution. Due to homogeneity, we may assume that a1 + a2 + ... + an = 1. Write the
inequality as follows:
‹
X
a1
(a2k + a3k + · · · + ank ) ≤ n(a1k + a2k + · · · + ank );
1+
a2 + a3 + · · · + an
X a1 (a k + a k + · · · + a k )
2
3
n
≤ a1k + a2k + · · · + ank ;
a2 + a3 + · · · + an
Œ
‚
k
k
k
X
a
+
a
+
·
·
·
+
a
2
3
n
≥ 0;
a1 a1k−1 −
a2 + a3 + · · · + an
X a1 a2 (a k−1 − a k−1 ) + a1 a3 (a k−1 − a k−1 ) + · · · + a1 an (a k−1 − a k−1 )
1
2
1
3
1
a2 + a3 + · · · + an
X
ai a j
1≤i< j≤n
aik−1 − a k−1
j
1 − ai
+
a k−1
− aik−1
j
1 − aj
X
ai a j (aik−1 − a k−1
)(ai − a j )
j
1≤i< j≤n
(1 − ai )(1 − a j )
n
≥ 0;
!
≥ 0;
≥ 0.
Since the last inequality is true for k > 1, the proof is finished. The equality holds for
a1 = a2 = · · · = an .
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Vasile Cîrtoaje
Chapter 3
Symmetric Power-Exponential
Inequalities
3.1
Applications
3.1. If a, b are positive real numbers such that a + b = a4 + b4 , then
3
3
aa b b ≤ 1 ≤ aa b b .
3.2. If a, b are positive real numbers, then
a2a + b2b ≥ a a+b + b a+b .
3.3. If a, b are positive real numbers, then
aa + b b ≥ a b + ba .
3.4. If a, b are positive real numbers, then
a2a + b2b ≥ a2b + b2a .
3.5. If a, b are nonnegative real numbers such that a + b = 2, then
(a)
a b + b a ≤ 1 + a b;
(b)
a2b + b2a ≤ 1 + a b.
419
420
Vasile Cîrtoaje
3.6. If a, b are nonnegative real numbers such that a2 + b2 = 2, then
a2b + b2a ≤ 1 + a b.
3.7. If a, b are positive real numbers, then
a a b b ≤ (a2 − a b + b2 )
3.8. If a, b ∈ (0, 1], then
a+b
2
.
a a b b ≤ 1 − a b + a2 b2 .
3.9. If a, b are positive real numbers such that a + b ≤ 2, then
a b  b ‹a
≤ 2.
+
b
a
3.10. If a, b are positive real numbers such that a + b = 2, then
1
2a a b b ≥ a2b + b2a + (a − b)2 .
2
3.11. If a, b ∈ (0, 1] or a, b ∈ [1, ∞), then
2a a b b ≥ a2 + b2 .
3.12. If a, b are positive real numbers, then
2a a b b ≥ a2 + b2 .
3.13. If a, b are positive real numbers, then
a b
a b ≥
a2 + b2
2
a+b
2
.
Symmetric Power-Exponential Inequalities
421
3.14. If a, b are positive real numbers such that a2 + b2 = 2, then
1
2a a b b ≥ a2b + b2a + (a − b)2 .
2
3.15. If a, b ∈ (0, 1], then
1
1
(a + b ) 2a + 2b
a
b
2
2

‹
≤ 4.
3.16. If a, b are positive real numbers such that a + b = 2, then
a b b a + 2 ≥ 3a b.
3.17. If a, b ∈ [0, 1], then
a b−a + b a−b + (a − b)2 ≤ 2.
3.18. If a, b are nonnegative real numbers such that a + b ≤ 2, then
a b−a + b a−b +
7
(a − b)2 ≤ 2.
16
3.19. If a, b are nonnegative real numbers such that a + b ≤ 4, then
a b−a + b a−b ≤ 2.
3.20. Let a, b be positive real numbers such that a + b = 2. If k ≥
kb
ka
a a b b ≥ 1.
3.21. If a, b are positive real numbers such that a + b = 2, then
a
p
a
p
b
b
≥ 1.
1
, then
2
422
Vasile Cîrtoaje
3.22. If a, b are positive real numbers such that a + b = 2, then
1 − a a+1 b b+1 ≥
1
(1 − a b)2 .
3
3.23. If a, b are positive real numbers such that a + b = 2, then
a−a + b−b ≤ 2.
3.24. If a, b are nonnegative real numbers such that a + b = 2, then
a2b + b2a ≥ a b + b a ≥ a2 b2 + 1.
3.25. If a, b are positive real numbers such that a + b = 2, then
a3b + b3a ≤ 2.
3.26. If a, b are nonnegative real numbers such that a + b = 2, then
a3b + b3a +

a−b
2
‹4
≤ 2.
3.27. If a, b are positive real numbers such that a + b = 2, then
2
2
a a + b b ≤ 2.
3.28. If a, b are positive real numbers such that a + b = 2, then
3
3
a a + b b ≥ 2.
3.29. If a, b are positive real numbers such that a + b = 2, then
2
2
a5b + b5a ≤ 2.
Symmetric Power-Exponential Inequalities
423
3.30. If a, b are positive real numbers such that a + b = 2, then
a2
p
b
+ b2
p
a
≤ 2.
3.31. If a, b are nonnegative real numbers such that a + b = 2, then
a b(1 − a b)2
a b(1 − a b)2
≤ a b+1 + b a+1 − 2 ≤
.
2
3
3.32. If a, b are nonnegative real numbers such that a + b = 1, then
a2b + b2a ≤ 1.
3.33. If a, b are positive real numbers such that a + b = 1, then
2a a b b ≥ a2b + b2a .
3.34. If a, b are positive real numbers such that a + b = 1, then
a−2a + b−2b ≤ 4.
424
Vasile Cîrtoaje
Symmetric Power-Exponential Inequalities
3.2
425
Solutions
P 3.1. If a, b are positive real numbers such that a + b = a4 + b4 , then
3
3
aa b b ≤ 1 ≤ aa b b .
(Vasile Cîrtoaje, 2008)
Solution. We will use the inequality
ln x ≤ x − 1,
x > 0.
To prove this inequality, let us denote
f (x) = x − 1 − ln x,
x > 0.
From
x −1
,
x
it follows that f (x) is decreasing on (0, 1] and increasing on [1, ∞). Therefore,
f 0 (x) =
f (x) ≥ f (1) = 0.
Using this inequality, we have
ln a a b b = a ln a + b ln b ≤ a(a − 1) + b(b − 1) = a2 + b2 − (a + b).
Therefore, the left inequality a a b b ≤ 1 is true if a2 + b2 ≤ a + b. We write this inequality
in the homogeneous form
(a2 + b2 )3 ≤ (a + b)2 (a4 + b4 ),
which is equivalent to the obvious inequality
a b(a − b)(a3 − b3 ) ≥ 0.
Taking now x =
1
in the inequality ln x ≤ x − 1 yields
a
a ln a ≥ a − 1.
Similarly,
b ln b ≥ b − 1,
and hence
3
3
ln a a b b = a3 ln a + b3 ln b ≤ a2 (a − 1) + b2 (b − 1) = a3 + b3 − (a2 + b2 ).
426
Vasile Cîrtoaje
3
3
Thus, to prove the right inequality a a b b ≥ 1, it suffices to show that a3 + b3 ≥ a2 + b2 ,
which is equivalent to the homogeneous inequality
(a + b)(a3 + b3 )3 ≥ (a4 + b4 )(a2 + b2 )3 .
We can write this inequality as
A − 3B ≥ 0,
where
A = (a + b)(a9 + b9 ) − (a4 + b4 )(a6 + b6 ),
B = a2 b2 (a2 + b2 )(a4 + b4 ) − a3 b3 (a + b)(a3 + b3 ).
Since
A = a b(a3 − b3 )(a5 − b5 ),
B = a2 b2 (a − b)(a5 − b5 ),
we get
A − 3B = a b(a − b)3 (a5 − b5 ) ≥ 0.
Both inequalities become equalities for a = b = 1.
P 3.2. If a, b are positive real numbers, then
a2a + b2b ≥ a a+b + b a+b .
(Vasile Cîrtoaje, 2010)
Solution. Assume that a ≥ b and consider the following two cases.
Case 1: a ≥ 1. Write the inequality as
a a+b (a a−b − 1) ≥ b2b (b a−b − 1).
For b ≤ 1, we have
a a+b (a a−b − 1) ≥ 0 ≥ b2b (b a−b − 1).
For b ≥ 1, the inequality is also true since
a a+b ≥ a2b ≥ b2b ,
a a−b − 1 ≥ b a−b − 1 ≥ 0.
Case 2: a ≤ 1. Since
a2a + b2b ≥ 2a a b b ,
it suffices to show that
2a a b b ≥ a a+b + b a+b ,
Symmetric Power-Exponential Inequalities
which can be written as
a b
b
427
 ‹a
b
+
≤ 2.
a
By Bernoulli’s inequality, we get
a b
b
+
 ‹a 
‹

‹
b
a−b b
b−a a
b(a − b)
a(b − a)
= 1+
+ 1+
≤1+
+1+
= 2.
a
b
a
b
a
The equality holds for a = b.
Conjecture 1. If a, b are positive real numbers, then
a4a + b4b ≥ a2a+2b + b2a+2b .
Conjecture 2. If a, b, c are positive real numbers, then
a3a + b3b + c 3c ≥ a a+b+c + b a+b+c + c a+b+c .
Conjecture 3. If a, b, c, d are positive real numbers, then
a4a + b4b + c 4c + d 4d ≥ a a+b+c+d + b a+b+c+d + c a+b+c+d + d a+b+c+d .
P 3.3. If a, b are positive real numbers, then
aa + b b ≥ a b + ba .
(M. Laub, 1985)
Solution. Assume that a ≥ b. We will show that if a ≥ 1, then the inequality is true.
From
a a−b ≥ b a−b ,
we get
bb ≥
a b ba
.
aa
Therefore,
aa + b b − a b − ba ≥ aa +
a b ba
(a a − a b )(a a − b a )
b
a
−
a
−
b
=
≥ 0.
aa
aa
Consider further the case 0 < b ≤ a < 1.
First Solution. Denoting
c = ab,
d = bb,
k=
a
,
b
428
Vasile Cîrtoaje
where c ≥ d and k ≥ 1, the inequality becomes
c k − d k ≥ c − d.
Since the function f (x) = x k is convex for x ≥ 0, from the well-known inequality
f (c) − f (d) ≥ f 0 (d)(c − d),
we get
c k − d k ≥ kd k−1 (c − d).
Thus, it suffices to show that
kd k−1 ≥ 1,
which is equivalent to
b1−a+b ≤ a.
Indeed, since 0 < 1 − a + b ≤ 1, by Bernoulli’s inequality, we get
b1−a+b = [1 + (b − 1)]1−a+b ≤ 1 + (1 − a + b)(b − 1) = a − b(a − b) ≤ a.
The equality holds for a = b.
Second Solution. Denoting
c=
ba
,
a b + ba
ab
,
a b + ba
d=
k=
a
,
b
where c + d = 1 and k ≥ 1, the inequality becomes
ck a + d k−b ≥ 1.
By the weighted AM-GM inequality, we have
ck a + d k−b ≥ k ac · k−bd = k ac−bd .
Thus, it suffices to show that ac ≥ bd; that is,
a1−b ≥ b1−a ,
which is equivalent to f (a) ≥ f (b), where
f (x) =
ln x
.
1− x
It is enough to prove that f (x) is an increasing function. Since
f 0 (x) =
g(x)
,
(1 − x)2
g(x) =
1
− 1 + ln x.
x
Symmetric Power-Exponential Inequalities
429
we need to show that g(x) ≥ 0 for x ∈ (0, 1). Indeed, from
g 0 (x) =
x −1
< 0,
x2
it follows that g(x) is strictly decreasing, hence g(x) > g(1) = 0.
P 3.4. If a, b are positive real numbers, then
a2a + b2b ≥ a2b + b2a .
Solution. Assume that a > b, and denote
c = ab,
d = bb,
k=
a
,
b
where c > d and k > 1. The inequality becomes
c 2k − d 2k ≥ c 2 − d 2 .
We shall show that
c 2k − d 2k > k(cd)k−1 (c 2 + d 2 ) > c 2 − d 2 .
The left inequality follows from Lemma below for x = (c/d)2 . The right inequality is
equivalent to
k(cd)k−1 > 1,
(a b)a−b >
b
,
a
1+a− b
ln a > ln b.
1−a+ b
For fixed a, let us define
f (b) =
1+a− b
ln a − ln b, 0 < b < a.
1−a+ b
If f 0 (b) < 0, then f (b) is strictly decreasing, and hence f (b) > f (a) = 0. Since
f 0 (b) =
−2
1
ln a − ,
2
(1 − a + b)
b
we need to show that g(a) > 0, where
g(a) = 2 ln a +
(1 − a + b)2
.
b
430
Vasile Cîrtoaje
From
2 2(1 − a + b) 2(a − b)(a − 1)
−
=
< 0,
a
b
ab
it follows that g(a) is strictly decreasing on [b, 1), therefore g(a) > g(1) = b > 0. This
completes the proof. The equality holds for a = b.
g 0 (a) =
Lemma. Let k and x be positive real numbers. If either k > 1 and x ≥ 1, or 0 < k < 1
and 0 < x ≤ 1, then
k−1
x k − 1 ≥ k x 2 (x − 1).
Proof. We need to show that f (x) ≥ 0, where
f (x) = x k − 1 − k x
We have
f 0 (x) =
Since
1 k−3
k x 2 g(x),
2
k−1
2
g(x) = 2x
(x − 1).
k+1
2
− (k + 1)x + k − 1.
Š
€ k−1
g 0 (x) = (k + 1) x 2 − 1 ≥ 0,
g(x) is increasing. If x ≥ 1, then g(x) ≥ g(1) = 0, f (x) is increasing, hence f (x) ≥
f (1) = 0. If 0 < x ≤ 1, then g(x) ≤ g(1) = 0, f (x) is decreasing, hence f (x) ≥ f (1) =
0. The equality holds for x = 1.
Remark. The following more general inequality holds for 0 < k ≤ e and any positive
numbers a and b (Vasile Cîrtoaje, 2006):
a ka + b kb ≥ a kb + b ka .
Conjecture 1. If 0 < k ≤ e and a, b ∈ (0, 4], then
p
2 a ka b kb ≥ a kb + b ka .
Conjecture 2. If a, b ∈ (0, 5], then
2a a b b ≥ a2b + b2a .
Conjecture 3. If a, b ∈ [0, 5], then
a2 + b2
2
a+b
2
≥ a2b + b2a .
Note that Conjecture 1 has been proved for a, b ∈ (0, 1] by A. Coronel and F. Huancas
(2014), and also by L. Matejicka (2014).
Symmetric Power-Exponential Inequalities
431
P 3.5. If a, b are nonnegative real numbers such that a + b = 2, then
(a)
a b + b a ≤ 1 + a b;
(b)
a2b + b2a ≤ 1 + a b.
(Vasile Cîrtoaje, 2007)
Solution. Without loss of generality, assume that a ≥ b. Since 0 ≤ b ≤ 1 and 0 ≤
a − 1 ≤ 1, by Bernoulli’s inequality, we have
a b ≤ 1 + b(a − 1) = 1 + b − b2
and
b a = b · b a−1 ≤ b[1 + (a − 1)(b − 1)] = b2 (2 − b).
(a) We have
a b + b a − 1 − a b ≤ (1 + b − b2 ) + b2 (2 − b) − 1 − (2 − b)b = −b(b − 1)2 ≤ 0.
The equality holds for a = b = 1, for a = 2 and b = 0, and for a = 0 and b = 2.
(b) We have
a2b + b2a − 1 − a b ≤ (1 + b − b2 )2 + b4 (2 − b)2 − 1 − (2 − b)b
= b3 (b − 1)2 (b − 2) = −a b3 (b − 1)2 ≤ 0.
The equality holds for a = b = 1, for a = 2 and b = 0, and for a = 0 and b = 2.
P 3.6. If a, b are nonnegative real numbers such that a2 + b2 = 2, then
a2b + b2a ≤ 1 + a b.
(Vasile Cîrtoaje, 2007)
Solution. Without loss of generality, assume that a ≥ 1 ≥ b. Applying Bernoulli’s inequality gives
a b ≤ 1 + b(a − 1),
hence
a2b ≤ (1 + a b − b)2 .
Also, since 0 ≤ b ≤ 1 and 2a ≥ 2, we have
b2a ≤ b2 .
432
Vasile Cîrtoaje
Therefore, it suffices to show that
(1 + a b − b)2 + b2 ≤ 1 + a b,
which can be written as
b(2 + 2a b − a − 2b − a2 b) ≥ 0.
So, we need to show that
2 + 2a b − a − 2b − a2 b ≥ 0,
which is equivalent to
2(1 − a)(1 − b) + a(1 − a b) ≥ 0,
4(1 − a)(1 − b) + a(a − b)2 ≥ 0.
Since a ≥ 1, it suffices to show that
4(1 − a)(1 − b) + (a − b)2 ≥ 0.
Indeed,
4(1 − a)(1 − b) + (a − b)2 = −4(a − 1)(1 − b) + [(a − 1) + (1 − b)]2
= [(a − 1) − (1 − b)]2 = (a + b − 2)2 ≥ 0.
The equality holds for a = b = 1, for a =
p
p
2 and b = 0, and for a = 0 and b = 2.
P 3.7. If a, b are positive real numbers, then
a a b b ≤ (a2 − a b + b2 )
a+b
2
.
Solution. By the weighted AM-GM inequality, we have
a
b
a · a + b · b ≥ (a + b)a a+b b a+b ,
a2 + b2
a+b
a+b
≥ aa b b .
Thus, it suffices to show that
2
2
a − ab + b ≥
a2 + b2
a+b
2
,
Symmetric Power-Exponential Inequalities
433
which is equivalent to
(a + b)(a3 + b3 ) ≥ (a2 + b2 )2 ,
a b(a − b)2 ≥ 0.
The equality holds for a = b.
P 3.8. If a, b ∈ (0, 1], then
a a b b ≤ 1 − a b + a2 b2 .
(Vasile Cîrtoaje, 2010)
Solution. We claim that
x x ≤ x2 − x + 1
for all x ∈ (0, 1]. If this is true, then
1 − a b + a2 b2 − a a b b ≥ 1 − a b + a2 b2 − (a2 − a + 1)(b2 − b + 1)
= (a + b)(1 − a)(1 − b) ≥ 0.
To prove the inequality x x ≤ x 2 − x + 1, we show that f (x) ≤ 0 for x ∈ (0, 1], where
f (x) = x ln x − ln(x 2 − x + 1).
We have
f 0 (x) = ln x + 1 −
f 00 (x) =
2x − 1
,
− x +1
x2
(1 − x)(1 − 2x − x 2 − x 4 )
.
x(x 2 − x + 1)2
Let x 1 ∈ (0, 1) be the positive root of the equation x 4 + x 2 + 2x = 1. Then, f 00 (x) > 0
for x ∈ (0, x 1 ) and f 00 (x) < 0 for x ∈ (x 1 , 1), hence f 0 is strictly increasing on (0, x 1 ]
and strictly decreasing on [x 1 , 1]. Since lim x→0 f 0 (x) = −∞ and f 0 (1) = 0, there is
x 2 ∈ (0, x 1 ) such that f 0 (x 2 ) = 0, f 0 (x) < 0 for x ∈ (0, x 2 ) and f 0 (x) > 0 for x ∈ (x 2 , 1).
Therefore, f is decreasing on (0, x 2 ] and increasing on [x 2 , 1]. Since lim x→0 f (x) = 0
and f (1) = 0, it follows that f (x) ≤ 0 for x ∈ (0, 1]. The proof is completed. The
equality holds for a = b = 1.
P 3.9. If a, b are positive real numbers such that a + b ≤ 2, then
a b  b ‹a
+
≤ 2.
b
a
(Vasile Cîrtoaje, 2010)
434
Vasile Cîrtoaje
Solution. Using the substitution a = t c and b = t d, where c, d, t are positive real
numbers such that c + d = 2 and t ≤ 1, we need to show that
c t d
d
 ‹t c
d
+
≤ 2.
c
Write this inequality as f (t) ≤ 2, where
t
t
f (t) = A + B , A =
c d
d
 ‹c
d
, B=
.
c
Since f (t) is a convex function, we have
f (t) ≤ max{ f (0), f (1)} = max{2, f (1)}.
Therefore, it suffices to show that f (1) ≤ 2; that is,
2c c d d ≥ c 2 + d 2 .
Setting c = 1 + x and d = 1 − x, where 0 ≤ x < 1, this inequality turns into
(1 + x)1+x (1 − x)1−x ≥ 1 + x 2 ,
which is equivalent to f (x) ≥ 0, where
f (x) = (1 + x) ln(1 + x) + (1 − x) ln(1 − x) − ln(1 + x 2 ).
We have
f 0 (x) = ln(1 + x) − ln(1 − x) −
f 00 (x) =
2x
,
1 + x2
1
1
2(1 − x 2 )
8x 2
+
−
=
.
1 + x 1 − x (1 + x 2 )2
(1 − x 2 )(1 + x 2 )2
Since f 00 (x) ≥ 0 for x ∈ [0, 1), it follows that f 0 is increasing, f 0 (x) ≥ f 0 (0) = 0, f (x)
is increasing, hence f (x) ≥ f (0) = 0. The proof is completed. The equality holds for
a = b.
P 3.10. If a, b are positive real numbers such that a + b = 2, then
1
2a a b b ≥ a2b + b2a + (a − b)2 .
2
(Vasile Cîrtoaje, 2010)
Symmetric Power-Exponential Inequalities
435
Solution. According to the inequalities in P 3.5-(b) and P 3.9, we have
a2b + b2a ≤ 1 + a b
and
2a a b b ≥ a2 + b2 .
Therefore, it suffices to show that
1
a2 + b2 ≥ 1 + a b + (a − b)2 .
2
which is an identity. The equality holds for a = b = 1.
P 3.11. If a, b ∈ (0, 1] or a, b ∈ [1, ∞), then
2a a b b ≥ a2 + b2 .
Solution. For a = x and b = 1, the desired inequality becomes
2x x ≥ x 2 + 1,
x > 0.
If this inequality is true, then
4a a b b − 2(a2 + b2 ) ≥ (a2 + 1)(b2 + 1) − 2(a2 + b2 ) = (a2 − 1)(b2 − 1) ≥ 0.
To prove the inequality 2x x ≥ x 2 + 1, we show that f (x) ≥ 0, where
f (x) = ln 2 + x ln x − ln(x 2 + 1).
We have
f 0 (x) = ln x + 1 −
f 00 (x) =
2x
,
+1
x2
x 2 (x + 1)2 + (x − 1)2
.
x(x 2 + 1)2
Since f 00 (x) > 0 for x > 0, f 0 is strictly increasing. Since f 0 (1) = 0, it follows that
f 0 (x) < 0 for x ∈ (0, 1) and f 0 (x) > 0 for x ∈ (1, ∞). Therefore, f is decreasing on
(0, 1] and increasing on [1, ∞), hence f (x) ≥ f (1) = 0 for x > 0. This completes the
proof. The equality holds for a = b = 1.
436
Vasile Cîrtoaje
P 3.12. If a, b are positive real numbers, then
2a a b b ≥ a2 + b2 .
(Vasile Cîrtoaje, 2014)
Solution. By Lemma below, it suffices to show that
(a4 − 2a3 + 4a2 − 2a + 3)(b4 − 2b3 + 4b2 − 2b + 3) ≥ 8(a2 + b2 ),
which is equivalent to A ≥ 0, where
A =a4 b4 − 2a3 b3 (a + b) + 4a2 b2 (a2 + b2 + a b) − [2a b(a3 + b3 ) + 8a2 b2 (a + b)]
+ [3(a4 + b4 ) + 4a b(a2 + b2 ) + 16a2 b2 ] − [6(a3 + b3 ) + 8a b(a + b)]
+ 4(a2 + b2 + a b) − 6(a + b) + 9.
We can check that
A = [a2 b2 − a b(a + b) + a2 + b2 − 1]2 + B,
where
B =a2 b2 (a + b)2 − 6a2 b2 (a + b) + [2(a4 + b4 ) + 4a b(a2 + b2 ) + 16a2 b2 ]
− [6(a3 + b3 ) + 10a b(a + b)] + [6(a2 + b2 ) + 4a b] − 6(a + b) + 8.
Also, we have
B = [a b(a + b) − 3a b + 1]2 + C,
where
C =[2(a4 + b4 ) + 4a b(a2 + b2 ) + 7a2 b2 ] − [6(a3 + b3 ) + 12a b(a + b)]
+ [6(a2 + b2 ) + 10a b] − 6(a + b) + 7,
and
C = (a b − 1)2 + 2D,
where
D =[a4 + b4 + 2a b(a2 + b2 ) + 3a2 b2 ] − [3(a3 + b3 ) + 6a b(a + b)]
+ 3(a + b)2 − 3(a + b) + 3,
It suffices to show that D ≥ 0. Indeed,
D =[(a + b)4 − 2a b(a + b)2 + a2 b2 ] − 3[(a + b)3 − a b(a + b)]
+ 3(a + b)2 − 3(a + b) + 3
=[(a + b)2 − a b]2 − 3(a + b)[(a + b)2 − a b] + 3(a + b)2 − 3(a + b) + 3
˜2

‹2
•
a+b
3
2
− 1 ≥ 0.
= (a + b) − a b − (a + b) + 3
2
2
Symmetric Power-Exponential Inequalities
437
This completes the proof. The equality holds for a = b = 1.
Lemma. If x > 0, then
1
x x ≥ x + (x − 1)2 (x 2 + 3).
4
Proof. We need to show that f (x) ≥ 0 for x > 0, where
g(x) = x 4 − 2x 3 + 4x 2 − 2x + 3.
f (x) = ln 4 + x ln x − ln g(x),
We have
f 0 (x) = 1 + ln x −
f 00 (x) =
2(2x 3 − 3x 2 + 4x − 1)
,
g(x)
x 8 + 6x 4 − 32x 3 + 48x 2 − 32x + 9 (x − 1)2 h(x)
=
,
g 2 (x)
g 2 (x)
where
h(x) = x 6 + 2x 5 + 3x 4 + 4x 3 + 11x 2 − 14x + 9.
Since
h(x) > 7x 2 − 14x + 7 = 7(x − 1)2 ≥ 0,
we have f 00 (x) ≥ 0, hence f 0 is strictly increasing on (0, ∞). Since f 0 (1) = 0, it follows
that f 0 (x) < 0 for x ∈ (0, 1) and f 0 (x) > 0 for x ∈ (1, ∞). Therefore, f is decreasing
on (0, 1] and increasing on [1, ∞), hence f (x) ≥ f (1) = 0 for x > 0.
P 3.13. If a, b are positive real numbers, then
aa b b ≥
a2 + b2
2
a+b
2
.
First Solution. Using the substitution a = b x, where x > 0, the inequality becomes as
follows:
2 2
bx+b
b x + b2 2
bx b
(b x) b ≥
,
2
x
(b x) b ≥
b2 x 2 + b2
2
b x+1 x x ≥ b x+1
x+1
2
x2 + 1
2
,
x+1
2
,
438
Vasile Cîrtoaje
x
x ≥
x2 + 1
2
x+1
2
.
It is true if f (x) ≥ 0 for all x > 0, where
f (x) =
We have
f 0 (x) =
x
1 x2 + 1
ln x − ln
.
x +1
2
2
g(x)
x
1
1
−
=
ln x +
,
(x + 1)2
x + 1 x 2 + 1 (x + 1)2
where
g(x) = ln x −
Since
g 0 (x) =
x2 − 1
.
x2 + 1
(x 2 − 1)2
≥ 0,
x(x 2 + 1)2
g is strictly increasing, therefore g(x) < 0 for x ∈ (0, 1), g(1) = 0, g(x) > 0 for x ∈
(1, ∞). Thus, f is decreasing on (0, 1] and increasing on [1, ∞), hence f (x) ≥ f (1) =
0. This completes the proof. The equality holds for a = b.
Second Solution. Write the inequality in the form
a ln a + b ln b ≥
a+b
a2 + b2
log
.
2
2
Without loss of generality, consider that a + b = 2k, k > 0, and denote
a = k + x,
b = k − x, 0 ≤ x < k.
We need to show that f (x) ≥ 0, where
f (x) = (k + x) ln(k + x) + (k − x) ln(k − x) − k ln(x 2 + k2 ).
We have
f 0 (x) = ln(k + x) − ln(k − x) −
2k x
,
+ k2
x2
1
1
2k(x 2 − k2 )
+
+
k+x k−x
(x 2 + k2 )2
2 2
8k x
= 2
.
(k − x 2 )(x 2 + k2 )2
f 00 (x) =
Since f 00 (x) ≥ 0 for x ≥ 0, f 0 is increasing, hence f 0 (x) ≥ f 0 (0) = 0. Therefore, f is
increasing on [0, k), hence f (x) ≥ f (0) = 0.
Symmetric Power-Exponential Inequalities
439
Remark. For a + b = 2, this inequality can be rewritten as
2a a b b ≥ a2 + b2 ,
2≥
a b
b
 ‹a
b
+
.
a
Also, for a + b = 1, the inequality becomes
2a2a b2b ≥ a2 + b2 ,
2≥
a 2b
b
+
 ‹2a
b
.
a
P 3.14. If a, b are positive real numbers such that a2 + b2 = 2, then
1
2a a b b ≥ a2b + b2a + (a − b)2 .
2
(Vasile Cîrtoaje, 2010)
Solution. According to the inequalities in P 3.6 and P 3.12, we have
a2b + b2a ≤ 1 + a b
and
2a a b b ≥ a2 + b2 .
Therefore, it suffices to show that
1
a2 + b2 ≥ 1 + a b + (a − b)2 ,
2
which is an identity. The equality holds for a = b = 1.
P 3.15. If a, b ∈ (0, 1], then
(a2 + b2 )

1
1
+ 2b
2a
a
b
‹
≤ 4.
(Vasile Cîrtoaje, 2014)
440
Vasile Cîrtoaje
Solution. For a = x and b = 1, the desired inequality becomes
x 2x ≥
1 + x2
,
3 − x2
x ∈ (0, 1].
If this inequality is true, it suffices to show that
3 − a2 3 − b2
2
2
≤ 4,
+
(a + b )
1 + a2 1 + b2
which is equivalent to
a2 b2 (2 + a2 + b2 ) + 2 − (a2 + b2 ) − (a2 + b2 )2 ≥ 0,
(2 + a2 + b2 )(1 − a2 )(1 − b2 ) ≥ 0.
To prove the inequality x 2x ≥
1 + x2
for 0 < x ≤ 1 , we show that f (x) ≥ 0, where
3 − x2
f (x) = x ln x +
1
1
ln(3 − x 2 ) − ln(1 + x 2 ).
2
2
We have
f 0 (x) = 1 + ln x −
x
x
−
,
2
3− x
1 + x2
1
3 + x2
1 − x2
−
−
x (3 − x 2 )2 (1 + x 2 )2
1 − x2
(1 − x)(9 + 6x − x 3 )
−
.
=
x(3 − x)2
(1 + x 2 )2
f 00 (x) =
We will show that f 00 (x) > 0 for 0 < x < 1. This is true if
9 + 6x − x 3
1+ x
−
> 0.
2
x(3 − x)
(1 + x 2 )2
Indeed,
9 + 6x − x 3
1+ x
9
1+ x
1
−
>
−
=
> 0.
2
2
2
2
x(3 − x)
(1 + x )
9x x(1 + x)
1+ x
Since f 00 (x) > 0, f 0 is strictly increasing on (0, 1]. Since f 0 (1) = 0, it follows that
f 0 (x) < 0 for x ∈ (0, 1), f is strictly decreasing on (0, 1], hence f (x) ≥ f (1) = 0. This
completes the proof. The equality holds for a = b = 1.
Symmetric Power-Exponential Inequalities
441
P 3.16. If a, b are positive real numbers such that a + b = 2, then
a b b a + 2 ≥ 3a b.
(Vasile Cîrtoaje, 2010)
Solution. Setting a = 1+ x and b = 1− x, where 0 ≤ x < 1, the inequality is equivalent
to
(1 + x)1−x (1 − x)1+x ≥ 1 − 3x 2 .
1
Consider further the non-trivial case 0 ≤ x < p , and write the desired inequality as
3
f (x) ≥ 0, where
f (x) = (1 − x) ln(1 + x) + (1 + x) ln(1 − x) − ln(1 − 3x 2 ).
We have
f 0 (x) = − ln(1 + x) + ln(1 − x) +
1− x 1+ x
6x
−
+
,
1 + x 1 − x 1 − 3x 2
1 00
−1
2(x 2 + 1) 3(3x 2 + 1)
f (x) =
−
+
.
2
1 − x 2 (1 − x 2 )2 (1 − 3x 2 )2
Making the substitution t = x 2 , 0 ≤ t <
1
, we get
3
3(3t + 1)
t +3
4t(5 − 9t)
1 00
f (x) =
−
=
> 0.
2
2
2
(3t − 1)
(t − 1)
(t − 1)2 (3t − 1)2
Therefore, f 0 (x) is strictly increasing, f 0 (x) ≥ f 0 (0) = 0, f (x) is strictly increasing, and
hence f (x) ≥ f (0) = 0. This completes the proof. The equality holds for a = b = 1.
P 3.17. If a, b ∈ [0, 1], then
a b−a + b a−b + (a − b)2 ≤ 2.
(Vasile Cîrtoaje, 2010)
Solution (by Vo Quoc Ba Can). Without loss of generality, assume that a ≥ b. Using the
substitution
c = a − b,
we need to show that
(b + c)−c + b c + c 2 ≤ 2
for
0 ≤ b ≤ 1 − c, 0 ≤ c ≤ 1.
442
Vasile Cîrtoaje
If c = 1, then b = 0, and the inequality is an equality. Also, for c = 0, the inequality is
an equality. Consider further that 0 < c < 1. We need to show that f (x) ≤ 0, where
f (x) = (x + c)−c + x c + c 2 − 2,
x ∈ [0, 1 − c].
We claim that f 0 (x) > 0 for x > 0. Then, f (x) is strictly increasing on [0, 1 − c], hence
f (x) ≤ f (1 − c) = (1 − c)c − (1 − c 2 ).
In addition, by Bernoulli’s inequality,
f (x) ≤ 1 + c(−c) − (1 − c 2 ) = 0.
Since
f 0 (x) =
c[(x + c)1+c − x 1−c ]
,
(x + c)1+c x 1−c
we have f 0 (x) > 0 for x > 0 if and only if
1−c
x + c > x 1+c .
Let d =
2c
1−c .
Using the weighted AM-GM inequality, we have

‹ 1

‹ 1+c
1−c
1 + c 1 − c 1−c 1−c
1 − c d 1+d
x 1+c .
x +c =1· x +d ·
=
≥ (1 + d) x
2
2
2
2
Thus, it suffices to show that

‹ 1−c
1+c
1 − c 1+c
≥
.
2
2
Indeed, using Bernoulli’s inequality, we get

1−c
2
1−c
‹ 1+c

‹ 1−c
1 + c 1+c
1−c 1+c
1+c
= 1−
≤1−
·
=
.
2
1+c
2
2
The equality holds for a = b, for a = 1 and b = 0, and for a = 0 and b = 1.
P 3.18. If a, b are nonnegative real numbers such that a + b ≤ 2, then
a b−a + b a−b +
7
(a − b)2 ≤ 2.
16
(Vasile Cîrtoaje, 2010)
Symmetric Power-Exponential Inequalities
443
Solution. Assume that a > b and use the substitution
c = a − b.
We need to show that
a−c + (a − c)c +
7 2
c ≤2
16
for
c
c ≤ a ≤1+ .
2
Equivalently, we need to show that f (x) ≤ 0 for
h
ci
x ∈ c, 1 +
, 0<c≤2,
2
0 < c ≤ 2,
where
7 2
c − 2.
16
As we have shown
preceding proof, f 0 (x) > 0 for x > c. Therefore, f (x) is strictly
h in the
i
c
increasing on c, 1 + , and hence
2
c c −c c c
7 2
f (x) ≤ f 1 +
= 1+
+ 1−
+
c − 2.
2
2
2
16
f (x) = x −c + (x − c)c +
To end the proof we need to show that
c c
7 2
c −c + 1−
+
c ≤2
1+
2
2
16
for c ∈ [0, 2]. Indeed, by Lemma 1 and Lemma 2 below, we have
c −c
3 2
1+
+
c ≤ 1,
2
16
c c 1 2
1−
+ c ≤ 1.
2
4
Adding these inequalities yields the desired inequality. The equality holds for a = b, for
a = 2 and b = 0, and for a = 0 and b = 2.
Lemma 1. If 0 ≤ x ≤ 2, then
x −x
3 2
1+
+
x ≤ 1,
2
16
with equality for x = 0 and x = 2.
Proof. We need to show that f (x) ≤ 0 for 0 ≤ x ≤ 2, where
‹

x
3 2
f (x) = −x ln 1 +
− ln 1 −
x .
2
16
444
We have
Vasile Cîrtoaje
x(3x 2 + 6x − 4)
x
+
,
f 0 (x) = − ln 1 +
2
(2 + x)(16 − 3x 2 )
f 00 (x) =
g(x)
(2 +
x)2 (16 − 3x 2 )2
,
where
g(x) = −9x 5 − 18x 4 + 168x 3 + 552x 2 + 128x − 640.
Since g(x 1 ) = 0 for x 1 ≈ 0, 88067, g(x) < 0 for x ∈ [0, x 1 ) and g(x) > 0 for x ∈ (x 1 , 2],
f 0 is strictly decreasing on [0, x 1 ] and strictly increasing on [x 1 , 2]. Since f 0 (0) = 0
5
and f 0 (2) = − ln 2 + > 0, there is x 2 ∈ (x 1 , 2) such that f 0 (x 2 ) = 0, f 0 (x) < 0 for
2
x ∈ (0, x 2 ), and f 0 (x) > 0 for x ∈ (x 2 , 2]. Therefore, f is decreasing on [0, x 2 ] and
increasing on [x 2 , 2]. Since f (0) = f (2) = 0, it follows that f (x) ≤ 0 for x ∈ [0, 2].
Lemma 2. If 0 ≤ x ≤ 2, then
x x 1 2
1−
+ x ≤ 1,
2
4
with equality for x = 0 and x = 2.
Proof. We need to show that f (x) ≤ 0, where
‹

x
1 2
f (x) = x ln 1 −
− ln 1 − x .
2
4
We have
x
x2
f 0 (x) = ln 1 −
−
,
2
4 − x2
−1
8x
f 00 (x) =
−
.
2 − x (4 − x 2 )2
Since f 00 < 0 for 0 ≤ x < 2, f 0 is strictly decreasing, hence f 0 (x) ≤ f 0 (0) = 0, f is
strictly decreasing, therefore f (x) ≤ f (0) = 0 for x ∈ [0, 2].
Conjecture. If a, b are nonnegative real numbers such that a + b =
1
, then
4
a2(b−a) + b2(a−b) ≤ 2.
P 3.19. If a, b are nonnegative real numbers such that a + b ≤ 4, then
a b−a + b a−b ≤ 2.
(Vasile Cîrtoaje, 2010)
Symmetric Power-Exponential Inequalities
445
Solution. Without loss of generality, assume that a ≥ b. Consider first that a − b ≥ 2.
We have
a ≥ a − b ≥ 2,
and from
4 ≥ a + b = (a − b) + 2b ≥ 2 + 2b,
we get b ≤ 1. Clearly, the desired inequality is true because
a b−a < 1,
b a−b ≤ 1.
Since the case a − b = 0 is trivial, consider further that 0 < a − b < 2 and use the
substitution
c = a − b.
So, we need to show that
a−c + (a − c)c ≤ 2
for
c
c ≤ a ≤2+ .
2
Equivalently, we need to show that f (x) ≤ 0 for
h
ci
x ∈ c, 2 +
, 0<c<2,
2
0 < c < 2,
where
f (x) = x −c + (x − c)c − 2.
The derivative
f 0 (x) =
c[x 1+c − (x − c)1−c ]
x 1+c (x − c)1−c
has the same sign as
g(x) = (1 + c) ln x − (1 − c) ln(x − c).
We have
c(2x − 1 − c)
.
x(x − c)
•
˜
5
Case 1: c = 1. We need to show that f (x) ≤ 0 for x ∈ 1, , where
2
g 0 (x) =
f (x) =
Indeed, we have
f (x) =
x 2 − 3x + 1
.
x
(x − 1)(2x − 5) + (x − 3)
< 0.
2x
446
Vasile Cîrtoaje
Case 2: 1 < c < 2. Since
2x − 1 − c ≥ 2c − 1 − c > 0,
we have g 0 (x) > 0, hence g(x) is strictly increasing. Since g(x) → −∞ when x → c
and
c
c
c
g 2+
= (1 + c) ln 2 +
+ (c − 1) ln 2 −
2
2
2
c
c
c2
> (c − 1) ln 2 +
+ (c − 1) ln 2 −
= (c − 1) ln 4 −
> 0,
2
2
4
c
there exists x 1 ∈ c, 2 +
such that g(x 1 ) = 0, g(x) < 0 for x ∈ (c, x 1 ) and g(x) > 0
2
h
ci
c
. Thus, f (x) is decreasing on [c, x 1 ] and increasing on x 1 , 2 + .
for x ∈ x 1 , 2 +
2
2
c
Then, it suffices to show that f (c) ≤ 0 and f 2 +
≤ 0. We have
2
f (c) = c −c − 2 < 1 − 2 < 0.
c
Also, the inequality f 2 +
≤ 0 has the form
2
c −c c c
2+
+ 2−
≤ 2,
2
2
which follows immediately from Lemma 1 below.
h
ci
Case 3: 0 < c < 1. We claim that g(x) > 0 for x ∈ c, 2 + . From
2
c(2x − 1 − c)
,
x(x − c)

˜
•
˜
1+c
1+c
c
it follows that g(x) is decreasing on c,
, and increasing on
, 2 + . Then,
2
2
2

‹
1+c
1−c
1+c
g(x) ≥ g
= (1 + c) ln
− (1 − c) ln
,
2
2
2
g 0 (x) =
and it suffices to show that

1+c
2
‹1+c
>

1−c
2
‹1−c
.
This inequality follows from Bernoulli’s inequality, as follows

1+c
2
‹1+c

‹
1 − c 1+c
(1 + c)(1 − c) 1 + c 2
= 1−
>1−
=
2
2
2
Symmetric Power-Exponential Inequalities
and

1−c
2
‹1−c
447

‹
1 + c 1−c
(1 − c)(1 + c) 1 + c 2
= 1−
<1−
=
.
2
2
2
h
ci
Since g(x) > 0 involves f 0 (x) > 0, it follows that f (x) is strictly increasing on c, 2 + ,
2
and hence
c
f (x) ≤ f 2 +
.
2
c
So, we need to show that f 2 +
≤ 0 for 0 < c < 2, which follows immediately from
2
Lemma 1.
The proof is completed. The equality holds for a = b.
Lemma 1. If 0 ≤ c ≤ 2, then
c c
c −c + 2−
≤ 2.
2+
2
2
Proof. According to Lemma 2, the following inequalities hold for c ∈ [0, 2]:
c −c
c3
2+
≤ 1 − c ln 2 + ,
2
9
c3
c c
≤ 1 + c ln 2 − .
2−
2
9
Summing these inequalities, the desired inequality follows.
Lemma 2. If
−2 ≤ x ≤ 2,
then
x x
x3
2−
≤ 1 + x ln 2 −
.
2
9
Proof. We have
ln 2 ≈ 0.693 < 7/9.
If x ∈ [0, 2], then
1 + x ln 2 −
x3
8
x3
≥1−
≥ 1 − > 0.
9
9
9
Also, if x ∈ [−2, 0], then
1 + x ln 2 −
x3
7x x 3
8 + 7x − x 3
≥1+
−
>
9
9
9
9
2(x + 2)2 + (−x)(x + 1)2
=
> 0.
9
448
Vasile Cîrtoaje
So, we can write the desired inequality as f (x) ≥ 0, where
x3
x
f (x) = ln 1 + d x −
− x ln 2 −
, d = ln 2.
9
2
We have
f 0 (x) =
9d − 3x 2
x
x
+
−
ln
2
−
.
9 + 9d x − x 3 4 − x
2
Since f (0) = 0, it suffices to show that f 0 (x) ≤ 0 for x ∈ [−2, 0], and f 0 (x) ≥ 0 for
x ∈ [0, 2]; that is, x f 0 (x) ≤ 0 for x ∈ [−2, 2]. According to Lemma 3, the inequality
x f 0 (x) ≤ 0 is true if x g(x) ≥ 0, where
‹

9d − 3x 2
x
x
1 2
g(x) =
+
− d− −
x .
9 + 9d x − x 3 4 − x
4 32
We have

‹
9d − 3x 2
x
x
1 2
g(x) =
−d +
+ +
x
9 + 9d x − x 3
4 − x 4 32
2
d x − 3x − 9d 2 64 − 4x − x 2
=x
+
,
9 + 9d x − x 3
32(4 − x)
hence
x g(x) =
x 2 g1 (x)
,
32(4 − x)(9 + 9d x − x 3 )
where
g1 (x) =32(4 − x)(d x 2 − 3x − 9d 2 ) + (64 − 4x − x 2 )(9 + 9d x − x 3 )
=x 5 + 4x 4 − (64 + 41d)x 3 + (87 + 92d)x 2 + 12(24d 2 + 48d − 35)x
+ 576(1 − 2d 2 ).
Since g1 (x) ≥ 0 for x ∈ [a1 , b1 ], where a1 ≈ −12.384 and b1 =≈ 2.652, we have
g1 (x) ≥ 0 for x ∈ [−2, 2].
Lemma 3. If x < 4, then
xh(x) ≤ 0,
where

‹
x
x
1 2
− ln 2 − −
x .
h(x) = ln 2 −
2
4 32
Proof. From
h0 (x) =
−x 2
≤ 0,
16(4 − x)
Symmetric Power-Exponential Inequalities
449
it follows that h(x) is decreasing. Since h(0) = 0, we have h(x) ≥ 0 for x ≤ 0, and
h(x) ≤ 0 for x ∈ [0, 4); that is, xh(x) ≤ 0 for x < 4.
P 3.20. Let a, b be positive real numbers such that a + b = 2. If k ≥
kb
1
, then
2
ka
a a b b ≥ 1.
(Vasile Cîrtoaje, 2010)
Solution. Setting a = 1+ x and b = 1− x, where 0 ≤ x < 1, the inequality is equivalent
to
(1 + x)k(1−x) ln(1 + x) + (1 − x)k(1+x) ln(1 − x) ≥ 0.
Consider further the non-trivial case 0 < x < 1, and write the desired inequality as
f (x) ≥ 0, where
f (x) = k(1 − x) ln(1 + x) − k(1 + x) ln(1 − x) + ln ln(1 + x) − ln(− ln(1 − x)).
We have
2k(1 + x 2 )
1
1
− k ln(1 − x 2 ) +
+
1 − x2
(1 + x) ln(1 + x) (1 − x) ln(1 − x)
1
1
2k
+
+
>
2
1− x
(1 + x) ln(1 + x) (1 − x) ln(1 − x)
1
1
1
≥
+
+
2
1− x
(1 + x) ln(1 + x) (1 − x) ln(1 − x)
g(x)
=
,
2
(1 − x ) ln(1 + x) ln(1 − x)
f 0 (x) =
where
g(x) = ln(1 + x) ln(1 − x) + (1 + x) ln(1 + x) + (1 − x) ln(1 − x).
It suffices to show that f 0 (x) > 0. Indeed, if this is true, then f (x) is strictly increasing,
hence
f (x) > lim f (x) = 0.
x→0
Thus, we need to prove that g(x) < 0. We have
g 0 (x) =
−x
h(x),
1 − x2
where
h(x) = (1 + x) ln(1 + x) + (1 − x) ln(1 − x).
450
Vasile Cîrtoaje
Since
h0 (x) = ln
1+ x
> 0,
1− x
h(x) is strictly increasing, h(x) > h(0) = 0, g 0 (x) < 0, g(x) is strictly decreasing, and
hence g(x) < g(0) = 0. This completes the proof. The equality holds for a = b = 1.
P 3.21. If a, b are positive real numbers such that a + b = 2, then
a
p
a
p
b
b
≥ 1.
(Vasile Cîrtoaje, 2010)
Solution. For a = b = 1, the equality holds. In what follows, we will assume that
a > 1 > b. Taking logarithms of both sides, the inequality becomes in succession
p
p
a ln a + b ln b ≥ 0,
p
p
a ln a ≥ b(− ln b),
1
1
ln a + ln ln a ≥ ln b + ln(− ln b).
2
2
Setting a = 1 + x and b = 1 − x, we need to show that f (x) ≥ 0 for 0 < x < 1, where
f (x) =
1
1
ln(1 + x) − ln(1 − x) + ln ln(1 + x) − ln(− ln(1 − x)).
2
2
We have
f 0 (x) =
1
1
1
+
+
.
2
1− x
(1 + x) ln(1 + x) (1 − x) ln(1 − x)
As shown in the proof of the preceding P 3.20, we have f 0 (x) > 0. Therefore, f (x) is
strictly increasing and
f (x) > lim f (x) = 0.
x→0
The equality holds for a = b = 1.
P 3.22. If a, b are positive real numbers such that a + b = 2, then
1 − a a+1 b b+1 ≥
1
(1 − a b)2 .
3
(Vasile Cîrtoaje, 2010)
Symmetric Power-Exponential Inequalities
451
Solution. Putting a = 1 + x and b = 1 − x, 0 ≤ x < 1, the inequality becomes
(1 + x)2+x (1 − x)2−x ≤ 1 −
1 4
x .
3
Write this inequality as f (x) ≤ 0, where
‹

1
f (x) = (2 + x) ln(1 + x) + (2 − x) ln(1 − x) − ln 1 − x 4 .
3
We have
f 0 (x) = ln(1 + x) − ln(1 − x) −
f 00 (x) =
=
2x
4x 3
+
,
1 − x2 3 − x4
2
2(1 + x 2 ) 4x 2 (x 4 + 9)
−
+
1 − x 2 (1 − x 2 )2
(3 − x 4 )2
4x 2 (x 4 + 9) −8x 4 [x 4 + 1 + 8(1 − x 2 )]
−4x 2
+
=
≤ 0.
(1 − x 2 )2
(3 − x 4 )2
(1 − x 2 )2 (3 − x 4 )2
Therefore, f 0 (x) is decreasing, f 0 (x) ≤ f 0 (0) = 0, f (x) is decreasing, f (x) ≤ f (0) = 0.
The equality holds for a = b = 1.
P 3.23. If a, b are positive real numbers such that a + b = 2, then
a−a + b−b ≤ 2.
(Vasile Cîrtoaje, 2010)
Solution. Consider that a ≥ b, when
0 < b ≤ 1 ≤ a < 2,
and write the inequality as
aa − 1 b b − 1
+
≥ 0.
aa
bb
According to Lemma from the proof of P 3.4, we have
aa − 1 ≥ a
a+1
2
(a − 1),
bb − 1 ≥ b
b+1
2
(b − 1).
Therefore, it suffices to show that
a
1−a
2
(a − 1) + b
1−b
2
(b − 1) ≥ 0,
which is equivalent to
a
1−a
2
≥b
1−b
2
,
452
Vasile Cîrtoaje
(a b)
1−b
2
≤ 1,
a b ≤ 1.
The last inequality follows immediately from the AM-GM inequality
ab ≤
1
(a + b)2 = 1.
4
The equality holds for a = b = 1.
P 3.24. If a, b are nonnegative real numbers such that a + b = 2, then
a2b + b2a ≥ a b + b a ≥ a2 b2 + 1.
(Vasile Cîrtoaje, 2010)
Solution. Since a, b ∈ [0, 2] and (1 − a)(1 − b) ≤ 0, from Lemma below, we have
ab − 1 ≥
b(a b + 1)(a − 1)
b(a b + 3 − a − b)(a − 1)
=
2
2
and
ba − 1 ≥
a(a b + 1)(b − 1)
.
2
Based on these inequalities, we get
a b + b a − a2 b2 − 1 = (a b − 1) + (b a − 1) + 1 − a2 b2
b(a b + 1)(a − 1) a(a b + 1)(b − 1)
+
+ 1 − a2 b2
2
2
= (ab + 1)(a b − 1) + 1 − a2 b2 = 0
≥
and
a2b + b2a − a b − b a = a b (a b − 1) + b a (b a − 1)
a b b(a b + 1)(a − 1) b a a(a b + 1)(b − 1)
+
2
2
a b(a b + 1)(a − b)(a b−1 − b a−1 )
=
.
4
≥
On the valid assumption a ≥ b, we only need to show that a b−1 ≥ b a−1 , which is
equivalent to
a
b−a
2
≥b
a−b
2
, 1 ≥ (a b)
a−b
2
, 1 ≥ a b, (a − b)2 ≥ 0.
Symmetric Power-Exponential Inequalities
453
For both inequalities, the equality holds when a = b = 1, when a = 0 and b = 2, and
when a = 2 and b = 0.
Lemma. If x, y ∈ [0, 2] such that (1 − x)(1 − y) ≤ 0, then
xy −1≥
y(x y + 3 − x − y)(x − 1)
,
2
with equality for x = 1, for y = 0, for y = 1, and for x = 0 and y = 2.
Proof. For y = 0, y = 1 and y = 2, the inequality is an identity. For fixed y ∈ (0, 1) ∪
(1, 2), let us define
f (x) = x y − 1 −
We have
0
•
f (x) = y x
y−1
y(x y + 3 − x − y)(x − 1)
.
2
˜
x y + 3 − x − y (x − 1)( y − 1)
−
,
−
2
2
f 00 (x) = y( y − 1)(x y−2 − 1).
Since f 00 (x) ≥ 0 for x ∈ (0, 2], f 0 is increasing. There are two cases to consider.
Case 1: x ≥ 1 > y. We have f 0 (x) ≥ f 0 (1) = 0, f (x) is increasing, hence f (x) ≥ f (1) =
0.
Case 2: y > 1 ≥ x. We have f 0 (x) ≤ f 0 (1) = 0, f (x) is decreasing, hence f (x) ≥
f (1) = 0.
P 3.25. If a, b are positive real numbers such that a + b = 2, then
a3b + b3a ≤ 2.
(Vasile Cîrtoaje, 2007)
Solution. Without loss of generality, assume that a ≥ b. Using the substitution a = 1+ x
and b = 1 − x, 0 ≤ x < 1, we can write the inequality as
e3(1−x) ln(1+x) + e3(1+x) ln(1−x) ≤ 2.
Applying Lemma below, it suffices to show that f (x) ≤ 2, where
f (x) = e3(1−x)(x−
x2
x3
2 + 3 )
+ e−3(1+x)(x+
x2
x3
2 + 3 )
.
454
Vasile Cîrtoaje
Since f (0) = 2, it suffices to show that f 0 (x) ≤ 0 for x ∈ [0, 1). From
9x 2 5x 3
4
15 2
x − 4x 3 )e3x− 2 + 2 −x
2
9x 2 5x 3
4
15 2
−(3 + 9x +
x + 4x 3 )e−3x− 2 − 2 −x ,
2
f 0 (x) =(3 − 9x +
it follows that f 0 (x) ≤ 0 is equivalent to
3
e−6x−5x ≥
6 − 18x + 15x 2 − 8x 3
.
6 + 18x + 15x 2 + 8x 3
For the nontrivial case 6 − 18x + 15x 2 − 8x 3 > 0, we rewrite this inequality as g(x) ≥ 0,
where
g(x) = −6x − 5x 3 − ln(6 − 18x + 15x 2 − 8x 3 ) + ln(6 + 18x + 15x 2 + 8x 3 ).
Since g(0) = 0, it suffices to show that g 0 (x) ≥ 0 for x ∈ [0, 1). From
(6 + 8x 2 ) − 10x
1 0
(6 + 8x 2 ) + 10x
g (x) = −2 − 5x 2 +
+
,
3
6 + 15x 2 − (18x + 8x 3 ) 6 + 15x 2 + (18x + 8x 3 )
it follows that g 0 (x) ≥ 0 is equivalent to
2(6 + 8x 2 )(6 + 15x 2 ) − 20x(18x + 8x 3 ) ≥ (2 + 5x 2 )[(6 + 15x 2 )2 − (18x + 8x 3 )2 ].
Since
(6 + 15x 2 )2 − (18x + 8x 3 )2 ≤ (6 + 15x 2 )2 − 324x 2 − 288x 4 ≤ 4(9 − 36x 2 ),
it suffices to show that
(3 + 4x 2 )(6 + 15x 2 ) − 5x(18x + 8x 3 ) ≥ (2 + 5x 2 )(9 − 36x 2 ).
This reduces to 6x 2 +200x 4 ≥ 0, which is clearly true. The equality holds for a = b = 1.
Lemma. If t > −1, then
ln(1 + t) ≤ t −
t2 t3
+ .
2
3
Proof. We need to prove that f (t) ≥ 0, where
f (t) = t −
t2 t3
+
− ln(1 + t).
2
3
Since
t3
,
t +1
f (t) is decreasing on (−1, 0] and increasing on [0, ∞). Therefore, f (t) ≥ f (0) = 0.
f 0 (t) =
Symmetric Power-Exponential Inequalities
455
P 3.26. If a, b are nonnegative real numbers such that a + b = 2, then
a3b + b3a +

a−b
2
‹4
≤ 2.
(Vasile Cîrtoaje, 2007)
Solution (by M. Miyagi and Y. Nishizawa). We may assume that a = 1+ x and b = 1− x,
where 0 ≤ x ≤ 1. The desired inequality is equivalent to
(1 + x)3(1−x) + (1 − x)3(1+x) + x 4 ≤ 2.
By Lemma below, we have
(1 + x 1−x ≤
1
(1 + x)2 (2 − x 2 )(2 − 2x + x 2 ),
4
(1 − x 1+x ≤
1
(1 − x)2 (2 − x 2 )(2 + 2x + x 2 ).
4
Therefore, it suffices to show that
(1 + x)6 (2 − x 2 )3 (2 − 2x + x 2 )3 + (1 − x)6 (2 − x 2 )3 (2 + 2x + x 2 )3 + 64x 2 ≤ 128,
which is equivalent to
x 4 (1 − x 2 )[x 6 (x 6 − 8x 4 + 31x 2 − 34) − 2(17x 6 − 38x 4 + 16x 2 + 8)] ≤ 0.
Clearly, it suffices to show that
t 3 − 8t 2 + 31t − 34 < 0
and
17t 3 − 38t 2 + 16t + 8 > 0
for all t ∈ [0, 1]. Indeed, we have
t 3 − 8t 2 + 31t − 34 < t 3 − 8t 2 + 31t − 24 = (t − 1)(t 2 − 7t + 24) ≤ 0,
17t 3 − 38t 2 + 16t + 8 = 17t(t − 1)2 + (−4t 2 − t + 8) > 0.
Lemma. If −1 ≤ t ≤ 1, then
(1 + t)1−t ≤
1
(1 + t)2 (2 − t 2 )(2 − 2t + t 2 ),
4
with equality for t = −1, t = 0 and t = 1.
456
Vasile Cîrtoaje
Proof. We need to prove that f (t) ≥ 0 for −1 < t ≤ 1, where
f (t) = (1 + t) ln(1 + t) + ln(2 − t 2 ) + ln(2 − 2t + t 2 ) − ln 4.
We have
f 0 (t) = 1 + ln(1 + t) −
f 00 (t) =
2(t − 1)
2t
+
,
2 − t 2 2 − 2t + t 2
t 2 g(t)
,
(1 + t)(2 − t 2 )2 (2 − 2t + t 2 )2
where
g(t) = t 6 − 8t 5 + 12t 4 + 8t 3 − 20t 2 − 16t + 16.
Case 1: 0 ≤ t ≤ 1. From
g 0 (t) = 6t 5 − 40t 4 + 48t 3 + 24t 2 − 40t − 16 = 6t 5 − 8t − 16 − 8t(5t 3 − 6t 2 − 3t + 4)
= (6t 5 − 8t − 16) − 8t(t − 1)2 (5t + 4) < 0,
it follows that g is strictly decreasing on [0, 1]. Since g(0) = 16 and g(1) = −7, there
exists a number c ∈ (0, 1) such that g(c) = 0, g(t) > 0 for 0 ≤ t < c and g(t) < 0
for c < t ≤ 1. Therefore, f 0 is strictly increasing on [0, c] and strictly decreasing on
[c, 1]. From f 0 (0) = 0 and f 0 (1) = ln 2 − 1 < 0, it follows that there exists a number
d ∈ (0, 1) such that f 0 (d) = 0, f 0 (t) > 0 for 0 < t < d and f 0 (t) < 0 for c < t ≤ 1. As
a consequence, f is strictly increasing on [0, d] and strictly decreasing on [d, 1]. Since
f (0) = 0 and f (1) = 0, we have f (t) ≥ 0 for 0 ≤ t ≤ 1.
Case 2: −1 < t ≤ 0. From
g(t) = t 4 (t − 2)(t − 6) + 4(t + 1)(2t 2 − 7t + 3) + 4 > 0,
it follows that f 0 is strictly increasing on (−1, 0]. Since f 0 (0) = 0, we have f 0 (t) < 0
for −1 < t < 0, hence f is strictly decreasing on (−1, 0]. From f (0) = 0, it follows that
f (t) ≥ 0 for −1 < t ≤ 0.
Conjecture. If a, b are nonnegative real numbers such that a + b = 2, then

‹
a−b 2
a3b + b3a +
≥ 2.
2
P 3.27. If a, b are positive real numbers such that a + b = 2, then
2
2
a a + b b ≤ 2.
(Vasile Cîrtoaje, 2008)
Symmetric Power-Exponential Inequalities
457
Solution. Without loss of generality, assume that
0 < a ≤ 1 ≤ b < 2,
and write the inequality as
1
1 1/a
2
a
+
1
1 2/b
b
≤ 2.
By Bernoulli’s inequality, we have
‹

‹
1 1/a
1 1
a3 − a2 + 1
≥
1
+
−
1
=
,
a2
a a2
a3
 ‹2/b

‹
1
2 1
b2 − 2b + 2
≥1+
−1 =
.
b
b b
b2

Therefore, it suffices to show that
a3
b2
+
≤ 2,
a3 − a2 + 1 b2 − 2b + 2
which is equivalent to
a3
(2 − b)2
≤
,
a3 − a2 + 1
b2 − 2b + 2
a3
a2
≤
,
a3 − a2 + 1
a2 − 2a + 2
a2 (a − 1)2 ≥ 0.
The equality happens for a = b = 1.
P 3.28. If a, b are positive real numbers such that a + b = 2, then
3
3
a a + b b ≥ 2.
(Vasile Cîrtoaje, 2008)
Solution. Assume that a ≤ b; that is,
0 < a ≤ 1 ≤ b < 2.
3
3
and ≤ a ≤ 1.
5
5
3
7
Case 1: 0 < a ≤ . From a + b = 2, we get
≤ b < 2. Let
5
5
There are two cases to consider: 0 < a ≤
3
f (x) = x x , 0 < x < 2.
458
Vasile Cîrtoaje
Since
3
f 0 (x) = 3x x −2 (1 − ln x) > 0,
‹
 ‹
•
7
7
, 2 , and hence f (b) ≥ f
; that is,
f (x) is increasing on
5
5
3
bb ≥
 ‹15/7
7
.
5
Using Bernoulli’s inequality gives
 ‹15/7

‹

‹
7
7
2 8/7 7
16
51
b ≥
=
1+
>
1+
=
> 2.
5
5
5
5
35
25
3
b
hence
3
3
a a + b b > 2.
Case 2:
3
≤ a ≤ 1. By Lemma below, we have
5
3
2a a ≥ 3 − 15a + 21a2 − 7a3
and
3
2b b ≥ 3 − 15b + 21b2 − 7b3 .
Summing these inequalities, we get
€ 3
Š
3
2 a a + b b ≥ 6 − 15(a + b) + 21(a2 + b2 ) − 7(a3 + b3 )
= 6 − 15(a + b) + 21(a + b)2 − 7(a + b)3 = 4.
This completes the proof. The equality holds for a = b = 1.
Lemma. If
3
≤ x ≤ 2, then
5
3
2x x ≥ 3 − 15x + 21x 2 − 7x 3 ,
with equality for x = 1.
Proof. We first show that h(x) > 0, where
h(x) = 3 − 15x + 21x 2 − 7x 3 .
From
h0 (x) = 3(−5 + 14x − 7x 2 ),
Symmetric Power-Exponential Inequalities
s
2
,1 +
7
459
s s
2
2
, and decreasing on 1 +
,∞ .
7
7
it follows that h(x) is increasing on 1 −
 ‹
3
Then, it suffices to show that f
≥ 0 and f (2) ≥ 0. Indeed
5
 ‹
3
6
f
=
, f (2) = 1.
5
125
Write now the desired inequality as f (x) ≥ 0, where
f (x) = ln 2 +
We have
3
ln x − ln(3 − 15x + 21x 2 − 7x 3 ),
x
x2 0
f (x) = g(x),
3
g(x) = 1 − ln x +
g 0 (x) =
3
≤ x ≤ 2.
5
x 2 (7x 2 − 14x + 5)
,
3 − 15x + 21x 2 − 7x 3
g1 (x)
,
(3 − 15x + 21x 2 − 7x 3 )2
where
g1 (x) = −49x 7 + 245x 6 − 280x 5 − 147x 4 + 471x 3 − 321x 2 + 90x − 9.
In addition,
g1 (x = (x − 1)2 g2 (x),
g2 (x) = 11x 5 + 3g3 (x),
g2 (x) = −49x 5 + 147x 4 + 63x 3 − 168x 2 + 72x − 9,
g3 (x) = −20x 5 + 49x 4 + 21x 3 − 56x 2 + 24x − 3,
g3 (x) = (4x − 1)g4 (x),
g4 (x) = x 5 + g5 (x),
g4 (x) = −5x 4 + 11x 3 + 8x 2 − 12x + 3,
g5 (x) = −6x 4 + 11x 3 + 8x 2 − 12x + 3,
g5 (x) = (2x − 1)g6 (x),
g6 (x) = −3x 3 + 4x 2 + 6x − 3,
g6 (x) = 1 + (2 − x)(3x 2 + 2x − 2).
Therefore, we get in succession g6 (x) > 0, g5 (x) > 0, g4 (x) > 0, g3 (x) > 0, g2 (x)
‹
• > 0,
3
0
g1 (x) ≥ 0, g (x) ≥ 0, g(x) is increasing. Since g(1) = 0, we have g(x) < 0 on
,1
5
•
˜
3
and g(x) > 0 on (1, 2]. Then, f (x) is decreasing on
, 1 and increasing on [1, 2],
5
hence f (x) ≥ f (1) = 0.
460
Vasile Cîrtoaje
P 3.29. If a, b are positive real numbers such that a + b = 2, then
2
2
a5b + b5a ≤ 2.
(Vasile Cîrtoaje, 2010)
Solution. Assume that a ≥ b. For a = 2 and b = 0, the inequality is obvious. Otherwise,
using the substitution a = 1 + x and b = 1 − x, 0 ≤ x < 1, we can write the desired
inequality as
2
2
e5(1−x) ln(1+x) + e5(1+x) ln(1−x) ≤ 2.
According to Lemma below, it suffices to show that f (x) ≤ 2, where
f (x) = e5(u−v) + e−5(u+v) ,
7 3 31 5
5
17 4
9 6
x +
x , v = x2 +
x +
x .
3
30
2
12
20
If f 0 (x) ≤ 0, then f (x) is decreasing, hence f (x) ≤ f (0) = 2. Since
u= x+
f 0 (x) = 5(u0 − v 0 )e5(u−v) − 5(u0 + v 0 )e−5(u+v) ,
u0 = 1 + 7x 2 +
31 4
x ,
6
v 0 = 5x +
17 3 27 5
x +
x ,
3
10
the inequality f 0 (x) ≤ 0 becomes
e−10u (u0 + v 0 ) ≥ u0 − v 0
For the nontrivial case u0 − v 0 > 0, we rewrite this inequality as g(x) ≥ 0, where
g(x) = −10u + ln(u0 + v 0 ) − ln(u0 − v 0 ).
If g 0 (x) ≥ 0, then g(x) is increasing, hence g(x) ≥ f (0) = 2. We have
g 0 (x) = −10u0 +
u00 + v 00 u00 − v 00
− 0
,
u0 + v 0
u − v0
where
62 3
27 4
x , v 00 = 5 + 17x 2 +
x .
3
2
Thus, the inequality g 0 (x) ≥ 0 is equivalent to
u00 = 14x +
u0 v 00 − v 0 u00 ≥ 5u0 (u0 + v 0 )(u0 − v 0 ),
a1 t + a2 t 2 + a3 t 3 + a4 t 4 + a5 t 5 + a6 t 6 + a7 t 7 ≥ 0,
where t = x 2 , 0 ≤ t < 1, and
a1 = 2,
a2 = 321.5,
a3 ≈ 152.1,
a4 ≈ −498.2,
Symmetric Power-Exponential Inequalities
a5 ≈ −168.5,
461
a6 ≈ 356.0,
a7 ≈ 188.3.
This inequality is true if
300t 2 + 150t 3 − 500t 4 − 200t 5 + 250t 6 ≥ 0.
Since the last inequality is equivalent to the obvious inequality
50t 2 (1 − t)(6 + 9t − t 2 − 5t 3 ) ≥ 0,
the proof is completed. The equality holds for a = b = 1.
Lemma. If −1 < t < 1, then
(1 − t)2 ln(1 + t) ≤ t −
9 6
5 2 7 3 17 4 31 5
t + t −
t +
t −
t .
2
3
12
30
20
Proof. We shall show that
‹

1
1
1
1
(1 − t)2 ln(1 + t) ≤ (1 − t)2 t − t 2 + t 3 − t 4 + t 5
2
3
4
5
5 2 7 3 17 4 31 5
9 6
≤t− t + t −
t +
t −
t .
2
3
12
30
20
The left inequality is equivalent to f (t) ≥ 0, where
f (t) = t −
1 2 1 3 1 4 1 5
t + t − t + t − ln(1 + t).
2
3
4
5
Since
t5
,
1+ t
f (t) is decreasing on (−1, 0] and increasing on [0, 1); therefore, f (t) ≥ f (0) = 0.
The right inequality is equivalent to t 6 (t − 1) ≤ 0, which is clearly true.
f 0 (t) =
P 3.30. If a, b are positive real numbers such that a + b = 2, then
a2
p
b
+ b2
p
a
≤ 2.
(Vasile Cîrtoaje, 2010)
Solution. Assume that a ≥ b. For a = 2 and b = 0, the inequality is obvious. Otherwise,
using the substitution a = 1 + x and b = 1 − x, 0 ≤ x < 1, we can write the desired
inequality as f (x) ≤ 2, where
p
1−x ln(1+x)
f (x) = e2
p
+ e2
1+x ln(1−x)
.
462
Vasile Cîrtoaje
There are two cases to consider.
13
Case 1:
≤ x < 1. If f 0 (x) ≤ 0, then f (x) is decreasing, and hence
20
f (x) ≤ f

13
33
=
20
20
‹

q
‹
7
5
q
+

7
20
‹
33
5
<
 ‹5/4  ‹2
5
1
+
< 2.
3
4
Since
p
2 1 − x ln(1 + x) 2p1−x ln(1+x)
− p
e
f (x) =
1+ x
1− x
p
2 1 + x ln(1 − x) 2p1+x ln(1−x)
e
− p
−
1− x
1+ x
p
2 1 − x ln(1 + x) 2p1−x ln(1+x)
<
− p
e
,
1+ x
1− x
0
it is enough to show that g(x) ≤ 0,where
g(x) =
2(1 − x)
− ln(1 + x).
1+ x
Clearly, g is decreasing, and hence
‹
13
14
33
g(x) ≤ g
=
− ln
< 0.
20
33
20

Case 2: 0 ≤ x ≤
where
13
. According to Lemma below, it suffices to show that f (x) ≤ 2,
20
f (x) = e2x−2x
2
11 3 1 4
+ 12
x −2 x
+ e−(2x+2x
2
3 1 4
+ 11
12 x + 2 x )
.
If f 0 (x) ≤ 0, then f (x) is decreasing, hence f (x) ≤ f (0) = 2. Since
f 0 (x) = (2 − 4x +
2 11 3 1 4
11 2
x − 2x 3 )e2x−2x + 12 x − 2 x
4
2 11 3 1 4
11 2
x + 2x 3 )e−(2x+2x + 12 x + 2 x ) ,
4
0
the inequality f (x) ≤ 0 is equivalent to
−(2 + 4x +
11
e−4x− 6
x3
≥
8 − 16x + 11x 2 − 8x 3
.
8 + 16x + 11x 2 + 8x 3
For the non-trivial case 8 − 16x + 11x 2 − 8x 3 > 0, rewrite this inequality as g(x) ≥ 0,
where
g(x) = −4x −
11 3
x − ln(8 − 16x + 11x 2 − 8x 3 ) + ln(8 + 16x + 11x 2 + 8x 3 ).
6
Symmetric Power-Exponential Inequalities
463
If g 0 (x) ≥ 0, then g(x) is increasing, hence g(x) ≥ g(0) = 0. From
g 0 (x) = −4 −
11 2
(16 + 24x 2 ) − 22x
(16 + 24x 2 ) + 22x
x +
+
,
2
8 + 11x 2 − (16x + 8x 3 ) 8 + 11x 2 + (16x + 8x 3 )
it follows that g 0 (x) ≥ 0 is equivalent to
(16 + 24x 2 )(8 + 11x 2 ) − 22x(16x + 8x 3 ) ≥
1
(8 + 11x 2 )[(8 + 11x 2 )2 − (16x + 8x 3 )2 ].
4
Since
(8 + 11x 2 )2 − (16x + 8x 3 )2 ≤ (8 + 11x 2 )2 − 256x 2 − 256x 4 ≤ 16(4 − 5x 2 ),
it suffices to show that
(4 + 6x 2 )(8 + 11x 2 ) − 11x(8x + 4x 3 ) ≥ (8 + 11x 2 )(4 − 5x 2 ).
This reduces to 77x 4 ≥ 0. The proof is completed. The equality holds for a = b = 1.
Lemma. If −1 < t ≤
13
, then
20
p
11 3 1 4
1 − t ln(1 + t) ≤ t − t 2 +
t − t .
24
4
Proof. We consider two cases.
Case 1: 0 ≤ t ≤
inequalities
13
. We can prove the desired inequality by multiplying the following
20
p
1
1 3
1
t ,
1 − t ≤ 1 − t − t2 −
2
8
16
1
1
1
1
ln(1 + t) ≤ t − t 2 + t 3 − t 4 + t 5 ,
2
3
4
5

‹
‹
1
1
1 3
1
1
1
1
11 3 1 4
1 − t − t2 −
t
t − t2 + t3 − t4 + t5 ≤ t − t2 +
t − t .
2
8
16
2
3
4
5
24
4
The first inequality is equivalent to f (t) ≥ 0, where

‹
1
1 2
1 3
1
t − ln(1 − t).
f (t) = ln 1 − t − t −
2
8
16
2
Since
f 0 (t) =
1
8 + 4t + 3t 2
5t 3
−
=
≥ 0,
2(1 − t) 16 − 8t − 2t 2 − t 3
2(1 − t)(16 − 8t − 2t 2 − t 3 )
464
Vasile Cîrtoaje
f (t) is increasing, and hence f (t) ≥ f (0) = 0.
The second inequality is equivalent to f (t) ≥ 0, where
f (t) = t −
1 2 1 3 1 4 1 5
t + t − t + t − ln(1 + t).
2
3
4
5
Since
f 0 (t) = 1 − t + t 2 − t 3 + t 4 −
1
t5
=
≥ 0,
1+ t
1+ t
f (t) is increasing, and hence f (t) ≥ f (0) = 0.
The third inequality is equivalent to
t 4 (160 − 302t + 86t 2 + 9t 3 + 12t 4 ) ≥ 0.
This is true since
160 − 302t + 86t 2 + 9t 3 + 12t 4 ≥ 2(80 − 151t + 43t 2 ) > 0.
Case 2: −1 < t ≤ 0. Write the desired inequality as
p
11 3 1 4
− 1 − t ln(1 + t) ≥ −t + t 2 −
t + t .
24
4
This is true if
p
1
1
1 − t ≥ 1 − t − t 2,
2
8
1
1
− ln(1 + t) ≥ −t + t 2 − t 3 + t 4 ,
3
4

‹
‹
1
1
1
1
11 3 1 4
1 − t − t 2 −t + t 2 − t 3 + t 4 ≥ −t + t 2 −
t + t .
2
8
3
4
24
4
The first inequality is equivalent to f (t) ≥ 0, where

‹
1
1 2
1
f (t) = ln(1 − t) − ln 1 − t − t .
2
2
8
Since
f 0 (t) =
−1
2(2 + t)
−3t 2
+
=
≤ 0,
2(1 − t) 8 − 4t − t 2
2(1 − t)(8 − 4t − t 2 )
f (t) is decreasing, and hence f (t) ≥ f (0) = 0.
The second inequality is equivalent to f (t) ≥ 0, where
f (t) = t −
1 2 1 3 1 4
t + t − t − ln(1 + t).
2
3
4
Symmetric Power-Exponential Inequalities
Since
f 0 (t) = 1 − t + t 2 − t 3 −
465
−t 4
1
=
≤ 0,
1+ t
1+ t
f (t) is decreasing, and hence f (t) ≥ f (0) = 0.
The third inequality reduces to the obvious inequality
t 4 (10 − 8t − 3t 2 ) ≥ 0.
P 3.31. If a, b are nonnegative real numbers such that a + b = 2, then
a b(1 − a b)2
a b(1 − a b)2
≤ a b+1 + b a+1 − 2 ≤
.
2
3
(Vasile Cîrtoaje, 2010)
Solution. Assume that a ≥ b, which yields 1 ≤ a ≤ 2 and 0 ≤ b ≤ 1.
(a) To prove the left inequality we apply Lemma 1 below. For x = a and k = b, we
have
a b+1 ≥ 1 + (1 + b)(a − 1) +
a b+1 ≥ a − b + a b +
b(1 + b)(1 − b)
b(1 + b)
(a − 1)2 −
(a − 1)3 ,
2
6
b(1 + b)
b(1 + b)
(a − 1)2 −
(a − 1)4 .
2
6
(*)
Also, for x = b and k = a − 1, we have
b a ≥ 1 + a(b − 1) +
a(a − 1)
a(a − 1)(2 − a)
(b − 1)2 −
(b − 1)3 ,
2
6
a
ab
b a ≥ 1 − a + a b + (a − 1)3 +
(a − 1)4 ,
2
6
b a+1 ≥ b − a b + a b2 +
ab
a b2
(a − 1)3 +
(a − 1)4 .
2
6
Summing up (*) and (**) gives
a b+1 + b a+1 − 2 ≥ −b(a − 1)2 +
Since
−b(a − 1)2 +
b(3 − a b)
b(1 + b − a b)
(a − 1)2 −
(a − 1)4 .
2
6
b(3 − a b)
b
(a − 1)2 = (a − 1)4 ,
2
2
(**)
466
Vasile Cîrtoaje
we have
a b+1 + b a+1 − 2 ≥
=
b
b(1 + b − a b)
(a − 1)4 −
(a − 1)4
2
6
a b(1 + b)
ab
a b(1 − a b)2
(a − 1)4 ≥
(a − 1)4 =
.
6
6
6
The equality holds for a = b = 1, for a = 2 and b = 0, and for a = 0 and b = 2.
(b) To prove the right inequality we apply Lemma 2 below. For x = a and k = b,
we have
a b+1 ≤ 1 + (b + 1)(a − 1) +
+
a b+1 ≤ 1 + (b + 1)(a − 1) +
(b + 1)b
(b + 1)b(b − 1)
(a − 1)2 +
(a − 1)3
2
6
(b + 1)b(b − 1)(b − 2)
(a − 1)4 ,
24
b(b + 1)
b(b + 1)
a b(b + 1)
(a − 1)2 −
(a − 1)4 +
(a − 1)5 .
2
6
24
Also, for x = b and k = a, we have
b a+1 ≤ 1 + (a + 1)(b − 1) +
a(a + 1)
a b(a + 1)
a(a + 1)
(b − 1)2 −
(b − 1)4 +
(b − 1)5 .
2
6
24
Summing these inequalities gives
a b+1 + b a+1 − 2 ≤ −2(a − 1)2 +
≤
a2 + b2 + 2
a2 + b2 + 2
ab
(a − 1)2 −
(a − 1)4 −
(a − 1)6
2
6
12
a2 + b2 + 2
a2 + b2 + 2
a2 + b2 − 2
(a − 1)2 −
(a − 1)4 = (a − 1)4 −
(a − 1)4
2
6
6
=
ab
a b(1 − a b)2
(a − 1)4 =
.
3
3
The equality holds for a = b = 1, for a = 2 and b = 0, and for a = 0 and b = 2.
Lemma 1. If x ≥ 0 and 0 ≤ k ≤ 1, then
x k+1 ≥ 1 + (1 + k)(x − 1) +
k(1 + k)(1 − k)
k(1 + k)
(x − 1)2 −
(x − 1)3 ,
2
6
with equality for x = 1, for k = 0, for k = 1.
Proof. For k = 0 and k = 1, the inequality is an identity. For fixed k, 0 < k < 1, let us
define
f (x) = x k+1 − 1 − (1 + k)(x − 1) −
k(1 + k)
k(1 + k)(1 − k)
(x − 1)2 +
(x − 1)3 .
2
6
Symmetric Power-Exponential Inequalities
467
We need to show that f (x) ≥ 0. We have
1
k(1 − k)
f 0 (x) = x k − 1 − k(x − 1) +
(x − 1)2 ,
1+k
2
1
f 00 (x) = x k−1 − 1 + (1 − k)(x − 1),
k(1 + k)
1
f 000 (x) = −x k−2 + 1.
k(1 + k)(1 − k)
Case 1: 0 ≤ x ≤ 1. Since f 000 ≤ 0, f 00 is decreasing, f 00 (x) ≥ f 00 (1) = 0, f 0 is increasing,
f 0 (x) ≤ f 0 (1) = 0, f is decreasing, f (x) ≥ f (1) = 0.
Case 2: x ≥ 1. Since f 000 ≥ 0, f 00 is increasing, f 00 (x) ≥ f 00 (1) = 0, f 0 is increasing,
f 0 (x) ≥ f 0 (1) = 0, f is increasing, f (x) ≥ f (1) = 0.
Lemma 2. If either x ≥ 1 and 0 ≤ k ≤ 1, or 0 ≤ x ≤ 1 and 1 ≤ k ≤ 2, then
x k+1 ≤ 1 + (k + 1)(x − 1) +
(k + 1)k
(k + 1)k(k − 1)
(x − 1)2 +
(x − 1)3
2
6
(k + 1)k(k − 1)(k − 2)
(x − 1)4 ,
24
with equality for x = 1, for k = 0, for k = 1, for k = 2.
+
Proof. For k = 0, k = 1 and k = 2, the inequality is an identity. For fixed k, k ∈
(0, 1) ∪ (1, 2), let us define
f (x) = x k+1 − 1 − (k + 1)(x − 1) −
−
(k + 1)k
(k + 1)k(k − 1)
(x − 1)2 −
(x − 1)3
2
6
(k + 1)k(k − 1)(k − 2)
(x − 1)4 .
24
We need to show that f (x) ≤ 0. We have
1
k(k − 1)
k(k − 1)(k − 2)
f 0 (x) = x k − 1 − k(x − 1) −
(x − 1)2 −
(x − 1)3 ,
k+1
2
6
1
(k − 1)(k − 2)
f 00 (x) = x k−1 − 1 − (k − 1)(x − 1) −
(x − 1)2 ,
k(k + 1)
2
1
f 000 (x) = x k−2 − 1 − (k − 2)(x − 1),
k(k + 1)(k − 1)
1
f (4) (x) = x k−3 − 1.
k(k + 1)(k − 1)(k − 2)
468
Vasile Cîrtoaje
Case 1: x ≥ 1, 0 < k < 1. Since f (4) (x) ≤ 0, f 000 (x) is decreasing, f 000 (x) ≤ f 000 (1) =
0, f 00 is decreasing, f 00 (x) ≤ f 00 (1) = 0, f 0 is decreasing, f 0 (x) ≤ f 0 (1) = 0, f is
decreasing, f (x) ≤ f (1) = 0.
Case 2: 0 ≤ x ≤ 1, 1 < k < 2. Since f (4) ≤ 0, f 000 is decreasing, f 000 (x) ≥ f 000 (1) = 0, f 00
is increasing, f 00 (x) ≤ f 00 (1) = 0, f 0 is decreasing, f 0 (x) ≥ f 0 (1) = 0, f is increasing,
f (x) ≤ f (1) = 0.
P 3.32. If a, b are nonnegative real numbers such that a + b = 1, then
a2b + b2a ≤ 1.
(Vasile Cîrtoaje, 2007)
Solution. Without loss of generality, assume that
0≤ b≤
1
≤ a ≤ 1.
2
Applying Lemma 1 below for c = 2b, 0 ≤ c ≤ 1, we get
a2b ≤ (1 − 2b)2 + 4a b(1 − b) − 2a b(1 − 2b) ln a,
which is equivalent to
a2b ≤ 1 − 4a b2 − 2a b(a − b) ln a.
Similarly, applying Lemma 2 below for d = 2a − 1, d ≥ 0, we get
b2a−1 ≤ 4a(1 − a) + 2a(2a − 1) ln(2a + b − 1),
which is equivalent to
b2a ≤ 4a b2 + 2a b(a − b) ln a.
Adding up these inequalities, the desired inequality follows. The equality holds for
a = b = 1/2, for a = 0 and b = 1, and for a = 1 and b = 0.
Lemma 1. If 0 < a ≤ 1 and c ≥ 0, then
a c ≤ (1 − c)2 + ac(2 − c) − ac(1 − c) ln a,
with equality for a = 1, for c = 0, and for c = 1.
Proof. Making the substitution a = e−x , x ≥ 0, we need to prove that f (x) ≥ 0, where
f (x) = (1 − c)2 e x + c(2 − c) + c(1 − c)x − e(1−c)x ,
Symmetric Power-Exponential Inequalities
469
f 0 (x) = (1 − c)[(1 − c)e x + c − e(1−c)x ].
If f 0 ≥ 0 on [0, ∞), then f is increasing, and hence f (x) ≥ f (0) = 0. In order to prove
that f 0 ≥ 0, we consider two cases.
Case 1: 0 ≤ c ≤ 1. By the weighted AM-GM inequality, we have
(1 − c)e x + c ≥ e(1−c)x ,
and hence f 0 (x) ≥ 0.
Case 2: c ≥ 1. By the weighted AM-GM inequality, we have
(c − 1)e x + e(1−c)x ≥ c,
which yields
f 0 (x) = (c − 1)[(c − 1)e x + e(1−c)x − c] ≥ 0.
Lemma 2. If 0 ≤ b ≤ 1 and d ≥ 0, then
b d ≤ 1 − d 2 + d(1 + d) ln(b + d),
with equality for d = 0, and for b = 0 and d = 1.
Proof. Write the inequality as
(1 + d)[1 − d + d ln(b + d)] ≥ b d .
Excepting the equality cases, since
1 − d + d ln(b + d) ≥ 1 − d + d ln d > 0,
we can rewrite the inequality in the form
ln(1 + d) + ln[1 − d + d ln(b + d)] ≥ d ln b.
Using the substitution b = e−x − d, where − ln(1 + d) ≤ x ≤ − ln d, we need to prove
that f (x) ≥ 0, where
f (x) = ln(1 + d) + ln(1 − d − d x) + d x − d ln(1 − d e x ).
Since
f 0 (x) =
d 2 (e x − 1 − x)
≥ 0,
(1 − d − d x)(1 − d e x )
f is increasing, and hence
f (x) ≥ f (− ln(1 + d)) = ln[1 − d 2 + d(1 + d) ln(1 + d)].
470
Vasile Cîrtoaje
To complete the proof, we only need to show that −d 2 + d(1 + d) ln(1 + d) ≥ 0; that is,
(1 + d) ln(1 + d) ≥ d.
This inequality follows from e x ≥ 1 + x, where x =
−d
.
1+d
Conjecture. If a, b are nonnegative real numbers such that 1 ≤ a + b ≤ 15, then
a2b + b2a ≤ a a+b + b a+b .
P 3.33. If a, b are positive real numbers such that a + b = 1, then
2a a b b ≥ a2b + b2a .
Solution. Taking into account the inequality a2b + b2a ≤ 1 in the preceding P 3.32, it
suffices to show that
2a a b b ≥ 1.
Write this inequality as
2a a b b ≥ a a+b + b a+b ,
a b  b ‹a
2≥
+
.
b
a
Since a < 1 and b < 1, we apply Bernoulli’s inequality as follows
a b
b
 ‹a

‹
a
b
b
+
−1 +1+a
− 1 = 2.
≤1+ b
a
b
a
Thus, the proof is completed. The equality holds for a = b = 1/2.
P 3.34. If a, b are positive real numbers such that a + b = 1, then
a−2a + b−2b ≤ 4.
Symmetric Power-Exponential Inequalities
471
Solution. Applying Lemma below, we have
a−2a ≤ 4 − 2 ln 2 − 4(1 − ln 2)a,
b−2b ≤ 4 − 2 ln 2 − 4(1 − ln 2)b.
Adding these inequalities, the desired inequality follows. The equality holds for a = b =
1/2.
Lemma. If x ∈ (0, 1], then
x −2x ≤ 4 − 2 ln 2 − 4(1 − ln 2)x,
with equality for x = 1/2.
Proof. Write the inequality as
1 −2x
x
≤ 1 − c − (1 − 2c)x,
4
c=
1
ln 2 ≈ 0.346.
2
This is true if f (x) ≤ 0, where
f (x) = −2 ln 2 − 2x ln x − ln[1 − c − (1 − 2c)x].
We have
f 0 (x) = −2 − 2 ln x +
f 00 (x) = −
1 − 2c
,
1 − c − (1 − 2c)x
g(x)
2
(1 − 2c)2
+
=
,
x [1 − c − (1 − 2c)x]2
x[1 − c − (1 − 2c)x]2
where
g(x) = 2(1 − 2c)2 x 2 − (1 − 2c)(5 − 6c)x + 2(1 − c)2 .
Since
g 0 (x) = (1 − 2c)[4(1 − 2c)x − 5 + 6c] ≤ (1 − 2c)[4(1 − 2c) − 5 + 6c]
= (1 − 2c)(−1 − 2c) < 0,
g is decreasing on (0, 1], hence g(x) ≥ g(1) = −2c 2 + 4c − 1 > 0, f 00 (x) > 0 for
x ∈ (0, 1]. Since f 0 is increasing and f 0 (1/2) = 0, we have f 0 (x) ≤ 0 for x ∈ (0, 1/2],
and f 0 (x) ≥ 0 for x ∈ [1/2, 1]. Therefore, f is decreasing on (0, 1/2] and increasing on
[1/2, 1], hence f (x) ≥ f (1/2) = 0.
Remark. According to the inequalities in P 3.32 and P 3.34, the following inequality
holds for all positive numbers a, b such that a + b = 1:
‹
 1
1
a2b + b2a
+
≤ 4.
a2a b2b
Actually, this inequality holds for all a, b ∈ (0, 1]. In this case, it is sharper than the
inequality in P 3.15.
472
Vasile Cîrtoaje
Chapter 4
Bibliography
473