Inequalities of Olympiad
Caliber
José Luis Dı́az-Barrero
RSME Olympiad Committee
BARCELONA TECH
José Luis Dı́az-Barrero
RSME Olympic Committee
UPC–BARCELONA TECH
[email protected]
Basic facts to prove
inequalities
Hereafter, some useful facts for proving inequalities are presented:
1. If x ≥ y and y ≥ z then holds x ≥ z for any x, y, z ∈ R.
2. If x ≥ y and a ≥ b then x + a ≥ y + b for any x, y, a, b ∈ R.
3. If x ≥ y then x + z ≥ y + z for any x, y, z ∈ R.
4. If a ≥ b and c > 0 then ac ≥ bc for any a, b ∈ R.
5. If a ≥ b and c < 0 then bc ≥ ac for any a, b ∈ R.
6. If x ≥ y and a ≥ b then xa ≥ yb for any x, y ∈ R+ or a, b ∈ R+ .
7. Let a, b, c ∈ R such that a ≥ b ≥ c then a + b ≥ a + c ≥ b + c.
8. Let x, y, z, t ∈ R such that x ≥ y ≥ z ≥ t then x + y + z ≥ x + y + t ≥ x + z + t ≥
y + z + t.
9. If x ∈ R then x2 ≥ 0 with equality if and only if x = 0.
10. If Ai ∈ R+ and xi ∈ R, (1 ≤ i ≤ n) then holds
A1 x21 + A2 x22 + A3 x23 + . . . + An x2n ≥ 0,
with equality if and only if x1 = x2 = x3 = . . . = xn = 0.
1
José Luis Dı́az-Barrero
RSME Olympic Committee
UPC–BARCELONA TECH
[email protected]
Inequalities Warm-up
1. Prove the following statements:
1. If ab > 0, then holds
a
+
b
a
≥ 2. If ab < 0, then holds
b
a
2. Let a > 1, b < 1. Then, a + b > 1 + ab holds.
r
3. Let a, b be positive numbers, then holds (a + b)
4. Let a, b be positive numbers, then holds
5. If a > 1, then
1
a−1
+
1
a
+
1
a+1
>
3
a
1
2
a+b
2
(a + b) +
holds.
b
1
4
+
≥a
r
≥
b
≤ −2.
a
√
b+b
a+b
2
√
a.
.
(Pietro Mengoli 1625–1686)
6. Let a, b be positive numbers such that a + b = 1, then holds
1 2
1 2
25
a+
+ b+
≥
a
b
2
2. Prove the following statements:
1. Let a, b, c be nonnegative real numbers, then
√ holds√
√
6a + 4b + 5c ≥ 5 ab + 3 bc + 7 ca.
2. Let a, b, c be positive numbers such that a ≤ b ≤ c, then s
holds
√
3
a+b+c
a2 + b2 + c2
3
a≤
≤ abc ≤
≤
≤c
1/a + 1/b + 1/c
3
3
3. Let√
a, b, c be the length
a triangle ABC. Then,
holds
p
√ of the sides of√
a(a + c − b) + b(a + b − c) + c(b + c − a) ≤ (a2 + b2 + c2 )(a + b + c).
4. Let a, b, c be positive numbers
that a +
such
b
+ c =1, then holds
1
1
1
1+
1+
1+
≥ 64.
a
b
c
5. Let a, b, c be positive numbers, then holds
a
b+c
+
b
c+a
+
c
a+b
≥
3
(Nesbitt 1903)
2
6. Let a, b, c be positive numbers lying in the interval (0, 1]. Then holds
a
1 + b + ca
+
b
1 + c + ab
+
c
1 + a + bc
≤1
I hope you enjoy solving the preceding proposals and be sure that I will be very
pleased checking and discussing your nice solutions.
2
José Luis Dı́az-Barrero
RSME Olympic Committee
UPC–BARCELONA TECH
[email protected]
SET 1
1. Let x, y, z be strictly positive real numbers. Prove that
x
y
+ √
3
2
z
y
+
xyz
z
2
x
+ √
3
xyz
+
z
y
x
2
≥ 12
+ √
3
xyz
2. Let x, y, and z be three distinct positive real numbers such that
q
x+
y+
√
z=z+
q
√
y+ x
Prove that 40xz < 1.
3. Let a, b, and c be positive real numbers. Prove that
5a − b
2
+
b+c
5b − c
2
+
c+a
5c − a
2
≥ 12
a+b
4. Let a, b, c be positive real numbers such that a + b + c = 2. Prove that
√
4
ca
ab
2√
4
+ √
+ √
≤
3
4
4
3
3a2 + 4
3b2 + 4
3c2 + 4
bc
5. Let a, b, c be three positive numbers such that ab + bc + ca = 1. Prove that
2
2
a +b +c
2
4
2
a
r
3
s
a
b+c
+b
2
3
b
c+a
+c
2
r
c
3
!3
≥
a+b
1
2
6. Let a, b, c be three positive numbers such that a2 + b2 + c2 = 1. Prove that
1
a3 (b
+
c)5
+
1
b3 (c
+
a)5
+
1/5
1
c3 (a
+
b)5
≥
3
2
I hope you enjoy solving the preceding proposals and be sure that I will be very
pleased checking and discussing your nice solutions.
3
José Luis Dı́az-Barrero
RSME Olympic Committee
UPC–BARCELONA TECH
[email protected]
Solutions Warm-up
1. Prove the following statements:
1. If ab > 0, then holds
a
+
b
a
≥ 2. If ab < 0, then holds
b
a
2. Let a > 1, b < 1. Then, a + b > 1 + ab holds.
r
3. Let a, b be positive numbers, then holds (a + b)
4. Let a, b be positive numbers, then holds
5. If a > 1, then
1
a−1
+
1
a
+
1
a+1
>
3
1
2
a+b
2
(a + b) +
holds.
a
b
1
4
+
≥a
r
≥
b
≤ −2.
a
√
b+b
a+b
2
√
a.
.
(Pietro Mengoli 1625–1686)
6. Let a, b be positive numbers such that a + b = 1, then holds
1 2
25
1 2
+ b+
≥
a+
a
b
2
Solution 1.1 From the inequality (x − 1)2 ≥ 0 or x2 + 1 ≥ 2x immediately follows
x + 1/x ≥ 2 after division by x > 0. Equality holds when x = 1. Likewise, from
(x + 1)2 ≥ 0 or x2 + 1 ≥ −2x we get x + 1/x ≤ −2 after division by x < 0 (fact
5). Equality holds when x = −1. An alternative proof can be obtained finding the
maximum and minimum of the function f (x) = x + 1/x. Finally, putting x = a/b in
the preceding the inequalities claimed are proven.
Solution 1.2 From a > 1 and b < 1 or a − 1 > 0 and 1 − b > 0, we have that
a + b − 1 − ab = a(1 − b) + b − 1 = (a − 1)(1 − b) > 0
and we are done.
√
Solution 1.3 We have a + b ≥ 2 ab and
r
a+b
s
√
√
( a)2 + ( b)2
√
√
a+ b
=
≥
2
2
2
on account of mean inequalities. Multiplying up the preceding we get the inequality
claimed. Equality holds when a = b.
r
1
a+b
1
Solution 1.4 We observe that (a + b) + ≥
or equivalently
2
4
2
!2
r
r
a+b
a+b
1
1
a+b
−
+ =
−
≥0
2
2
4
2
2
4
which trivially holds with equality when a + b = 1/2.
Solution 1.5 Note that the inequality given is equivalent to
1
+
1
>
a−1
a+1
applying HM-AM inequality to the positive numbers a − 1 and a + 1 yields
2
1
a−1
+
1
a+1
≤
(a − 1) + (a + 1)
2
2
a
. Now
=a
with equality if and only if a − 1 = a + 1 which is impossible. So, the inequality given
is strict.
2
Solution 1.6 On account that x2 + y 2 ≥ 2 x+y
, we have
2
a+
1
2
a
1 2
+ b+
b
1
1 2
a+
+ b+
2
a
b
2
1
1
1
a+b+ +
2
a
b
1
1 2
1+
2
ab
25
1
≥
=
=
≥
2
because from a + b = 1 immediately follows
1
=
a+b
≥
√
1
≥ 4. Equality
ab and
ab
2
2
holds when a = b = 1/2 and we are done.
Notice that this inequality is a generalization of the well-known inequality
2 2
1
1
25
2
sin2 x +
+
cos
x
+
≥
cos2 x
2
sin2 x
2. Prove the following statements:
1. Let a, b, c be nonnegative real numbers, then
√ holds√
√
6a + 4b + 5c ≥ 5 ab + 3 bc + 7 ca.
2. Let a, b, c be positive numbers such that a ≤ b ≤ c, then s
holds
√
a
+
b
+
c
a2 + b2 + c2
3
3
≤ abc ≤
≤
≤c
a≤
1/a + 1/b + 1/c
3
3
3. Let√
a, b, c be the length
a triangle ABC. Then,
holds
p
√ of the sides of√
a(a + c − b) + b(a + b − c) + c(b + c − a) ≤ (a2 + b2 + c2 )(a + b + c).
4. Let a, b, c be positive numbers
that a +
such
b
+ c =1, then holds
1
1
1
1+
1+
1+
≥ 64.
a
b
c
5. Let a, b, c be positive numbers, then holds
a
b+c
+
b
c+a
+
c
a+b
≥
3
(Nesbitt 1903)
2
6. Let a, b, c be positive numbers lying in the interval (0, 1]. Then holds
a
1 + b + ca
+
b
1 + c + ab
5
+
c
1 + a + bc
≤1
Solution 2.1 Applying AM-GM inequality, we have
√
√
√
a+b
b+c
c+a
5 ab + 3 bc + 7 ca ≤ 5
+3
+7
= 6a + 4b + 5c
2
2
2
Equality holds when a = b = c and we are done.
Solution 2.2 We begin proving that a ≤
3
. Indeed, this inequality is equi+ 1b + 1c
valent to ab + ac ≤ 2 which trivially holds on account of the fact that 0 < a ≤ b ≤ c.
Next we prove that
√
a+b+c
3
abc ≤
3
1
a
To do it, we need the inequality a2 + b2 + c2 ≥ ab + bc + ca which follows immediately
adding up the trivial inequalities a2 + b2 ≥ 2ab, b2 + c2 ≥ 2bc, c2 + a2 ≥ 2ca, and
the identity a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − ab − bc − ca) (that can
be easily checked). Combining the preceding results, we get a3 + b3 + c3 − 3abc ≥ 0.
Now putting a = x3 , b = y 3 , c = z 3 , we have
x3 + y 3 + z 3 ≥ 3xyz ⇔
x3 + y 3 + z 3
3
≥ xyz ⇔
a+b+c
3
≥
√
3
abc
Applying this inequality to the positive numbers 1/a, 1/b and 1/c yields
r
1
+ 1b + 1c
1
3
a
≥
3
abc
√
3
a+b+c
3
Inverting terms immediately follows 1
abc. To prove that
≤
1
1 ≤
3
+b+c
a
q
a2 +b2 +c2
we square both sides and we get 3(a2 + b2 + c2 ) ≥ (a + b + c)2 or
3
2
2
2
equivalently,
q a + b + c ≥ ab + bc + ca. It holds as was proven before. Finally, the
inequality
terms.
a2 +b2 +c2
3
≤ c trivially holds after squaring its both sides and rearranging
Solution 2.3 Since a, b, c are the lengths of the sides of a triangle, then a + b >
c, b + c > a and a + c > b. Then, we can write
p
p
√
a(a + c − b) =
a(a + c − b) a + c − b,
p
p
√
b(a + b − c) =
b(a + b − c) a + b − c,
p
p
√
c(b + c − a) =
c(b + c − a) b + c − a.
p
p
p
Applying CBS inequality to the vectors u
~ =
a(a + c − b), b(a + b − c), c(b + c − a)
√
√
√
and ~
v=
a + c − b, a + b − c, b + c − a yields
√
2
√
√
a(a + c − b) + b(a + b − c) + c(b + c − a)
≤ a(a + c − b) + b(a + b − c) + c(b + c − a) a + c − b + a + b − c + b + c − a
= (a2 + b2 + c2 )(a + b + c)
from which the statement follows. Equality holds when a = b = c and we are done.
6
Solution 2.4 We have
1
1
1
1
1
1
1
1
1
1
1+
1+
1+
=1+ + + +
+
+
+
a
b
c
a
b
c
ab
bc
ca
abc
≥ 1 + 9 + 27 + 27 = 64
on account that from a + b + c = 1, and applying mean inequalities, immediately
1
1
1
1
1
1
c
a
b
1
1
≥ 27, + + ≥ 9 and
+
+
=
+
+
=
≥ 27.
follows
abc
a
b
c
ab
bc
ca
abc
abc
abc
abc
Equality holds when a = b = c = 1/3.
Solution 2.5 We have
a+b
b+c
+
b+c
a+b
or
1+
+
2a
b+c
b+c
c+a
+1+
+
c+a
b+c
2b
c+a
+
c+a
a+b
+1+
+
2c
a+b
a+b
c+a
≥6
≥6
from which immediately follows
a
b+c
+
b
c+a
+
c
a+b
≥
3
2
Equality holds when a = b = c.
Solution 2.6 Since 1 + ab = (1 − a)(1 − b) + a + b, then
1 + c + ab = (1 − a)(1 − b) + a + b + c ≥ a + b + c
Likewise, we have
1 + a + bc
=
(1 − b)(1 − c) + a + b + c ≥ a + b + c,
1 + b + ca
=
(1 − c)(1 − a) + a + b + c ≥ a + b + c.
Therefore,
a
1 + b + ca
+
b
1 + c + ab
+
c
1 + a + bc
≤
a
a+b+c
+
b
a+b+c
+
c
a+b+c
=1
Equality holds when at least two of the a, b, c are equal to one, and we are done.
7
José Luis Dı́az-Barrero
RSME Olympic Committee
UPC–BARCELONA TECH
[email protected]
Solutions
1. Let x, y, z be strictly positive real numbers. Prove that
x
y
+ √
3
2
z
+
xyz
y
z
x
2
+
+ √
3
xyz
z
x
y
2
+ √
3
xyz
≥ 12
Solution 1. Applying AM-QM inequality, yields
"
2 2 2 #
1
y
z
x
z
x
y
+
+
+ √
+ √
+ √
3
3
3
3
y
xyz
z
xyz
x
xyz
≥
1
9
from which follows
x
z
+ √
3
y
xyz
≥
1
3
x
y
2
+
+
y
z
y
z
+
z
x
+ √
3
+
x+y+z
√
3
xyz
x
xyz
2
+
z
x
2
+ √
3
y
2
xyz
r
x+y+z 2
1
x y z
x+y+z 2
3
+ + + √
≥
3
· · + √
3
3
y
z
x
xyz
3
y z x
xyz
2
1
x+y+z
=
3+ √
≥ 12
3
3
xyz
x
y
z
x+y+z
≥ 3 on account of AM-GM inequality. Equality holds when x =
√
3
xyz
y = z and we are done.
because
x
y
z
Solution 2. Setting a = √
,b = √
and c = √
into the statement yields
3
3
3
xyz
xyz
xyz
c 2
a 2
+ b+
+ c+
≥ 12
c
a
b
b
c
a
To prove the preceding inequality we set u
~ = a + ,b + ,c +
,~
v = (1, 1, 1)
c
a
b
into the CBS inequality and we obtain
a+
b
2
b 2
c 2
a 2
a+
+ b+
+ c+
c
a
b
8
2 2 #
c
a
+ b+
+ c+
= (12 + 12 + 12 )
a+
3
c
a
b
1
b
c
a 2
≥
a+
+ b+
+ c+
3
c
a
b
"
1
b
2
Now, taking into account that abc = 1 and applying the AM-GM inequality twice, we
get
c
a
b
+ b+
+ c+
a+
c
a
b
s r
s
c
a
b
ab bc ca
3
≥33 a+
b+
c+
≥ 3 23
=6
c
a
b
c a b
Therefore, on account of the preceding, we have
1
3
a+
b
c
c
a 2
≥ 12
+ b+
+ c+
a
b
and we are done. Note that equality holds when x = y = z.
2. Let x, y, and z be three distinct positive real numbers such that
q
x+
y+
√
z=z+
q
√
y+ x
Prove that 40xz < 1.
Solution. From x +
p
p
√
√
y + z = z + y + x we obtain
q
q
√
√
x−z = y+ x− y+ z
and
q
(x−z)
y+
√
q
x+
√
y+ z
q
=
√
y+ x+
=
√
x−
√
q
q
q
√
√
√
y+ z
y+ x− y+ z
z
Since x 6= z, dividing by both sides of the preceding expression by
q
q
√
√ √
√
x+ z
y+ x+ y+ z =1
√
x−
√
z yields
p
√
√
√
4
4
On
p the√other hand, since x, y, and z are positive, then x < y + x and z <
y + z. Therefore,
q
q
√
√ √
√ √
√ √
√
x+ z 4x+ 4z <
x+ z
y+ x+ y+ z =1
√
√
√
√
√
√
Applying AM-GM inequality, we obtain 2 4 xz ≤ x + z, and 2 8 xz ≤ 4 x + 4 z.
Multiplying up the last two inequalities, we get
p
√
√ √
√ 4 8 (xz)3 ≤
x+ z 4x+ 4z <1
9
Thus, (xz)3/8 <
1
4
and
xz <
8/3
1
4
1
1
1
= √
< √
=
3
3
40
65536
64000
from which follows 40xz < 1.
3. Let a, b, and c be positive real numbers. Prove that
5a − b
b+c
2
+
5b − c
2
c+a
+
5c − a
a+b
2
≥ 12
Solution. WLOG we can assume that a ≥ b ≥ c from which immediately follows
1
1
1
that a + b ≥ a + c ≥ b + c and
≥
≥
. Since the first and the last
b+c
c+a
a+b
sequences are sorted in the same way, by applying rearrangement inequality, we get
a
b
c
a
b
c
2
+
+
≥ 2
+
+
b+c
c+a
a+b
a+b
b+c
c+a
a
b
c
b
c
a
2
+
+
≥ 2
+
+
b+c
c+a
a+b
a+b
b+c
c+a
b
c
a
b
c
a
+
+
≥
+
+
b+c
c+a
a+b
a+b
b+c
c+a
Adding up the preceding inequalities, yields
a
b
c
a
b
c
5
+
+
≥6+
+
+
b+c
c+a
a+b
a+b
b+c
c+a
from which we obtain
1
3
5a − b
b+c
+
5b − c
c+a
+
5c − a
a+b
≥2
Taking into account AM-QM inequality, we have
v "
#
u
u 1 5a − b 2 5b − c 2 5c − a 2
1 5a − b
5b − c
5c − a
t
+
+
≥
+
+
≥2
3
b+c
c+a
a+b
3
b+c
c+a
a+b
from which the statement follows. Equality holds when a = b = c and we are done.
4. Let a, b, c be positive real numbers such that a + b + c = 2. Prove that
√
4
bc
3a2
ca
ab
2√
4
+ √
+ √
≤
3
4
4
2
2
3
+4
3b + 4
3c + 4
Solution. Squaring both terms, we get
2
bc
ca
ab
4√
+ √
+ √
≤
3
√
4
4
4
9
3a2 + 4
3b2 + 4
3c2 + 4
10
Applying CBS to the vectors u
~ =
s
bc
s
ca
r
,
√
3a2 + 4
~
v=
we obtain
√
√ √
bc, ca, ab and
,
√
3b2 + 4
√
!
ab
3c2 + 4
2
bc
ca
ab
+
+
√
√
√
4
4
4
3a2 + 4
3b2 + 4
3c2 + 4
bc
ca
ab
≤ (ab + bc + ca) √
+√
+√
3a2 + 4
3b2 + 4
3c2 + 4
From a + b + c = 2 and the well-known inequality (a + b + c)2 ≥ 3(ab + bc + ca)
immediately follows ab + bc + ca ≤ 4/3. Equality holds when a = b = c = 2/3.
On the other hand,
√
bc
=
3 √
3a2 + 4
bc
p
bc
≤ √
a2 + ab + bc + ca
1
bc
bc
p
≤
+
2 a+b
a+c
(a + b)(a + c)
=
a2 + 4/3
bc
The last inequality holds in account of HM-GM inequality. Likewise,
√
ca
1
ca
ca
3 √
≤
+
2 b+c
b+a
3b2 + 4
and
√
ab
1
ab
ab
≤
+
2 c+a
c+b
3c2 + 4
Adding the preceding inequalities, we obtain
√
bc
ca
ab
3 √
+√
+√
3a2 + 4
3b2 + 4
3c2 + 4
ca
ca
bc
bc
ab
ab
1
+
+
+
+
+
≤
2 b+c
b+a
a+b
a+c
c+a
c+b
1
= (a + b + c) = 1
2
Therefore,
√
bc
ca
ab
3
+√
+√
≤
√
2
2
2
3
3a + 4
3b + 4
3c + 4
Equality holds when a = b = c = 2/3. From the preceding, we have
√
2
bc
ca
ab
4
3
+
+
≤
·
√
√
√
4
4
4
3
3
3a2 + 4
3b2 + 4
3c2 + 4
3
√
and the statement follows. Equality holds when a = b = c = 2/3, and we are done.
5. Let a, b, c be three positive numbers such that ab + bc + ca = 1. Prove that
2
2
a +b +c
2
4
2
a
r
3
a
b+c
s
+b
2
11
3
b
c+a
+c
2
r
3
c
a+b
!3
≥
1
2
Solution 1. First we observe that from a2 + b2 + c2 ≥ ab + bc + ca and the constrain,
immediately follows that a2 + b2 + c2 ≥ 1. Multiplying and dividing the LHS of the
inequality claimed by (a2 + b2 + c2 )3 we have
 q
3
q
q
a
b
c
a2 3 b+c
+ b2 3 c+a
+ c2 3 a+b


(a2 + b2 + c2 )7 

a2 + b2 + c2
2
2
≥ (a + b + c )
=
!
a2 + b2 + c2
2 7
a2
(a2 + b2 + c2 )8
2(ab + bc + ca)
=
b+c
a
+ b2
c+a
b
+ c2
(a2 + b2 + c2 )8
2
a+b
c
≥
1
2
In the preceding we have used the inequality f (1/3) ≥ f (−1), where
f (α) =
w1 aα + w2 bα + w3 cα
w1 + w2 + w3
is the mean powered inequality with weights w1 =
and w3 =
c2
a2 + b2 + c2
1/α
a2
+
b2
+
c2
, w2 =
b2
,
+ b2 + c2
√
respectively. Equality holds when a = b = c = 1/ 3, and we
a2
a2
are done.
Solution 2. First, using the constrain, we write the inequality claimed in the most
convenient form
s
r
r
1
a
b
c
ab + bc + ca 4/3
2 3
2 3
2 3
a
+b
+c
≥ √
3
b+c
c+a
a+b
2 a2 + b2 + c2
1
ab + bc + ca 4/3
= √
(ab + bc + ca)
3
2 a2 + b2 + c2
To prove the preceding inequality, we consider the function f : [0, +∞) → R defined
by f (t) = t7/3 which is convex, as can be easily checked. Applying Jensen’s inequali(b + c)2
(c + a)2
(a + b)2
ty, with q1 =
, q2 =
, q3 =
, where A = (a + b)2 + (b + c)2 +
A
A
A
a
b
c
, x2 =
, x3 =
, we have
(c + a)2 ; and x1 =
b+c
c+a
a+b
a
b
c
q1 f
+ q2 f
+ q3 f
b+c
c+a
a+b
a
b
c
≥ f q1
+ q2
+ q3
,
b+c
c+a
a+b
or equivalently,
(b + c)2
A
7/3
a
+
b+c
≥
(c + a)2
A
b
c+a
7/3
+
(a + b)2
(a(b + c) + b(c + a) + c(a + b)
A
12
A
7/3
c
a+b
7/3
Rearranging terms, and after simplification, we obtain
a2
r
3
h
s
a
b+c
+ b2
b
3
+ c2
c+a
r
c
3
≥ h
a+b
2(ab + bc + ca)
i7/3
i4/3
2(a2 + b2 + c2 + ab + bc + ca)
h
i7/3
2(ab + bc + ca)
ab + bc + ca 4/3
1
=
(ab + bc + ca)
≥ h
√
i4/3
3
2 a2 + b2 + c2
4(a2 + b2 + c2 )
on account of the well-known
inequality a2 + b2 + c2 ≥ ab + bc + ca. Equality holds
√
when a = b = c = 1/ 3, and we are done.
6. Let a, b, c be three positive numbers such that a2 + b2 + c2 = 1. Prove that
1
a3 (b
+
c)5
+
1
b3 (c
+
a)5
+
1/5
1
c3 (a
+
≥
b)5
3
2
Solution 1. To prove the inequality claimed we consider the function f : R → R
defined by f (t) = t5 . This function is convex in [0, +∞) as can be easily checked.
Applying Jensen’s inequality, namely
!
3
3
X
X
qk f (xk ) ≥ f
qk x k
k=1
with
q1 =
a2
a2 + b2 + c
and
x1 =
k=1
,
2
q2 =
1
b2
a2 + b2 + c
,
x2 =
1
5
a(b + c)
1
b(c + a)
,
2
,
q3 =
x3 =
c2
a2 + b2 + c2
1
c(a + b)
,
yields
"
1
2
a2 + b2 + c2
a
+b
a(b + c)
≥
a
b+c
+
2
5
1
+c
b(c + a)
b
c+a
+
c
a+b
2
5 #
1
c(a + b)
5
(a2 + b2 + c2 )5
or equivalently,
1
a3 (b + c)5
+
1
b3 (c + a)5
+
1
c3 (a + b)5
≥
a
b+c
+
b
c+a
+
c
5
a+b
where we have used the constrain a2 + b2 + c2 = 1. Now, it is easy to see that
a
b+c
+
b
c+a
+
13
c
a+b
≥
3
2
Indeed, WLOG we can assume that a ≥ b ≥ c from which follows a+b ≥ a+c ≥ b+c
1
1
1
and b+c
≥ c+a
≥ a+b
. Now, applying rearrangement inequality, we get
a
b
+
b+c
a
c+a
b
+
b+c
c+a
+
+
c
a
≥
a+b
c
a+b
b
≥
a+b
+
+
a+b
b
b+c
c
b+c
+
+
c
c+a
a
c+a
Adding the preceding expressions, we obtain
a
b+c
+
b
+
c+a
c
3
≥
a+b
(Nesbitt’s Inequality)
2
Equality holds when a = b = c.
Finally, substituting this√result in the preceding the statement follows. Equality
holds when a = b = c = 3/3, and we are done.
Solution 2. We recall that Hölder’s inequality can be stated as follows: Let a1 , a2 , ..., an
and b1 , b2 , ..., bn , (n ≥ 2), be sequences of positive real numbers and let p, q ∈ R∗+ such
that p1 + q1 = 1, then the following inequality holds
! p1
n
X
n
X
api
! q1
bqi
≥
i=1
i=1
n
X
ai bi
i=1
Particularizing the preceding result for n = 3 with p = 5/4, q = 5 for which (1/p +
1/q = 1), we get
! 15
! 54
3
3
3
X
X
X
5/4
5
bi
≥
ai bi
ai
i=1
i=1
i=1
Putting in the above result a1 = a8/5 , a2 = b8/5 , a3 = c8/5 , b1 =
1
b3/5 (c
8/5
a
+ a)
5/4
, b3 =
+ b
8/5
1
c3/5 (a
5/4
+ b)
8/5
+ c
1
, b2 =
a3/5 (b
+ c)
5
1
c3/5 (a + b)
yields
"
5/4 4/5 ≥
a
b+c
1
a3/5 (b + c)
+
b
c+a
+
5
+
1
b3/5 (c + a)
+
5 #1/5
c
a+b
or equivalently,
2
2
2 4/5
(a + b + c )
1
a3 (b
+
c)5
+
1
b3 (c
+
a)5
+
1/5
1
c3 (a
+
b)5
≥
a
b+c
+
b
c+a
+
c
a+b
from which the statement follows on account
√of the constrain and Nesbitt’s inequality. Equality holds when when a = b = c = 3/3, and we are done.
Solution 3 by Emilio Fernández Moral. On account of mean inequalities we have
!1/5
1/5
1
1
1
3
+ 3
+ 3
≥ p
3
a3 (b + c)5
b (c + a)5
c (a + b)5
a3 b3 c3 (a + b)5 (b + c)5 (c + a)5
14
r
=
5
3
abc
1
p
3
(a + b)(b + c)(c + a)
≥
31/5+1/2
2 (abc)1/5
≥
3
2
p
2(a + b + c)
(a + b) + (b + c) + (c + a)
=
≤
because 3 (a + b)(b + c)(c + a) ≤
3
3
r
3/10
3/10
3/10
√
1
a2 + b2 + c2
1
3
2
a2 b2 c2
≤
and (abc)1/5 = (a2 b2 c2 )1/10 =
=
3
3
√ 3
and 31/5+1/2+3/10 = 3. Equality holds when a = b = c = 3/3.
15
José Luis Dı́az-Barrero
RSME Olympic Committee
UPC–BARCELONA TECH
[email protected]
Some Results
Hereafter, some classical discrete inequalities are stated and proven. We begin with
Theorem 1 (General Mean Inequalities) Let a1 , a2 , . . . , an be positive real numbers.
Then, the function f : R → R defined by
f (α) =
α
α
aα
1 + a2 + . . . + an
1/α
n
is nondecreasing and the following limits hold
n o
√
lim f (α) = min ai , lim f (α) = n a1 a2 . . . an ,
α→−∞
1≤i≤n
α→0
n o
lim f (α) = max ai
α→+∞
1≤i≤n
Proof. Let 0 < a < b. Consider the function g : [0, +∞) → R defined by g(α) = αb/a .
Since
b b
g 00 (α) =
− 1 αb/a−2 ≥ 0, (α ≥ 0),
a a
then g is convex in [0, +∞). Applying Jensen’s inequality, we get


n
n
1 X
1 X a
αj  ≤
g(αa
g
j)
n j=1
n j=1
or


n
1 X
n
b/a

αa
j
j=1
≤
n
1 X
n
αbj
j=1
from which follows f (a) ≤ f (b). For negative values of α consider the function
h(α) = (−α)b/a which is convex in (−∞, 0).
α
α 1/α
a1 + aα
2 + . . . + an
Let L1 = lim f (α). Then, ln L1 = lim ln
. That is,
α→−∞
α→−∞
n
α
α 1
a1 + aα
2 + . . . + an
ln L1 =
lim
ln
α→−∞ α
n
α
α
aα
ln
a
+
a
ln
a
1
2 + . . . + an ln an
1
2
=
lim
α
α
α→−∞
aα
1 + a2 + . . . + an
n o
= ln min ai
1≤i≤n
16
n o
from which follows L1 = min ai . Likewise, we obtain that L2 =
1≤i≤n
n o
max ai . Finally, denoting by L = lim f (α), we have
lim f (α) =
α→+∞
α→0
1≤i≤n
ln L =
=
lim
α→0
lim
1
ln
α
α
aα
1 + a2 + . . . + an
α
n
α
α
aα
ln
a
+
a
ln
a
1
2 + . . . + an ln an
1
2
α
α
aα
1 + a2 + . . . + an
ln a1 + ln a2 + . . . + ln an
√
=
= ln n a1 a2 . . . an
n
√
from which immediately follows lim f (α) = n a1 a2 . . . an and the proof is complete.
α→0
In 1821 Cauchy published his famous inequality as the second of the two notes
on the theory of inequalities that formed the final part of his book Cours d’Analyse
Algébrique. Namely,
α→0
Theorem 2 Let a1 , a2 , ..., an , b1 , b2 , ..., bn be any real numbers. Then, the following
inequality holds:
!2
!
!
n
n
n
X
X
X
2
2
ak bk
≤
ak
bk
k=1
k=1
k=1
Proof. Consider the quadratic polynomial A(x) =
n
X
(ak x − bk )2 . Then,
k=1
A(x) = x
2
n
X
a2k
n
X
− 2x
k=1
ak bk +
k=1
n
X
b2k ≥ 0
k=1
Since the preceding equation is nonnegative its discriminant must be less or equal
than zero. That is,
!2
!
!
n
n
n
X
X
X
2
2
ak bk
−
ak
bk ≤ 0
k=1
k=1
k=1
from which the statement follows. Equality holds when the n-tuples (a1 , a2 , . . . , an )
and (b1 , b2 , . . . , bn ) are proportional. This completes the proof.
A generalization of CBS inequality is the well known inequality of Hölder. It is stated
and proved in the following theorem.
Theorem 3 Let a1 , a2 , ..., an and b1 , b2 , ..., bn , (n ≥ 2) be sequences of positive real
numbers and let p, q ∈ R∗+ such that p1 + q1 = 1, then the following inequality holds
n
X
i=1
! p1
api
n
X
! q1
bqi
≥
i=1
n
X
ai bi .
i=1
Proof 1. We will argue by induction. Assume that the inequality holds for n and we
have to prove it for n + 1. In fact,
n+1
X
i=1
ai bi =
n
X
ai bi + an+1 bn+1 ≤
i=1
n
X
i=1
17
! p1
api
n
X
i=1
! q1
bqi
+ an+1 bn+1
≤
n
X
! p1
api
+
n
X
apn+1
! q1
bqi
+
bqn+1
n+1
X
=
i=1
i=1
! p1
n+1
X
api
! q1
bqi
.
i=1
i=1
To complete the inductive process, we must verify that the case when n = 2 also
holds. Indeed, the inequality
1
1
a1 b1 + a2 b2 ≤ (ap1 + ap2 ) p (bq1 + bq2 ) q ,
immediate follows from the fact that the function
1
1
f (x) = (ap1 + ap2 ) p (bq1 + xq ) q − a1 b1 − a2 x,
is strictly positive for all x > 0. This completes the proof.
Proof 2. First, we write the inequality in the most convenient form
Pn
i=1 ai bi
1 ≤ 1
1
Pn
q q
p p Pn
i=1 bi
i=1 ai
or equivalently,
n X
p1 api
Pn
i=1
i=1
Pn
api
q1
bqi
i=1
≤ 1.
bqi
Now, using the power mean inequality,
1
1
xα y β ≤
1
α
x+
1
β
y,
we have
n X
Pn
i=1
p1 api
i=1
Pn
api
=
i=1
1
p
q1
bqi
Pn
Pi=1
n
i=1
api
api
≤
bqi
+
n X
1
i=1
1
q
Pn
Pi=1
n
bqi
q
i=1 bi
p
ap
Pn i
=
i=1
1
p
+
api
1
q
+
1
q
bq
Pn i
i=1
bqi
= 1,
as desired. Equality holds if and only if the n-tuples (ap1 , ap2 , ..., apn ) and (bq1 , bq2 , ..., bqn )
are proportional.
Notice that for p = q = 2, we get the inequality
(a1 b1 + a2 b2 + ... + an bn )2 ≤ (a21 + a22 + ... + a2n )(b21 + b22 + ... + b2n )
This is the Cauchy-Bunyakowski-Schwarz inequality.
Using Hölder’s inequality can be easily proven the following inequality of Minkowski.
Theorem 4 Let a1 , a2 , ..., an and b1 , b2 , ..., bn , (n ≥ 2) be sequences of positive real
numbers and let p > 1, be a real number, then
n
X
! p1
(ai + bi )p
i=1
≤
n
X
i=1
When p < 1 inequality reverses.
18
! p1
api
+
n
X
i=1
! p1
bpi
Proof. Let be q =
p
,
p−1
n
X
then
1
p
(ai + bi )p =
i=1
≤
n
X
+
! p1
api
1
q
= 1, and applying Hölder’s inequality, we obtain
n
X
i=1
n
X

=
! q1
(p−1)q
(ai + bi )
+
n
X
! p1
bpi
i=1
n
X
! p1
api
+
! p1 
n
X
bpi
(ai + bi )
n
X
Pn
i=1 (ai
! p1
≤
i=1
! q1
(p−1)q
(ai + bi )
! p−1
p
(ai + bi )p
i=1
Multiplying the preceding inequality by
p
n
X
i=1

i=1
i=1
n
X
bi (ai + bi )p−1
i=1
i=1
i=1
n
X
ai (ai + bi )p−1 +
n
X
+ bi )p
! p1
api
i=1
+
− p−1
p
n
X
, we get
! p1
bpi
i=1
and the proof is complete. Equality holds if and only if (a1 , a2 , ..., an ) and (b1 , b2 , ..., bn )
are proportional.
A most useful result is presented in the following
Theorem 5 (Rearrangement) Let a1 , a2 , · · · , an and b1 , b2 , · · · , bn be sequences of
positive real numbers and let c1 , c2 , · · · , cn be a permutation of b1 , b2 , · · · , bn . The
sum S = a1 b1 + a2 b2 + · · · + an bn is maximal if the two sequences a1 , a2 , · · · , an and
b1 , b2 , · · · , bn are sorted in the same way and minimal if the two sequences are sorted
oppositely, one increasing and the other decreasing.
Proof. Let ai > aj . Consider the sums
S1
= a1 c1 + · · · + ai ci + · · · + aj cj + · · · + an cn
S2
= a1 c1 + · · · + ai cj + · · · + aj ci + · · · + an cn
We have obtained S2 from S1 by switching the positions of ci and cj . Then
S1 − S2 = ai ci + aj cj − ai cj − aj ci = (ai − aj )(ci − cj )
Therefore,
ci > cj ⇒ S1 > S2
and ci < cj ⇒ S1 < S2 .
This completes the proof.
In the sequel, after defining the concepts of convex and concave function, the inequality of Jensen is presented.
Definition. Let f : I ⊆ R → R be a real function. We say that f is convex (concave)
on I if for all a1 , a2 ∈ I and for all t ∈ [0, 1], we have
f (ta1 + (1 − t)a2 ) ≤ f (a1 ) + (1 − t)f (a2 )
When f is concave the inequality reverses.
19
Theorem 6 Let f : I ⊆ R → R be a convex function. Let q1 , q2 , . . . , qn be nonnegative
real numbers such that q1 + q2 + . . . + qn = 1. Then for all ak ∈ I (1 ≤ k ≤ n) holds
!
n
n
X
X
f
qk ak ≤
qk f (ak )
k=1
k=1
Equality holds when a1 = a2 = . . . = an . If f is concave the preceding inequality
reverses.
Proof. We will argue by mathematical induction. For n = 2, we have
f (q1 a1 + q2 a2 ) ≤ q1 f (a1 ) + q2 f (a2 )
The preceding inequality holds because f is convex and q1 + q2 = 1. Assume that
!
n
n
X
X
f
qk ak ≤
qk f (ak )
k=1
and we have to see that
f
n+1
X
k=1
!
qk ak
≤
k=1
n+1
X
qk f (ak )
k=1
when q1 + q2 + . . . + qn + qn+1 = 1. Let q =
Pn
k=1 qk . Then, we have
n
X
qk
k=1
n X
qk
q
= 1,
ak ∈ I and q + qn+1 = 1. Now, taking into account the case when n = 2
q
and the inductive hypothesis, yields
!
"
#
!
n+1
n X
X
qk
f
qk ak
= f q
ak + qn+1 an+1
q
k=1
k=1
!
n X
qk
≤ qf
ak + qn+1 f (an+1 )
q
k=1
n X
qk
f (ak ) + qn+1 f (an+1 )
≤ q
q
k=1
k=1
=
n+1
X
qk f (ak )
k=1
and by the principle of mathematical induction (PMI) the statement is proven. Equality holds when a1 = a2 = . . . = an .
20
Think about
“Pulchra dicuntur quae visa placent– beauty is that which, being seen, pleases”. This definition, applies well to Mathematical
beauty in which lack of understanding is so often responsible of
lack of pleasure.
Thomas Aquina’s definition of Beauty