INEQUALITIES
Ozgur Kircak
September 27, 2009
2
Contents
1 MEANS INEQUALITIES
1.1 EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
7
2 CAUCHY-SCHWARZ INEQUALITY
11
2.1 EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
3 REARRANGEMENT INEQUALITY
17
3.1 EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
4 CHEBYSHEV’S INEQUALITY
21
4.1 EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
5 MIXED PROBLEMS
25
6 PROBLEMS FROM OLYMPIADS
29
6.1 Years 1996 ∼ 2000 . . . . . . . . . . . . . . . . . . . . . . . . . . 36
6.2 Years 1990 ∼ 1995 . . . . . . . . . . . . . . . . . . . . . . . . . . 42
6.3 Supplementary Problems . . . . . . . . . . . . . . . . . . . . . . . 44
3
4
CONTENTS
Chapter 1
MEANS INEQUALITIES
Definition 1 Arithmetic mean of a1 , a2 , ..., an is AM =
Definition 2 Geometric mean of a1 , a2 , ..., an is GM =
Definition 3 Harmonic mean of a1 , a2 , ..., an is HM =
Definition 4 Quadratic mean of a1 , a2 , ..., an is QM =
a1 +a2 +...+an
n
√
n
1
a1
q
Definition 5 The rth power mean of a1 , a2 , ..., an is Pr =
a1 · a2 · ... · an
n
+ a1 +...+ a1n
2
a21 +a22 +...+a2n
n
q
r
ar1 +ar2 +...+arn
n
Theorem 1 Let ai ∈ R+
QM (a1 , a2 , ..., an ) ≥ AM (a1 , a2 , ..., an ) ≥ GM (a1 , a2 , ..., an ) ≥ HM (a1 , a2 , ..., an )
and equality holds if and only if a1 = a2 = ... = an .
Theorem 2 Let ai ∈ R+ then Pr1 ≥ Pr2 whenever r1 ≥ r2 .
Example 1 Let a, b, c > 0, prove that a2 + b2 + c2 ≥ ab + bc + ca.
Solution: By AM-GM inequality we have a2 + b2 ≥ 2ab. Similarly, b2 + c2 ≥ 2bc
and a2 + c2 ≥ 2ac. Adding these three inequalities we get
2(a2 + b2 + c2 ) ≥ 2(ab + bc + ca) =⇒ a2 + b2 + c2 ≥ ab + bc + ca.
Let’s link this result since we will use it in many problems.
a2 + b2 + c2 ≥ ab + bc + ca
5
(1.1)
6
CHAPTER 1. MEANS INEQUALITIES
a
b
Example 2 Prove that if a, b, c > 0, then
Solution: By AM-GM we have
a
b
c
b+c+a
3
≥
+
q
3
b
c
a·b·c
b·c·a
+
c
a
≥ 3.
= 1. So,
a
b
+
b
c
+
c
a
≥ 3.
Example 3 Prove that for any positive real numbers a, b, c we have
1
1
1
9
+
+
≥
.
a+b b+c c+a
2(a + b + c)
Solution: By AM-HM inequality, we have
1
a+b
+
1
b+c
+
1
c+a
3
3
≥
1
1
a+b
+
1
1
b+c
+
1
=
1
c+a
3
.
2(a + b + c)
Therefore,
1
1
9
1
+
+
≥
.
a+b b+c c+a
2(a + b + c)
Example 4 Prove that if a, b, c > 0 then
a
b
c
3
+
+
≥ .
b+c c+a a+b
2
Solution: By the previous example we have,
1
1
1
9
+
+
]≥
a+b b+c c+a
2
a+b+c a+b+c a+b+c
9
⇐⇒
+
+
≥
a+b
b+c
c+a
2
a
a
9
a
+1+
+1+
+1≥
⇐⇒
b+c
b+c
b+c
2
a
a
3
a
⇐⇒
+
+
≥ .
b+c b+c b+c
2
(a + b + c)[
This inequality is called Nesbitt’s inequality.
Example 5 Prove that
p
3
3+
√
3
3+
p
3
3−
√
3
√
3 < 2 3 3.
Solution: By AM − P3 we have,
s p
p
p
p
√
√
√
√
3
3
3
3
√
3 (
3+ 33+ 3− 33
3 + 3 3)3 + ( 3 − 3 3)3
3
≤
= 3.
2
2
Since
the terms are
not equal we have strict inequality.
p
p
√
√
√
3
3
So, 3 + 3 3 + 3 − 3 3 < 2 3 3.
(1.2)
1.1. EXERCISES
1.1
7
EXERCISES
1. Prove that for any positive real numbers a, b, c we have
√
(a + 2)(b + 3)(c + 6) ≥ 48 abc.
2. Prove that if a, b, c > 0, then (a + b)(b + c)(c + a) ≥ 8abc.
3. Prove that if a, b, c > 0, then
ab
bc
ca
+ 2 + 2 ≥ 3.
c2
a
b
4. Prove that if a, b, c > 0, then
ab bc ca
+
+
≥ a + b + c.
c
a
b
5. Prove that if x, y > 0 then,
y
1
x
.
+ 2
≤
x4 + y 2
x + y4
xy
6. Prove that (a + b − c)(b + c − a)(c + a − b) ≤ abc if
(a) a, b, c are sides of a triangle
(b) a, b, c are positive real numbers.
7. Prove that if a, b, c > 0 and a2 + b2 + c2 = 3, then
1
1
1
3
+
+
≥ .
1 + ab 1 + bc 1 + ca
2
8. Prove that if x, y, z are real numbers with z > 0, then
x2 + y 2 + 12z 2 + 1
≥ x + y + 1.
4z
9. Prove that the inequality (3a + b + c)2 ≥ 12a(b + c) holds for any real
numbers a, b, c.
10. Prove that if x, y, z > 0, then
(a)
√1
xy
x2
y2
+
√1
yz
y2
z2
+
z2
x2
√1
zx
≤
1
x
+
1
y
+ z1 .
+ yz + xy .
q
p
p
2
2
2
(c) xy + yz + zx ≥ x yz + y xz + z xy .
√
√
√
(d) x4 + y 4 + z 4 ≥ xyz( xy + yz + zx).
(b)
+
+
≥
x
z
8
CHAPTER 1. MEANS INEQUALITIES
11. Prove that if x, y > 0, then
1
1
1
≤
+ .
x+y
4x 4y
12. Let a, b, c ≥ 0 and a + b + c ≤ 3. Prove that
b
c
3
1
1
1
a
+
+
≤ ≤
+
+
.
2
2
2
1+a
1+b
1+c
2
1+a 1+b 1+c
13. Prove the inequality x4 + y 4 + 8 ≥ 8xy for positive real numbers x, y.
14. Prove that if a and b are positive real numbers, then
a
b
(1 + )n + (1 + )n ≥ 2n+1 .
b
a
15. Prove that if p, q > 0 and p + q = 1, then
1
25
1
.
(p + )2 + (q + )2 ≥
p
q
2
16. Prove that if x + y + z = 1 then,
1
1
1
1
8( − xy − yz − zx)
+
+
≥ 9.
2
(x + y)2
(y + z)2
(z + x)2
17. Prove that if a1 , a2 , ..., an are distinct positive real numbers and
a1 + a2 + ... + an = S, then
S
S
S
n2
.
+
+ ... +
>
S − a1
S − a2
S − an
n−1
18. Find the minimum value of
a1
a2
a2009
+
+...+
1 + a2 + a3 + ... + a2009 1 + a1 + a3 ... + a2009
1 + a1 + a2 + ... + a2008
where a1 , a2 , ..., a2009 > 0 and a1 + a2 + ... + a2009 = 1.
2
x
1
19. Prove that for any x ∈ R we have 1+x
4 ≤ 2.
√
√
20. Let x, y ≥ 1. Prove that x y − 1 + y x − 1 ≤ xy.
21. Prove the inequality
2
√x +2
x2 +1
≥ 2 for any x ∈ R.
22. Let x > y > 0 and xy = 1. Prove that
23. Prove that if x > y ≥ 0, then x +
x2 +y 2
x−y
4
(x−y)(y+1)2
√
> 2 2.
≥ 3.
1.1. EXERCISES
9
24. Prove that if a, b, c > 0 then
a3
bc
+
b3
ca
+
c3
ab
≥ a + b + c.
25. Prove that if a + b + c = 1, then a2 + b2 + c2 ≥ 31 .
26. Prove that if a + b + c = 3, then a2 + b2 + c2 ≥ 3 ≥ ab + bc + ca.
27. Let a, b, c be positive real numbers such that a2 + b2 + c2 = 3. Prove that
1
1
1
+
+
≥ 1.
a+b+1 b+c+1 c+a+1
28. Prove that the inequality
a2 + b2 + c2 ≥
(a + b + c)2
≥ ab + bc + ca
3
holds for any real numbers a, b, c.
29. Prove that if x, y, z > 0, then
y2
z2
x y
z
x2
+
+
≥ + + .
2
2
2
y
z
x
y
z
x
30. Prove that if x3 + y 3 = 2, then x + y ≤ 2.
31. Let a, b, c be positive real numbers with a3 + b3 + c3 = 24. Prove that
a + b + c ≤ 6.
32. Prove that if a, b, c > 0, then
a
b+c+d
+
b
a+c+d
+
c
a+b+d
+
d
a+b+c
≥ 34 .
33. Prove that if a + b ≥ 1, then a4 + b4 ≥ 81 .
34. Let a, b, c > 0. Prove that
a3 − a + 2 b3 − b + 2 c3 − c + 2
+
+
≥ 3.
b+c
c+a
a+b
35. Prove that for positive real numbers x, y, z we have
x
y
z
3
+
+
≥ .
x + 2y + 2z
y + 2z + 2x z + 2x + 2y
5
36. Prove that if a, b, c > 0, then
a4
b4
c4
a2
b2
c2
+
+
≥
+
+
.
b2 c 2
c2 a2
a2 b2
bc
ca ab
√
√
√
√
37. Prove the inequality 2 x + 1 − 2 x < √1x < 2 x − 2 x − 1 for x ≥ 1.
38. Prove that if a, b, c > 0 then
1
1
27
1
+
+
≥
.
b(a + b) c(b + c) a(c + a)
2(a + b + c)2
10
CHAPTER 1. MEANS INEQUALITIES
39. Prove that for x, y > 0,
(1 +
1
√
x)2
+
(1 +
1
√
y)2
≥
2
.
x+y+2
40. Prove that if a, b, c > 0 then
a3 + 2
b3 + 2
c3 + 2
3
+
+
≥ .
3b + 3c 3c + 3a 3a + 3b
2
41. Prove that if a, b, c > 0, then
1+
6
3
≥
.
ab + bc + ca
a+b+c
42. Solve the system in R+
a + b + c + d = 12
abcd = 27 + ab + ac + ad + bc + bd + cd.
43. Solve the system in the set of real numbers
4x2
= y,
4x2 + 1
4y 2
= z,
4y 2 + 1
4z 2
= x.
4z 2 + 1
44. Solve the system where x, y, z ∈ R
2
= 2y,
x
2
y + = 2z,
y
2
z + = 2x.
z
x+
45. Find the positive real numbers x, y, z, t satisfying the system
16xyzt = (x2 + y 2 + z 2 + t2 )(xyz + xyt + xzt + yzt),
8 = 2xy + 2zt + xz + xt + yz + yt.
Chapter 2
CAUCHY-SCHWARZ
INEQUALITY
Theorem 3 (Cauchy-Schwarz)If xi , yi are real numbers, then
(x21 + x22 + ... + x2n )(y12 + y22 + ... + yn2 ) ≥ (x1 y1 + x2 y2 + ... + xn yn )2
and equality holds if and only if
x1
y1
=
Example 6 Prove that a2 +b2 +c2 ≥
x2
y2
= ... =
(a+b+c)2
3
xn
yn .
for any real numbers a, b, c.
Solution: By Cauchy-Schwarz we have that
(12 + 12 + 12 )(a2 + b2 + c2 ) ≥ (1 · a + 1 · b + 1 · c)2 = (a + b + c)2 .
Therefore, a2 + b2 + c2 ≥
(a+b+c)2
.
3
Example 7 Let x, y, z be positive real numbers. Prove that
x2
y2
z2
x+y+z
+
+
≥
.
x+y y+z
z+x
2
Solution: By Cauchy-Schwarz inequality we have
√
√
√
x
y
z
2
2
2
2
2
2
[( x + y) + ( y + z) + ( z + x) ] ( √
) + (√
) + (√
) ≥
x+y
y+z
z+x
2
√
√
√
x
y
z
= (x + y + z)2
x+y· √
+ y+z· √
+ z+x· √
x+y
y+z
z+x
2
y2
x
z2
So, what we got is (2x + 2y + 2z) x+y
+ y+z
+ z+x
≥ (x + y + z)2 .
Therefore,
x2
x+y
+
y2
y+z
+
z2
z+x
≥
x+y+z
.
2
11
12
CHAPTER 2. CAUCHY-SCHWARZ INEQUALITY
This is a very useful trick that can be applied to many problems. We can
generalize this result as:
x21
x2
x2
(x1 + x2 + ... + xn )2
+ 2 + ... + n ≥
y1
y2
yn
y1 + y2 + ... + yn
Example 8 Let a, b, c > 0. Prove that
Solution: We can write
c3
a+b
a3
b+c
=
a4
ab+ac
a3
b+c
+
b3
c+a
+
c3
a+b
and similarly
≥
b3
c+a
(2.1)
a2 +b2 +c2
.
2
=
b4
bc+ba
and
c4
ca+cb .
=
So by (2.1) we have
LHS =
a4
b4
c4
(a2 + b2 + c2 )2
a2 + b2 + c2
+
+
≥
≥
ab + ac bc + ba ca + cb
2(ab + bc + ca)
2
by (1.1) as desired.
2.1. EXERCISES
2.1
13
EXERCISES
46. Prove that the inequality x2 +y 2 +z 2 ≥
(x+2y+3z)2
14
holds for any x, y, z ∈ R.
47. Prove that if a, b, c > 0, then
1 1 1
(a + b + c)( + + ) ≥ 9.
a b
c
48. Prove that if x, y > 0 and x + 3y = 1 then x2 + y 2 ≥
1
10 .
49. a, b, c, x, y, z are real numbers and a2 + b2 + c2 = 16, x2 + y 2 + z 2 = 25
a+b+c
.
and ax + by + cz = 20. Compute x+y+z
q
q
√
√
2
2
50. Prove that if a, b > 0, then a + b ≤ ab + ba .
51. Solve the system
x1 + x2 + ... + xk = 9
1
1
1
+
+ ... +
=1
x1
x2
xk
where xi ∈ R+ .
52. Prove that if a, b, c are positive real numbers with abc = 1, then
a2
b2
c2
3
+
+
≥ .
a+b b+c c+a
2
53. Prove that if x, y, z > 0 and x + y + z = 1, then
1
1
1
1
8( − xy − yz − zx)
+
+
≥ 9.
2
(x + y)2
(y + z)2
(z + x)2
54. Prove that if a, b, c > 0 then
a
b
c
9
+
+
≥
.
(b + c)2
(c + a)2
(a + b)2
4(a + b + c)
55. Prove that if a, b, c, d are positive real numbers, then
1 1 4 16
64
+ + +
≥
.
a b
c
d
a+b+c+d
56. Let a, b, c, d be positive real numbers. Prove that
p
√
√
(a + c)(b + d) ≥ ab + cd.
57. Let a > c > 0 and b > c > 0. Prove that
p
c(a − c) +
p
c(b − c) ≤
√
ab.
14
CHAPTER 2. CAUCHY-SCHWARZ INEQUALITY
58. Let a, b, c > 0 with a + b + c = 3. Prove that
a2
b2
c2
+
+
≥ 1.
a+b+1 b+c+1 c+a+1
59. Let a, b, c > 0 with abc = 1. Prove that
a2
b2
c2
+
+
≥ 1.
a+b+1 b+c+1 c+a+1
60. Let a, b, c be positive real numbers such that a2 + b2 + c2 = 3. Prove that
a3
b3
c3
+
+
≥ 1.
a + 2b b + 2c c + 2a
61. Let a, b, c, x, y, z be positive real numbers such that x + y + z = 1. Prove
that
p
ax + by + cz + 2 (xy + yz + zx)(ab + bc + ca) ≤ a + b + c.
62. Let a, b, c be positive real numbers. Prove that
a3
b2
c2
b3
c3
a2
+
+ .
+ 2+ 2 ≥
2
b
c
a
b
c
a
63. Prove that if a, b, c, x, y, z > 0, then
x4
y4
z4
(x + y + z)4
+
+
≥
.
a3
b3
c3
(a + b + c)3
64. Prove that if a, b, c are positive real numbers, then
a
b
c
+
+
≥ 1.
a + 2b b + 2c c + 2a
65. Prove that if a, b, c are positive real numbers, then
a
b
c
+
+
≥ 1.
b + 2c c + 2a a + 2b
66. Prove that if a, b, c > 0, then
b
c
a
+
+
≤ 1.
2a + b 2b + c 2c + a
67. Let a, b, c, d be positive real numbers such that (a2 + b2 )3 = c2 + d2 . Prove
that
a3
b3
+
≥ 1.
c
d
2.1. EXERCISES
15
68. Let a, b be positive real numbers. Prove that
a2 + b2
a4 + b4
≥
.
a3 + b3
a+b
69. Let a, b, c > 0 with abc = 1. Prove that
a2 + b2 + 1 b2 + c2 + 1 c2 + a2 + 1
+
+
≥ 3.
a+b+1
b+c+1
c+a+1
70. Prove that if a, b, c, x, y, z > 0 and (a2 + b2 + c2 )3 = x2 + y 2 + z 2 , then
a3
b3
c3
+
+
≥ 1.
x
y
z
71. Prove that if a, b, c are positive real numbers, then
27a2
(b + c)2
+
≥ 12b.
c
a
72. Let a, b, c, d ≥ 0 with ab + bc + cd + da = 1. Prove that
b3
c3
d3
1
a3
+
+
+
≥ .
b+c+d a+c+d a+b+d a+b+c
3
73. Let a1 , a2 , ..., an ; b1 , b2 , ..., bn be positive real numbers such that
a1 + a2 + ... + an = b1 + b2 + ... + bn . Show that
a21
a22
a2n
a1 + a2 + ... + an
+
+ ... +
≥
.
a1 + b1
a2 + b2
an + bn
2
74. Let a, b, c > 0. Prove that
a
b
c
9
+
+
≥
.
2
2
2
2
b(b + c)
c(c + a)
a(a + b)
4(a + b2 + c2 )
75. Let a, b, c > 0 with a + b + c = 3. Prove that
a2 (b + 1)
b2 (c + 1)
c2 (a + 1)
+
+
≥ 2.
a + b + ab b + c + bc c + a + ca
16
CHAPTER 2. CAUCHY-SCHWARZ INEQUALITY
Chapter 3
REARRANGEMENT
INEQUALITY
Definition 6 Let a1 ≤ a2 ≤ ... ≤ an and b1 ≤ b2 ≤ ... ≤ bn . The
number A = a1 b1 + a2 b2 + ... + an bn is called ordered sum and the number
B = a1 bn + a2 bn−1 + ... + an b1 is called reversed sum. And if x1 , x2 , ..., xn
is a permutation of b1 , b2 , ..., bn then X = a1 x1 + a2 x2 + ... + an xn is called
a mixed sum.
Theorem 4 Let a1 ≤ a2 ≤ ... ≤ an and b1 ≤ b2 ≤ ... ≤ bn be given. For
any mixed sum X we have B ≤ X ≤ A.
Example 9 Prove that for positive real numbers a, b, c the inequality
a2 + b2 + c2 ≥ ab + bc + ca holds.
Solution: Since the inequality is symmetric, WLOG we can assume that
a ≥ b ≥ c. So by Rearrangement inequality, being the ordered sum
a2 + b2 + c2 = a · a + b · b + c · c ≥ a · b + b · c + c · a.
Example 10 Prove that if a, b, c are positive real numbers, then
ab bc ca
+
+
≥ a + b + c.
c
a
b
Solution: WLOG, we can assume that a ≥ b ≥ c. Then ab ≥ ac ≥ bc and
1
1
1
c ≥ b ≥ a . By Rearrangement inequality we have,
LHS = ab ·
1
1
1
1
1
1
+ ac · + bc · ≥ ab · + ac · + bc · = a + b + c.
c
b
a
a
c
b
17
18
CHAPTER 3. REARRANGEMENT INEQUALITY
Example 11 Prove that if a, b, c are positive real numbers, then
a3
b3
c3
+
+
≥ a + b + c.
bc
ca ab
Solution: By symmetry assume that a ≥ b ≥ c then a3 ≥ b3 ≥ c3 and
1
1
1
bc ≥ ca ≥ ab . By Rearrangement inequality we have that
LHS = a3 ·
1
1
1
1
1
1
a2 b2 c2
+ b3 · + c 3 ·
≥ a3 · + b3 · + c3 ·
=
+ +
bc
ca
ab
ca
ab
bc
c
a
b
And again by Rearrangement inequality, being the reversed sum
a + b + c = a2 ·
1
1
1
1
1
1
+ b2 · + c2 · ≤ a2 · + b2 · + c2 ·
a
b
c
c
a
b
Combining these two inequalities we get that
b3
c3
a3
+
+
≥ a + b + c.
bc
ca ab
3.1
EXERCISES
76. Let a, b, c be positive real numbers. Prove that
(a) a3 + b3 ≥ ab(a + b)
(b) a5 + b5 ≥ ab(a3 + b3 )
(c) a3 + b3 + c3 ≥ a2 b + b2 c + c2 a
√
√
√
(d) ab + bc + ca ≥ a bc + b ca + c ab
(e)
(f)
a2
b
a2
b2
+
+
b2
c
b2
c2
+
+
c2
a
c2
a2
≥a+b+c
≥
b
a
+
c
a
b + c
3 2
(g) abc(ab + bc + ca) ≤ a b + b3 c2 + c3 a2
(h)
a
b+c
+
b
c+a
+
c
a+b
≥
a
a+b
+
b
b+c
+
c
c+a .
77. Prove that if a, b, c are positive real numbers, then
a
b
c
3
+
+
≥ .
b+c c+a a+b
2
78. Let xi ∈ R+ and y1 , y2 , ..., yn be a permutation of x1 , x2 , ..., xn . Prove that
x21
x2
x2
+ 2 + ... + n ≥ x1 + x2 + ... + xn .
y1
y2
yn
79. Let x1 , x2 , ..., xn be positive real numbers. Prove that
x21
x2
x2
+ 2 + ... + n ≥ x1 + x2 + ... + xn .
x2
x3
x1
3.1. EXERCISES
19
80. Let a1 , a2 , ..., an be distinct positive integers. Prove that
a2
an
1 1
a1
1
+ 2 + ... + 2 ≥ + + ... + .
12
2
n
1 2
n
81. Prove that if for any a, b, c we have
a2
b2
c2
a2
b2
c2
a2
b2
c2
+
+
≥
+
+
≥
+
+
a+b b+c c+a
b+c c+a a+b
c+a a+b b+c
then a = b = c.
82. Let a1 ≥ a2 ≥ a3 and b1 ≥ b2 ≥ b3 . Prove that
a1 b1 + a2 b2 + a3 b3 ≥
1
(a1 + a2 + a3 )(b1 + b2 + b3 )
3
20
CHAPTER 3. REARRANGEMENT INEQUALITY
Chapter 4
CHEBYSHEV’S
INEQUALITY
Theorem 5 Let a1 ≥ a2 ≥ ... ≥ an and b1 ≥ b2 ≥ ... ≥ bn . Then
a1 b1 +a2 b2 +...+an bn ≥
1
(a1 +a2 +...+an )(b1 +b2 +...+bn ) ≥ a1 bn +a2 bn−1 +...+an b1 .
n
Example 12 Prove that a2 + b2 + c2 ≥ 31 (a + b + c)2 .
Solution: By Chebyshev’s inequality,
a2 + b2 + c2 = a · a + b · b + c · c ≥
1
1
(a + b + c)(a + b + c) = (a + b + c)2 .
3
3
Example 13 Let a, b, c > 0. Prove that
a
b
c
3
+
+
≥ .
b+c c+a a+b
2
Solution: Due to symmetry, we can assume that a ≥ b ≥ c then
1
1
1
b+c ≥ c+a ≥ a+b . So by Chebyshev’s inequality we have
a
b
c
1
1
1
1
+
+
≥ (a + b + c)(
+
+
)
b+c c+a a+b
3
b+c c+a a+b
(4.1)
And by AM-HM we have
1 1
1
1
3
3
(
+
+
)≥
=
(4.2)
3 b+c c+a a+b
b+c+c+a+a+b
2(a + b + c)
Combining (4.1) and (4.2) we get,
a
b
c
3
+
+
≥ .
b+c c+a a+b
2
21
22
CHAPTER 4. CHEBYSHEV’S INEQUALITY
Example 14 Prove that if a, b, c > 0 and abc = 1, then
a2
b2
c2
3
+
+
≥ .
b+c c+a a+b
2
Solution: WLOG, suppose that a ≥ b ≥ c. Then a2 ≥ b2 ≥ c2 and
1
1
1
b+c ≥ c+a ≥ a+b . So by Chebyshev’s inequality we have,
a2
b2
c2
1
1
1
1
+
+
≥ (a2 + b2 + c2 )(
+
+
)
b+c c+a a+b
3
b+c c+a a+b
By Example 11 and (4.2) we have that
1
1
1
(a + b + c)2
3
1 2
(a + b2 + c2 )(
+
+
)≥
·
=
3
b+c c+a a+b
3
2(a + b + c)
√
3
a+b+c
abc
=
≥
2
2
4.1
EXERCISES
83. Prove that if a, b, c > 0. then
(a) a3 + b3 ≥
(b)
(a+b)(a2 +b2 )
2
an+1 +bn+1
≥ a+b
an +bn
2
4
4
4
(c) a + b + c ≥ abc(a + b + c)
(d) a3 + b3 + c3 ≥
(a+b+c)3
9
84. Prove that if a, b, c > 0, then
b
c
3
a
+
+
≥ .
a + 2b + 2c b + 2c + 2a c + 2a + 2b
5
85. Prove that if a, b, c > 0, then
a
b
c
3
+
+
≥ .
a + 3b + 3c b + 3c + 3a c + 3a + 3b
7
86. Prove that if a, b, c > 0 and abc = 1, then
a2
b2
c2
3
+
+
≥ .
a + 2b + 2c b + 2c + 2a c + 2a + 2b
5
87. Prove that if a, b, c > 0, then
a3
b3
c3
ab + bc + ca
+
+
≥
.
b+c c+a a+b
2
4.1. EXERCISES
23
88. Prove that if a, b, c > 0 with abc = 1, then
a3
b3
c3
+
+
≥ 1.
b+c+1 c+a+1 a+b+1
89. Let a, b, c, d ≥ 0 with ab + bc + cd + da = 1. Prove that
b3
c3
d3
1
a3
+
+
+
≥ .
b+c+d a+c+d a+b+d a+b+c
3
90. Prove that if a, b, c > 0, then
b+c
c+a
a+b
6
+
+
≤ .
a + 3b + 3c b + 3c + 3a c + 3a + 3b
7
91. Prove that if a, b, c > 0, then
c+a
a+b
6
b+c
+
+
≤ .
a + 2b + 2c b + 2c + 2a c + 2a + 2b
5
92. Prove that if a, b, c > 0 and abc = 1 then
a
b
c
+
+
≥ 1.
b+c+1 c+a+1 a+b+1
93. Let x, y, z be positive real numbers with xyz = 1 and a ≥ 1. Prove that
ya
za
3
xa
+
+
≥ .
y+z
z+x x+y
2
94. Prove that if a, b, c > 0 with a + b + c = 1, then
a3
b3
c3
1
+
+
≥ .
2
2
2
2
2
2
b +c
c +a
a +b
2
95. Prove that if a, b, c > 0 then
b3
a5
b5
c5
(a + b + c)2
+ 3
+ 3
≥
.
3
3
3
+c
c +a
a +b
6
96. Let a, b, c > 0. Prove that
b3 + c3
c3 + a3
a3 + b3
+ 2
+ 2
≥ a + b + c.
2
2
2
a +b
b +c
c + a2
24
CHAPTER 4. CHEBYSHEV’S INEQUALITY
Chapter 5
MIXED PROBLEMS
97. Prove that if a, b, c > 0 and abc = 1, then
a3
1
1
1
+ 3
+ 3
≤ 1.
3
3
+ b + 1 b + c + 1 c + a3 + 1
98. Prove that if a, b, c > 0 and abc = 1, then
1
1
1
+
+
≤ 1.
a+b+1 b+c+1 c+a+1
99. Prove that if x, y, z > 0 then
xy
yz
zx
+ 2
+
≤ 1.
x2 + xy + yz
y + yz + zx z 2 + zx + xy
100. Prove that if x, y, z > 0 then
xy
yz
zx
1
+ 2
+ 2
≤ .
3x2 + 2y 2 + z 2
3y + 2z 2 + x2
3z + 2x2 + y 2
2
101. Prove that if a, b, c > 0 and abc = 1, then
a5
bc
ca
ab
+ 5
+ 5
≤ 1.
5
5
+ b + ab b + c + bc c + a5 + ca
102. Let a, b, c be real numbers such that a2 + b2 + c2 = 1. Prove that
a2
b2
c2
3
+
+
≥ .
1 + 2bc 1 + 2ca 1 + 2ab
5
103. (IMO95/2) Let a, b, and c be positive real numbers such that abc = 1.
Prove that
1
1
3
1
+ 3
+ 3
≥ .
3
a (b + c) b (c + a) c (a + b)
2
25
26
CHAPTER 5. MIXED PROBLEMS
104. Let a, b, c > 0 with a + b + c = 1. Prove that
b3
c3
1
a3
+
+
≥ .
2
2
2
2
2
2
a +b
b +c
c +a
2
105. (IMO2000/2) Let a, b, c be positive real numbers with product 1. Prove
that
1
1
1
(a − 1 + )(b − 1 + )(c − 1 + ) ≤ 1.
b
c
a
106. Prove that if a, b, c > 0 and abc = 1, then
1+
3
6
≥
.
a+b+c
ab + bc + ca
107. Let a, b, c be positive real numbers such that abc ≤ 1. Prove that
a b
c
+ + ≥ a + b + c.
b
c a
108. Let a, b, c be positive real numbers such that abc = 1. Prove that
√
√
b+c c+a a+b √
√ + √ + √ ≥ a + b + c + 3.
a
c
b
109. If a, b, c ∈ (0, 1), then prove that
p
√
abc + (1 − a)(1 − b)(1 − c) < 1.
p
p
p
110. Prove that a2 + (b − 1)2 + b2 + (c − 1)2 + c2 + (a − 1)2 ≥
arbitrary real numbers a, b, c.
√
3 2
2
for
111. Prove that 3(a2 − ab + b2 ) ≥ a2 + ab + b2 for any real numbers a and b.
112. (Russia2002) Let x, y, z be positive real numbers with sum 3. Prove that
√
√
√
x + y + z ≥ xy + yz + zx.
113. Let a, b, c be positive real numbers. Prove that
bc
ca
a+b+c
ab
+
+
≤
.
a + b + 2c b + c + 2a c + a + 2b
4
114. Prove that
a3
b3
c3
a+b+c
+
+
≥
2
2
2
2
2
2
a + ab + b
b + bc + c
c + ca + a
3
for positive real numbers a, b, c.
27
115. Prove that if a, b, c > 0, then
b3
c3
3(ab + bc + ca)
a3
+
+
≥
.
2
2
2
2
2
2
b − bc + c
c − ca + a
a − ab + b
a+b+c
116. (IMO2001/2)Let a, b, c be positive real numbers. Prove that
√
a2
b
c
a
+√
+√
≥ 1.
2
2
+ 8bc
b + 8ca
c + 8ab
117. Prove that if a, b, c are positive, then
b3 + c3
c3 + a3
a3 + b3
+
+
≥ a + b + c.
a2 + b2
b2 + c2
c2 + a2
118. Prove that for positive real numbers a, b, c, d the inequality holds
p
√
√
3
3
ab + cd ≤ 3 (a + c + b)(a + c + d).
p
119. Let x,
with x + y + z = 1 and let a = x2 + xy + y 2 ,
py, z be positive reals
√
b = y 2 + yz + z 2 , c = z 2 + zx + x2 . Prove that ab + bc + ca ≥ 1.
120. Let a, b, c > 0. Prove that
1
1
1
9
p
+
+
≥
.
a+b b+c c+a
a + b + c + 3(ab + bc + ca)
121. Let a, b, c > 0 with abc = 1. Prove that
b+c+1
c+a+1
(a + 1)(b + 1)(c + 1) + 1
a+b+1
+
+
≤
.
2
3
2
3
2
3
a+b +c
b+c +a
c+a +b
a+b+c
122. If x, y > 0 and x2 + y 3 ≥ x3 + y 4 . Prove that x3 + y 3 ≤ 2.
123. Let a, b, c be positive real numbers. Prove that
bc3
ca3
ab3
+
+
+ 3abc ≥ 2(a2 b + b2 c + c2 a).
c
a
b
124. Prove that if a, b, c are positive real numbers with a + b + c = 1, then
a5
b5
c5
1
+
+
≥ .
a4 + b4
b4 + c4
c4 + a 4
2
125. Let a, b, c > 0 with abc = 1. Prove that
a
b
c
+
+
≤ 1.
a2 + 2 b2 + 2 c2 + 2
28
CHAPTER 5. MIXED PROBLEMS
126. (Mathematical Excalibur) Let x, y, z > 1. Prove that
y4
z4
x4
+
+
≥ 48.
(y − 1)2
(z − 1)2
(x − 1)2
127. (MMO/2009) Let a, b, c > 0 such that ab + bc + ca = 31 . Prove that
a2
a
b
c
1
+ 2
+ 2
≥
.
− bc + 1 b − ca + 1 c − ab + 1
a+b+c
128. Prove that ∀x, y, z ∈ R+ we have:
X
3
x5
≥ .
x3 yz + yz 4
2
129. Let x, y, z > 0. Prove that
x3
y3
z3
+
+
≤ 1.
y(x2 + 2y 2 ) z(y 2 + 2z 2 ) x(z 2 + 2x2 )
130. Prove that if a, b, c > 0, then
b
c
a
+
+
≤ 1.
a2 + b + 1 b2 + c + 1 c2 + a + 1
131. Prove that if a, b, c > 0 with abc = 1, then
X a3 + b3 + 1
a2 + b2 + 1
≥ 3.
132. Prove that if x, y, z > 0, then
X x2 z 2 + y 4 + y 2 z 2
yz(xz + y 2 + yz)
≥ 3.
133. Let a, b, c > 0 with (a + b)(b + c)(c + a) = 8. Prove that
r
3
3
3
a+b+c
27 a + b + c
≥
.
3
3
134. Let x, y, z be positive reals. Prove that
yz
zx
xy
+ 2
+
≥ 1.
2x2 + yz
2y + zx 2z 2 + xy
135. Let x, y, z be positive reals. Prove that
x2
y2
z2
+
+
≤ 1.
2x2 + yz
2y 2 + zx 2z 2 + xy
136. Let a, b, x, y, z ∈ R+ . Prove that
x
y
z
3
+
+
≥
.
ay + bz
az + bx ax + by
a+b
Chapter 6
PROBLEMS FROM
OLYMPIADS
(From ’Inequalities Through Problems’-Hojoo Lee)
1 (BMO 2005, Proposed by Serbia and Montenegro) (a, b, c > 0)
b2
c2
4(a − b)2
a2
+
+
≥a+b+c+
b
c
a
a+b+c
2 (Romania 2005, Cezar Lupu) (a, b, c > 0)
b+c c+a a+b
1 1 1
+ 2 + 2 ≥ + +
a2
b
c
a b
c
3 (Romania 2005, Traian Tamaian) (a, b, c > 0)
a
b
c
d
+
+
+
≥1
b + 2c + d c + 2d + a d + 2a + b a + 2b + c
4 (Romania 2005, Cezar Lupu) a + b + c ≥
a+b+c≥
1
a
+
1
b
+ 1c , a, b, c > 0
3
abc
5 (Romania 2005, Cezar Lupu) (1 = (a + b)(b + c)(c + a), a, b, c > 0)
ab + bc + ca ≥
29
3
4
30
CHAPTER 6. PROBLEMS FROM OLYMPIADS
6 (Romania 2005, Robert Szasz) (a + b + c = 3, a, b, c > 0)
a2 b2 c2 ≥ (3 − 2a)(3 − 2b)(3 − 2c)
7 (Romania 2005) (abc ≥ 1, a, b, c > 0)
1
1
1
+
+
≤1
1+a+b 1+b+c 1+c+a
8 (Romania 2005, Unused) (abc = 1, a, b, c > 0)
a
b
c
3
+
+
≥
b2 (c + 1) c2 (a + 1) a2 (b + 1)
2
9 (Romania 2005, Unused) (a + b + c ≥
a
b
+
b
c
+ ac , a, b, c > 0)
b3 a
c3 b
3
a3 c
+
+
≥
b(c + a) c(a + b) a(b + c)
2
10 (Romania 2005, Unused) (a + b + c = 1, a, b, c > 0)
r
a
b
c
3
√
+√
+√
≥
2
c+a
b+c
a+b
11 (Romania 2005, Unused) (ab + bc + ca + 2abc = 1, a, b, c > 0)
√
√
ab +
bc +
√
ca ≥
3
2
12 (Chzech and Solvak 2005) (abc = 1, a, b, c > 0)
a
b
c
3
+
+
≥
(a + 1)(b + 1) (b + 1)(c + 1) (c + 1)(a + 1)
4
13 (Japan 2005) (a + b + c = 1, a, b, c > 0)
1
1
1
a (1 + b − c) 3 + b (1 + c − a) 3 + c (1 + a − b) 3 ≤ 1
31
14 (Germany 2005) (a + b + c = 1, a, b, c > 0)
b
c a
1+a 1+b 1+c
2
+ +
≥
+
+
a b
b
1−a 1−b 1−c
15 (Vietnam 2005) (a, b, c > 0)
a
a+b
3
+
b
b+c
3
+
c
c+a
3
≥
3
8
16 (China 2005) (a + b + c = 1, a, b, c > 0)
10(a3 + b3 + c3 ) − 9(a5 + b5 + c5 ) ≥ 1
17 (China 2005) (abcd = 1, a, b, c, d > 0)
1
1
1
1
+
+
+
≥1
2
2
2
(1 + a)
(1 + b)
(1 + c)
(1 + d)2
18 (China 2005) (ab + bc + ca = 13 , a, b, c ≥ 0)
a2
1
1
1
+ 2
+ 2
≤3
− bc + 1 b − ca + 1 c − ab + 1
19 (Poland 2005) (0 ≤ a, b, c ≤ 1)
a
b
c
+
+
≤2
bc + 1 ca + 1 ab + 1
20 (Poland 2005) (ab + bc + ca = 3, a, b, c > 0)
a3 + b3 + c3 + 6abc ≥ 9
21 (Baltic Way 2005) (abc = 1, a, b, c > 0)
b
c
a
+
+
≤1
a2 + 2 b2 + 2 c2 + 2
32
CHAPTER 6. PROBLEMS FROM OLYMPIADS
22 (Serbia and Montenegro 2005) (a, b, c > 0)
r
a
b
3
c
√
+√
≥
+√
(a + b + c)
2
c+a
b+c
a+b
23 (Serbia and Montenegro 2005) (a + b + c = 3, a, b, c > 0)
√
√
√
a + b + c ≥ ab + bc + ca
24 (Bosnia and Hercegovina 2005) (a + b + c = 1, a, b, c > 0)
√
√
√
1
a b+b c+c a≤ √
3
25 (Iran 2005) (a, b, c > 0)
2
a b
c
1 1 1
+ +
≥ (a + b + c)
+ +
b
c a
a b
c
26 (Austria 2005) (a, b, c, d > 0)
1
1
1
1
a+b+c+d
+ 3+ 3+ 3 ≥
3
a
b
c
d
abcd
27 (Moldova 2005) (a4 + b4 + c4 = 3, a, b, c > 0)
1
1
1
+
+
≤1
4 − ab 4 − bc 4 − ca
28 (APMO 2005) (abc = 8, a, b, c > 0)
a2
p
(1 + a3 )(1 + b3 )
+p
b2
(1 + b3 )(1 + c3 )
+p
c2
(1 + c3 )(1 + a3 )
29 (IMO 2005) (xyz ≥ 1, x, y, z > 0)
x5
y5 − y2
z5 − z2
x5 − x2
+ 5
+ 5
≥0
2
2
2
2
+y +z
y +z +x
z + x2 + y 2
≥
4
3
33
30 (Poland 2004) (a + b + c = 0, a, b, c ∈ R)
b2 c2 + c2 a2 + a2 b2 + 3 ≥ 6abc
31 (Baltic Way 2004) (abc = 1, a, b, c > 0, n ∈ N)
1
1
1
+
+
≤1
an + bn + 1 bn + cn + 1 cn + an + 1
32 (Junior Balkan 2004) ((x, y) ∈ R2 − {(0, 0)})
√
x+y
2 2
≥ 2
2
2
x +y
x − xy + y 2
33 (IMO Short List 2004) (ab + bc + ca = 1, a, b, c > 0)
r
r
r
1
3 1
3 1
3 1
+ 6b +
+ 6c +
+ 6a ≤
a
b
c
abc
34 (APMO 2004) (a, b, c > 0)
(a2 + 2)(b2 + 2)(c2 + 2) ≥ 9(ab + bc + ca)
35 (USA 2004) (a, b, c > 0)
(a5 − a2 + 3)(b5 − b2 + 3)(c5 − c2 + 3) ≥ (a + b + c)3
36 (Junior BMO 2003) (x, y, z > −1)
1 + x2
1 + y2
1 + z2
+
+
≥2
1 + y + z2
1 + z + x2
1 + x + y2
37 (USA 2003) (a, b, c > 0)
(2a + b + c)2
(2b + c + a)2
(2c + a + b)2
+ 2
+ 2
≤8
2
2
2
2a + (b + c)
2b + (c + a)
2c + (a + b)2
34
CHAPTER 6. PROBLEMS FROM OLYMPIADS
38 (Russia 2002) (x + y + z = 3, x, y, z > 0)
√
√
√
x + y + z ≥ xy + yz + zx
39 (Latvia 2002)
1
1+a4
+
1
1+b4
+
1
1+c4
+
1
1+d4
= 1, a, b, c, d > 0
abcd ≥ 3
40 (Albania 2002) (a, b, c > 0)
√
p
1+ 3 2
1 1 1
2
2
√ (a + b + c )
+ +
≥ a + b + c + a2 + b2 + c2
a b
c
3 3
41 (Belarus 2002) (a, b, c, d > 0)
p
p
p
p
2|ad − bc|
≥ a2 + b2 + c2 + d2 ≥ (a + c)2 + (b + d)2
(a + c)2 + (b + d)2 + p
(a + c)2 + (b + d)2
42 (Canada 2002) (a, b, c > 0)
b3
c3
a3
+
+
≥a+b+c
bc
ca ab
43 (Vietnam 2002, Dung Tran Nam) (a2 + b2 + c2 = 9, a, b, c ∈ R)
2(a + b + c) − abc ≤ 10
44 (Bosnia and Hercegovina 2002) (a2 + b2 + c2 = 1, a, b, c ∈ R)
a2
b2
c2
3
+
+
≥
1 + 2bc 1 + 2ca 1 + 2ab
5
45 (Junior BMO 2002) (a, b, c > 0)
1
1
1
27
+
+
≥
b(a + b) c(b + c) a(c + a)
2(a + b + c)2
35
46 (Greece 2002) (a2 + b2 + c2 = 1, a, b, c > 0)
√
√ 2
a
b
c
3 √
a
+
+
≥
a
+
b
b
+
c
c
b2 + 1 c2 + 1 a2 + 1
4
2
47 (Greece 2002) (bc 6= 0, 1−c
bc ≥ 0, a, b, c ∈ R)
10(a2 + b2 + c2 − bc3 ) ≥ 2ab + 5ac
48 (Taiwan 2002) a, b, c, d ∈ 0, 21
abcd
a4 + b4 + c4 + d4
≤
4
(1 − a)(1 − b)(1 − c)(1 − d)
(1 − a) + (1 − b)4 + (1 − c)4 + (1 − d)4
49 (APMO 2002) ( x1 +
√
x + yz +
√
1
y
+
1
z
y + zx +
= 1, x, y, z > 0)
√
z + xy ≥
√
xyz +
√
50 (Ireland 2001) (x + y = 2, x, y ≥ 0)
x2 y 2 (x2 + y 2 ) ≤ 2.
51 (BMO 2001) (a + b + c ≥ abc, a, b, c ≥ 0)
√
a2 + b2 + c2 ≥ 3abc
52 (USA 2001) (a2 + b2 + c2 + abc = 4, a, b, c ≥ 0)
0 ≤ ab + bc + ca − abc ≤ 2
53 (Columbia 2001) (x, y ∈ R)
3(x + y + 1)2 + 1 ≥ 3xy
x+
√
y+
√
z
36
CHAPTER 6. PROBLEMS FROM OLYMPIADS
54 (KMO Winter Program Test 2001) (a, b, c > 0)
p
p
(a2 b + b2 c + c2 a) (ab2 + bc2 + ca2 ) ≥ abc + 3 (a3 + abc) (b3 + abc) (c3 + abc)
55 (KMO Summer Program Test 2001) (a, b, c > 0)
p
p
p
p
a4 + b4 + c4 + a2 b2 + b2 c2 + c2 a2 ≥ a3 b + b3 c + c3 a + ab3 + bc3 + ca3
56 (IMO 2001) (a, b, c > 0)
√
6.1
a2
b
c
a
+√
+√
≥1
2
2
+ 8bc
b + 8ca
c + 8ab
Years 1996 ∼ 2000
57 (IMO 2000, Titu Andreescu) (abc = 1, a, b, c > 0)
1
1
1
a−1+
b−1+
c−1+
≤1
b
c
a
58 (Czech and Slovakia 2000) (a, b > 0)
s
r
r
1 1
a 3 b
3
3
2(a + b)
+
≥
+
a b
b
a
59 (Hong Kong 2000) (abc = 1, a, b, c > 0)
1 + ab2
1 + bc2
1 + ca2
18
+
+
≥ 3
3
3
3
c
a
b
a + b3 + c3
60 (Czech Republic 2000) (m, n ∈ N, x ∈ [0, 1])
(1 − xn )m + (1 − (1 − x)m )n ≥ 1
6.1. YEARS 1996 ∼ 2000
37
61 (Macedonia 2000) (x, y, z > 0)
x2 + y 2 + z 2 ≥
√
2 (xy + yz)
62 (Russia 1999) (a, b, c > 0)
a2 + 2bc b2 + 2ca c2 + 2ab
+ 2
+ 2
>3
b2 + c2
c + a2
a + b2
63 (Belarus 1999) (a2 + b2 + c2 = 3, a, b, c > 0)
1
1
1
3
+
+
≥
1 + ab 1 + bc 1 + ca
2
64 (Czech-Slovak Match 1999) (a, b, c > 0)
a
b
c
+
+
≥1
b + 2c c + 2a a + 2b
65 (Moldova 1999) (a, b, c > 0)
bc
ca
a
b
c
ab
+
+
≥
+
+
c(c + a) a(a + b) b(b + c)
c+a b+a c+b
66 (United Kingdom 1999) (p + q + r = 1, p, q, r > 0)
7(pq + qr + rp) ≤ 2 + 9pqr
67 (Canada 1999) (x + y + z = 1, x, y, z ≥ 0)
x2 y + y 2 z + z 2 x ≤
4
27
68 (Proposed for 1999 USAMO, [AB, pp.25]) (x, y, z > 1)
xx
2
+2yz y 2 +2zx z 2 +2xy
y
z
≥ (xyz)xy+yz+zx
38
CHAPTER 6. PROBLEMS FROM OLYMPIADS
69 (Turkey, 1999) (c ≥ b ≥ a ≥ 0)
(a + 3b)(b + 4c)(c + 2a) ≥ 60abc
70 (Macedonia 1999) (a2 + b2 + c2 = 1, a, b, c > 0)
a+b+c+
√
1
≥4 3
abc
71 (Poland 1999) (a + b + c = 1, a, b, c > 0)
√
a2 + b2 + c2 + 2 3abc ≤ 1
72 (Canda 1999) (x + y + z = 1, x, y, z ≥ 0)
x2 y + y 2 z + z 2 x ≤
73 (Iran 1998)
1
x
√
+
1
y
+
1
z
4
27
= 2, x, y, z > 1
x+y+z ≥
√
x−1+
p
√
y−1+ z−1
74 (Belarus 1998, I. Gorodnin) (a, b, c > 0)
c
a+b
b+c
a b
+ + ≥
+
+1
b
c a
b+c
a+b
75 (APMO 1998) (a, b, c > 0)
a+b+c
a
b c
1+
1+
1+
≥2 1+ √
3
b
c
a
abc
76 (Poland 1998) a + b + c + d + e + f = 1, ace + bdf ≥
abc + bcd + cde + def + ef a + f ab ≤
1
36
1
108
a, b, c, d, e, f > 0
6.1. YEARS 1996 ∼ 2000
39
77 (Korea 1998) (x + y + z = xyz, x, y, z > 0)
√
1
3
1
1
+p
≤
+√
2
2
2
1 + x2
1
+
z
1+y
78 (Hong Kong 1998) (a, b, c ≥ 1)
p
√
√
√
a − 1 + b − 1 + c − 1 ≤ c(ab + 1)
79 (IMO Short List 1998) (xyz = 1, x, y, z > 0)
x3
y3
z3
3
+
+
≥
(1 + y)(1 + z) (1 + z)(1 + x) (1 + x)(1 + y)
4
80 (Belarus 1997) (a, x, y, z > 0)
a+y
a+z
a+x
a+z
a+x
a+y
x+
y+
z ≥x+y+z ≥
x+
y+
z
a+x
a+x
a+y
a+z
a+y
a+z
81 (Ireland 1997) (a + b + c ≥ abc, a, b, c ≥ 0)
a2 + b2 + c2 ≥ abc
82 (Iran 1997) (x1 x2 x3 x4 = 1, x1 , x2 , x3 , x4 > 0)
1
1
1
1
3
3
3
3
x1 + x2 + x3 + x4 ≥ max x1 + x2 + x3 + x4 ,
+
+
+
x1
x2
x3
x4
83 (Hong Kong 1997) (x, y, z > 0)
p
√
xyz(x + y + z + x2 + y 2 + z 2 )
3+ 3
≥
9
(x2 + y 2 + z 2 )(xy + yz + zx)
84 (Belarus 1997) (a, b, c > 0)
a b
c
a+b
b+c
c+a
+ + ≥
+
+
b
c a
c+a a+b
b+c
40
CHAPTER 6. PROBLEMS FROM OLYMPIADS
85 (Bulgaria 1997) (abc = 1, a, b, c > 0)
1
1
1
1
1
1
+
+
≤
+
+
1+a+b 1+b+c 1+c+a
2+a 2+b 2+c
86 (Romania 1997) (xyz = 1, x, y, z > 0)
x9 + y 9
y9 + z9
z 9 + x9
+
+
≥2
x6 + x3 y 3 + y 6
y6 + y3 z3 + z6
z 6 + z 3 x3 + x6
87 (Romania 1997) (a, b, c > 0)
a2
a2
b2
c2
bc
ca
ab
+ 2
+ 2
≥1≥ 2
+
+
+ 2bc b + 2ca c + 2ab
a + 2bc b2 + 2ca c2 + 2ab
88 (USA 1997) (a, b, c > 0)
1
1
1
1
+
+
≤
.
a3 + b3 + abc b3 + c3 + abc c3 + a3 + abc
abc
89 (Japan 1997) (a, b, c > 0)
(b + c − a)2
(c + a − b)2
(a + b − c)2
3
+
+
≥
(b + c)2 + a2
(c + a)2 + b2
(a + b)2 + c2
5
90 (Estonia 1997) (x, y ∈ R)
p
p
x2 + y 2 + 1 > x y 2 + 1 + y x2 + 1
91 (APMC 1996) (x + y + z + t = 0, x2 + y 2 + z 2 + t2 = 1, x, y, z, t ∈ R)
−1 ≤ xy + yz + zt + tx ≤ 0
92 (Spain 1996) (a, b, c > 0)
a2 + b2 + c2 − ab − bc − ca ≥ 3(a − b)(b − c)
6.1. YEARS 1996 ∼ 2000
41
93 (IMO Short List 1996) (abc = 1, a, b, c > 0)
a5
ab
bc
ca
+ 5
+ 5
≤1
5
5
+ b + ab b + c + bc c + a5 + ca
94 (Poland 1996) a + b + c = 1, a, b, c ≥ − 34
a2
a
b
c
9
+ 2
+ 2
≤
+1 b +1 c +1
10
95 (Hungary 1996) (a + b = 1, a, b > 0)
b2
1
a2
+
≥
a+1 b+1
3
96 (Vietnam 1996) (a, b, c ∈ R)
(a + b)4 + (b + c)4 + (c + a)4 ≥
97 (Bearus 1996) (x + y + z =
√
4 4
a + b4 + c4
7
xyz, x, y, z > 0)
xy + yz + zx ≥ 9(x + y + z)
98 (Iran 1996) (a, b, c > 0)
1
1
1
9
(ab + bc + ca)
+
+
≥
2
2
2
(a + b)
(b + c)
(c + a)
4
99 (Vietnam 1996) (2(ab + ac + ad + bc + bd + cd) + abc + bcd + cda + dab =
16, a, b, c, d ≥ 0)
a+b+c+d≥
2
(ab + ac + ad + bc + bd + cd)
3
42
6.2
CHAPTER 6. PROBLEMS FROM OLYMPIADS
Years 1990 ∼ 1995
Any good idea can be stated in fifty words or less.
S. M. Ulam
100 (Baltic Way 1995) (a, b, c, d > 0)
a+c b+d c+a
d+b
+
+
+
≥4
a+b
b+c
c+d d+a
101 (Canda 1995) (a, b, c > 0)
aa bb cc ≥ abc
a+b+c
3
102 (IMO 1995, Nazar Agakhanov) (abc = 1, a, b, c > 0)
1
1
3
1
+
+
≥
a3 (b + c) b3 (c + a) c3 (a + b)
2
103 (Russia 1995) (x, y > 0)
1
x
y
≥ 4
+ 4
xy
x + y2
y + x2
104 (Macedonia 1995) (a, b, c > 0)
r
r
r
a
b
c
+
+
≥2
b+c
c+a
a+b
105 (APMC 1995) (m, n ∈ N, x, y > 0)
(n−1)(m−1)(xn+m +y n+m )+(n+m−1)(xn y m +xm y n ) ≥ nm(xn+m−1 y+xy n+m−1 )
106 (Hong Kong 1994) (xy + yz + zx = 1, x, y, z > 0)
√
4 3
x(1 − y )(1 − z ) + y(1 − z )(1 − x ) + z(1 − x )(1 − y ) ≤
9
2
2
2
2
2
2
6.2. YEARS 1990 ∼ 1995
43
107 (IMO Short List 1993) (a, b, c, d > 0)
a
b
c
d
2
+
+
+
≥
b + 2c + 3d c + 2d + 3a d + 2a + 3b a + 2b + 3c
3
108 (APMC 1993) (a, b ≥ 0)
v
!3
u √
√
√
√
√ !2
√
√
3
3
3
3
a + a2 b + ab2 + b
a + ab + b u
a+ b
a2 + b2
t
≤
≤
≤
2
4
3
2
109 (Poland 1993) (x, y, u, v > 0)
xy + xv + uy + uv
xy
uv
≥
+
x+y+u+v
x+y u+v
110 (IMO Short List 1993) (a + b + c + d = 1, a, b, c, d > 0)
abc + bcd + cda + dab ≤
1
176
+
abcd
27
27
111 (Italy 1993) (0 ≤ a, b, c ≤ 1)
a2 + b2 + c2 ≤ a2 b + b2 c + c2 a + 1
112 (Poland 1992) (a, b, c ∈ R)
(a + b − c)2 (b + c − a)2 (c + a − b)2 ≥ (a2 + b2 − c2 )(b2 + c2 − a2 )(c2 + a2 − b2 )
113 (Vietnam 1991) (x ≥ y ≥ z > 0)
z2x
x2 y y 2 z
+
+
≥ x2 + y 2 + z 2
z
x
y
114 (Poland 1991) (x2 + y 2 + z 2 = 2, x, y, z ∈ R)
x + y + z ≤ 2 + xyz
115 (Mongolia 1991) (a2 + b2 + c2 = 2, a, b, c ∈ R)
√
|a3 + b3 + c3 − abc| ≤ 2 2
116 (IMO Short List 1990) (ab + bc + cd + da = 1, a, b, c, d > 0)
b3
c3
d3
1
a3
+
+
+
≥
b+c+d c+d+a d+a+b a+b+c
3
44
6.3
CHAPTER 6. PROBLEMS FROM OLYMPIADS
Supplementary Problems
117 (Lithuania 1987) (x, y, z > 0)
x2
y3
z3
x+y+z
x3
+ 2
+ 2
≥
2
2
+ xy + y
y + yz + z
z + zx + x2
3
118 (Yugoslavia 1987) (a, b > 0)
√
√
1
1
(a + b)2 + (a + b) ≥ a b + b a
2
4
119 (Yugoslavia 1984) (a, b, c, d > 0)
a
b
c
d
+
+
+
≥2
b+c c+d d+a a+b
120 (IMO 1984) (x + y + z = 1, x, y, z ≥ 0)
0 ≤ xy + yz + zx − 2xyz ≤
7
27
121 (USA 1980) (a, b, c ∈ [0, 1])
a
b
c
+
+
+ (1 − a)(1 − b)(1 − c) ≤ 1.
b+c+1 c+a+1 a+b+1
122 (USA 1979) (x + y + z = 1, x, y, z > 0)
x3 + y 3 + z 3 + 6xyz ≥
1
.
4
123 (IMO 1974) (a, b, c, d > 0)
1<
a
b
c
d
+
+
+
<2
a+b+d b+c+a b+c+d a+c+d
6.3. SUPPLEMENTARY PROBLEMS
45
124 (IMO 1968) (x1 , x2 > 0, y1 , y2 , z1 , z2 ∈ R, x1 y1 > z1 2 , x2 y2 > z2 2 )
1
1
8
+
≥
x1 y1 − z1 2
x2 y2 − z2 2
(x1 + x2 )(y1 + y2 ) − (z1 + z2 )2
125 (Nesbitt’s inequality) (a, b, c > 0)
a
b
c
3
+
+
≥
b+c c+a a+b
2
126 (Polya’s inequality) (a 6= b, a, b > 0)
√
1
ln b − ln a
a+b
2 ab +
≥
3
2
b−a
127 (Klamkin’s inequality) (−1 < x, y, z < 1)
1
1
+
≥2
(1 − x)(1 − y)(1 − z) (1 + x)(1 + y)(1 + z)
128 (Carlson’s inequality) (a, b, c > 0)
r
r
ab + bc + ca
3 (a + b)(b + c)(c + a)
≥
8
3
129 ([ONI], Vasile Cirtoaje) (a, b, c > 0)
1
1
1
1
1
1
a+ −1
b+ −1 + b+ −1
c+ −1 + c+ −1
a+ −1 ≥3
b
c
c
a
a
b
130 ([ONI], Vasile Cirtoaje) (a, b, c, d > 0)
a−b
b−c
c−d
d−a
+
+
+
≥0
b+c
c+d d+a
a+b
46
CHAPTER 6. PROBLEMS FROM OLYMPIADS
131 (Elemente der Mathematik, Problem 1207, S̃efket Arslanagić)
(x, y, z > 0)
x y
z
x+y+z
+ + ≥ 3√
y
z
x
xyz
√
132 ( W U RZEL, Walther Janous) (x + y + z = 1, x, y, z > 0)
(1 + x)(1 + y)(1 + z) ≥ (1 − x2 )2 + (1 − y 2 )2 + (1 − z 2 )2
√
133 ( W U RZEL, Heinz-Jürgen Seiffert) (xy > 0, x, y ∈ R)
r
2xy
x2 + y 2
x+y
√
+
≥ xy +
x+y
2
2
√
134 ( W U RZEL, Šefket Arslanagić) (a, b, c > 0)
3
a3
b3
c3
(a + b + c)
+
+
≥
x
y
z
3 (x + y + z)
√
135 ( W U RZEL, Šefket Arslanagić) (abc = 1, a, b, c > 0)
a2
1
1
3
1
+ 2
+ 2
≥ .
(b + c) b (c + a) c (a + b)
2
√
136 ( W U RZEL, Peter Starek, Donauwörth) (abc = 1, a, b, c > 0)
1
1
1
1
+ 3 + 3 ≥ (a + b) (c + a) (b + c) − 1.
3
a
b
c
2
√
137 ( W U RZEL, Peter Starek, Donauwörth) (x+y+z = 3, x2 +y 2 +z 2 =
7, x, y, z > 0)
6
1 x y
z
1+
≥
+ +
xyz
3 z
x y
6.3. SUPPLEMENTARY PROBLEMS
47
√
138 ( W U RZEL, Šefket Arslanagić) (a, b, c > 0)
a
b
c
3 (a + b + c)
+
+
≥
.
b+1 c+1 a+1
a+b+c+3
139 (a, b ≥ 0)
a3 (b + 1) + b3 (a + 1) ≥ a2 (b + b2 ) + b2 (a + a2 )
140 (Latvia 1997) (n ∈ N, a, b, c > 0)
1
1
n
1
+
+ ··· +
<p
a + b a + 2b
a + nb
a(a + nb)
141 ([ONI], Gabriel Dospinescu, Mircea Lascu, Marian Tetiva) (a, b, c >
0)
a2 + b2 + c2 + 2abc + 3 ≥ (1 + a)(1 + b)(1 + c)
142 (Gazeta Matematicã) (a, b, c > 0)
p
p
p
p
p
p
a4 + a2 b2 + b4 + b4 + b2 c2 + c4 + c4 + c2 a2 + a4 ≥ a 2a2 + bc+b 2b2 + ca+c 2c2 + ab
143 (C1 2362, Mohammed Aassila) (a, b, c > 0)
b
c
3
a
+
+
≥
1+b 1+c 1+a
1 + abc
144 (C2580) (a, b, c > 0)
1 1 1
b+c
c+a
a+b
+ + ≥ 2
+
+
a b
c
a + bc b2 + ca c2 + ab
145 (C2581) (a, b, c > 0)
a2 + bc b2 + ca c2 + ab
+
+
≥a+b+c
b+c
c+a
a+b
1 CRUX
with MAYHEM
48
CHAPTER 6. PROBLEMS FROM OLYMPIADS
146 (C2532) (a2 + b2 + c2 = 1, a, b, c > 0)
1
1
2(a3 + b3 + c3 )
1
+
+
≥
3
+
a2
b2
c2
abc
147 (C3032, Vasile Cirtoaje) (a2 + b2 + c2 = 1, a, b, c > 0)
1
1
1
9
+
+
≤
1 − ab 1 − bc 1 − ca
2
148 (C2645) (a, b, c > 0)
2(a3 + b3 + c3 )
9(a + b + c)2
+ 2
≥ 33
abc
(a + b2 + c2 )
149 (x, y ∈ R)
−
1
(x + y)(1 − xy)
1
≤
≤
2
2
2
(1 + x )(1 + y )
2
150 (0 < x, y < 1)
xy + y x > 1
151 (x, y, z > 0)
√
3
xyz +
|x − y| + |y − z| + |z − x|
x+y+z
≥
3
3
152 (a, b, c, x, y, z > 0)
p
√
√
3
3
(a + x)(b + y)(c + z) ≥ abc + 3 xyz
153 (x, y, z > 0)
x
x+
p
(x + y)(x + z)
+
y
y+
p
(y + z)(y + x)
+
z+
z
p
≤1
(z + x)(z + y)
6.3. SUPPLEMENTARY PROBLEMS
49
154 (x + y + z = 1, x, y, z > 0)
z
x
y
√
+√
+√
≥
1−y
1−x
1−z
r
3
2
155 (a, b, c ∈ R)
√
p
p
p
3 2
2
2
2
2
2
2
a + (1 − b) + b + (1 − c) + c + (1 − a) ≥
2
156 (a, b, c > 0)
p
p
p
a2 − ab + b2 + b2 − bc + c2 ≥ a2 + ac + c2
157 (xy + yz + zx = 1, x, y, z > 0)
x
y
z
2x(1 − x2 ) 2y(1 − y 2 ) 2z(1 − z 2 )
+
+
≥
+
+
1 + x2
1 + y2
1 + z2
(1 + x2 )2
(1 + y 2 )2
(1 + z 2 )2
158 (x, y, z ≥ 0)
xyz ≥ (y + z − x)(z + x − y)(x + y − z)
159 (a, b, c > 0)
p
p
p
p
ab(a + b) + bc(b + c) + ca(c + a) ≥ 4abc + (a + b)(b + c)(c + a)
160 (Darij Grinberg) (x, y, z ≥ 0)
p
√
p
p
p
x (y + z) + y (z + x) + z (x + y) · x + y + z ≥ 2 (y + z) (z + x) (x + y).
161 (Darij Grinberg) (x, y, z > 0)
√
√
√
4 (x + y + z)
y+z
z+x
x+y
+
+
≥p
.
x
y
z
(y + z) (z + x) (x + y)
50
CHAPTER 6. PROBLEMS FROM OLYMPIADS
162 (Darij Grinberg) (a, b, c > 0)
a2 (b + c)
b2 (c + a)
c2 (a + b)
2
+
+
> .
2
2
2
2
2
2
(b + c ) (2a + b + c) (c + a ) (2b + c + a) (a + b ) (2c + a + b)
3
163 (Darij Grinberg) (a, b, c > 0)
a2
2a2
2
+ (b + c)
+
b2
2b2
+ (c + a)
2
+
c2
2c2
2
+ (a + b)
164 (Vasile Cirtoaje) (a, b, c ∈ R)
(a2 + b2 + c2 )2 ≥ 3(a3 b + b3 c + c3 a)
<
2
.
3