Inequalities of OlympiadCaliber

Jose Luis Dıaz-BarreroRSME Olympiad Committee

BARCELONA TECH

Jose Luis Dıaz-BarreroRSME Olympic CommitteeUPC–BARCELONA TECHjose.luis.diaz@upc.edu

 

Basic facts to proveinequalities

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

A1x21 +A2x

22 +A3x

23 + . . .+Anx

2n ≥ 0,

with equality if and only if x1 = x2 = x3 = . . . = xn = 0.

1

Jose Luis Dıaz-BarreroRSME Olympic CommitteeUPC–BARCELONA TECHjose.luis.diaz@upc.edu

 

Inequalities Warm-up1. Prove the following statements:

1. If ab > 0, then holdsa

b+b

a≥ 2. If ab < 0, then holds

a

b+b

a≤ −2.

2. Let a > 1, b < 1. Then, a+ b > 1 + ab holds.

3. Let a, b be positive numbers, then holds (a+ b)

√a+ b

2≥ a√b+ b

√a.

4. Let a, b be positive numbers, then holds1

2(a+ b) +

1

4≥√a+ b

2.

5. If a > 1, then1

a− 1+

1

a+

1

a+ 1>

3

aholds. (Pietro Mengoli 1625–1686)

6. Let a, b be positive numbers such that a+ b = 1, then holds(a+

1

a

)2

+

(b+

1

b

)2

≥25

2

2. Prove the following statements:

1. Let a, b, c be nonnegative real numbers, then holds6a+ 4b+ 5c ≥ 5

√ab+ 3

√bc+ 7

√ca.

2. Let a, b, c be positive numbers such that a ≤ b ≤ c, then holds

a ≤3

1/a+ 1/b+ 1/c≤ 3√abc ≤

a+ b+ c

3≤

√a2 + b2 + c2

3≤ c

3. Let a, b, c be the length of the sides of a triangle ABC. Then, holds√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 such that a+ b+ c = 1, then holds(1 +

1

a

)(1 +

1

b

)(1 +

1

c

)≥ 64.

5. Let a, b, c be positive numbers, then holds

a

b+ c+

b

c+ a+

c

a+ b≥

3

2(Nesbitt 1903)

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 verypleased checking and discussing your nice solutions.

2

Jose Luis Dıaz-BarreroRSME Olympic CommitteeUPC–BARCELONA TECHjose.luis.diaz@upc.edu

 

SET 1

1. Let x, y, z be strictly positive real numbers. Prove that(x

y+

z

3√xyz

)2

+

(y

z+

x

3√xyz

)2

+

(z

x+

y

3√xyz

)2

≥ 12

2. Let x, y, and z be three distinct positive real numbers such that

x+

√y +√z = z +

√y +√x

Prove that 40xz < 1.

3. Let a, b, and c be positive real numbers. Prove that(5a− bb+ c

)2

+

(5b− cc+ a

)2

+

(5c− aa+ b

)2

≥ 12

4. Let a, b, c be positive real numbers such that a+ b+ c = 2. Prove that

bc4√3a2 + 4

+ca

4√3b2 + 4

+ab

4√3c2 + 4

≤2

3

4√3

5. Let a, b, c be three positive numbers such that ab+ bc+ ca = 1. Prove that

(a2 + b2 + c2

)4 (a2 3

√a

b+ c+ b2 3

√b

c+ a+ c2 3

√c

a+ b

)3

≥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

c3(a+ b)5

)1/5

≥3

2

I hope you enjoy solving the preceding proposals and be sure that I will be verypleased checking and discussing your nice solutions.

3

Jose Luis Dıaz-BarreroRSME Olympic CommitteeUPC–BARCELONA TECHjose.luis.diaz@upc.edu

 

Solutions Warm-up1. Prove the following statements:

1. If ab > 0, then holdsa

b+b

a≥ 2. If ab < 0, then holds

a

b+b

a≤ −2.

2. Let a > 1, b < 1. Then, a+ b > 1 + ab holds.

3. Let a, b be positive numbers, then holds (a+ b)

√a+ b

2≥ a√b+ b

√a.

4. Let a, b be positive numbers, then holds1

2(a+ b) +

1

4≥√a+ b

2.

5. If a > 1, then1

a− 1+

1

a+

1

a+ 1>

3

aholds. (Pietro Mengoli 1625–1686)

6. Let a, b be positive numbers such that a+ b = 1, then holds(a+

1

a

)2

+

(b+

1

b

)2

≥25

2

Solution 1.1 From the inequality (x − 1)2 ≥ 0 or x2 + 1 ≥ 2x immediately followsx + 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 (fact5). Equality holds when x = −1. An alternative proof can be obtained finding themaximum and minimum of the function f(x) = x+ 1/x. Finally, putting x = a/b inthe 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

√a+ b

2=

√(√a)2 + (

√b)2

2≥√a+√b

2on account of mean inequalities. Multiplying up the preceding we get the inequalityclaimed. Equality holds when a = b.

Solution 1.4 We observe that1

2(a+ b) +

1

4≥√a+ b

2or equivalently

a+ b

2−√a+ b

2+

1

4=

(1

2−√a+ b

2

)2

≥ 0

4

which trivially holds with equality when a+ b = 1/2.

Solution 1.5 Note that the inequality given is equivalent to1

a− 1+

1

a+ 1>

2

a. Now

applying HM-AM inequality to the positive numbers a− 1 and a+ 1 yields

21

a−1+ 1

a+1

≤(a− 1) + (a+ 1)

2= a

with equality if and only if a−1 = a+1 which is impossible. So, the inequality givenis strict.

Solution 1.6 On account that x2 + y2 ≥ 2(x+y2

)2, we have(

a+1

a

)2

+

(b+

1

b

)2

≥1

2

[(a+

1

a

)+

(b+

1

b

)]2=

1

2

(a+ b+

1

a+

1

b

)2

=1

2

(1 +

1

ab

)2

≥25

2

because from a+ b = 1 immediately follows1

2=a+ b

2≥√ab and

1

ab≥ 4. Equality

holds when a = b = 1/2 and we are done.Notice that this inequality is a generalization of the well-known inequality(

sin2 x+1

sin2 x

)2

+

(cos2 x+

1

cos2 x

)2

≥25

2

2. Prove the following statements:

1. Let a, b, c be nonnegative real numbers, then holds6a+ 4b+ 5c ≥ 5

√ab+ 3

√bc+ 7

√ca.

2. Let a, b, c be positive numbers such that a ≤ b ≤ c, then holds

a ≤3

1/a+ 1/b+ 1/c≤ 3√abc ≤

a+ b+ c

3≤

√a2 + b2 + c2

3≤ c

3. Let a, b, c be the length of the sides of a triangle ABC. Then, holds√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 such that a+ b+ c = 1, then holds(1 +

1

a

)(1 +

1

b

)(1 +

1

c

)≥ 64.

5. Let a, b, c be positive numbers, then holds

a

b+ c+

b

c+ a+

c

a+ b≥

3

2(Nesbitt 1903)

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

5

Solution 2.1 Applying AM-GM inequality, we have

5√ab+ 3

√bc+ 7

√ca ≤ 5

(a+ b

2

)+ 3

(b+ c

2

)+ 7

(c+ a

2

)= 6a+ 4b+ 5c

Equality holds when a = b = c and we are done.

Solution 2.2 We begin proving that a ≤3

1a+ 1

b+ 1

c

. Indeed, this inequality is equi-

valent to ab+ a

c≤ 2 which trivially holds on account of the fact that 0 < a ≤ b ≤ c.

Next we prove that3√abc ≤

a+ b+ c

3

To do it, we need the inequality a2+b2+c2 ≥ ab+bc+ca which follows immediatelyadding up the trivial inequalities a2 + b2 ≥ 2ab, b2 + c2 ≥ 2bc, c2 + a2 ≥ 2ca, andthe identity a3 + b3 + c3 − 3abc = (a+ b+ c)(a2 + b2 + c2 − ab− bc− ca) (that canbe easily checked). Combining the preceding results, we get a3 + b3 + c3− 3abc ≥ 0.Now putting a = x3, b = y3, c = z3, we have

x3 + y3 + z3 ≥ 3xyz ⇔x3 + y3 + z3

3≥ xyz ⇔

a+ b+ c

3≥ 3√abc

Applying this inequality to the positive numbers 1/a, 1/b and 1/c yields

1a+ 1

b+ 1

c

3≥ 3

√1

abc

Inverting terms immediately follows3

1a+ 1

b+ 1

c

≤ 3√abc. To prove that

a+ b+ c

3≤√

a2+b2+c2

3we square both sides and we get 3(a2 + b2 + c2) ≥ (a + b + c)2 or

equivalently, a2 + b2 + c2 ≥ ab+ bc+ ca. It holds as was proven before. Finally, the

inequality√a2+b2+c2

3≤ c trivially holds after squaring its both sides and rearranging

terms.

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

√a(a+ c− b) =

√a(a+ c− b)

√a+ c− b,

√b(a+ b− c) =

√b(a+ b− c)

√a+ b− c,

√c(b+ c− a) =

√c(b+ c− a)

√b+ c− a.

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(√

a(a+ c− b) +√b(a+ b− c) +

√c(b+ c− a)

)2≤(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

a

)(1 +

1

b

)(1 +

1

c

)= 1 +

1

a+

1

b+

1

c+

1

ab+

1

bc+

1

ca+

1

abc

≥ 1 + 9 + 27 + 27 = 64

on account that from a + b + c = 1, and applying mean inequalities, immediately

follows1

abc≥ 27,

1

a+

1

b+

1

c≥ 9 and

1

ab+

1

bc+

1

ca=

c

abc+

a

abc+

b

abc=

1

abc≥ 27.

Equality holds when a = b = c = 1/3.

Solution 2.5 We have

a+ b

b+ c+b+ c

a+ b+b+ c

c+ a+c+ a

b+ c+c+ a

a+ b+a+ b

c+ a≥ 6

or

1 +2a

b+ c+ 1 +

2b

c+ a+ 1 +

2c

a+ b≥ 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

Jose Luis Dıaz-BarreroRSME Olympic CommitteeUPC–BARCELONA TECHjose.luis.diaz@upc.edu

 

Solutions1. Let x, y, z be strictly positive real numbers. Prove that(

x

y+

z

3√xyz

)2

+

(y

z+

x

3√xyz

)2

+

(z

x+

y

3√xyz

)2

≥ 12

Solution 1. Applying AM-QM inequality, yields

1

3

[(x

y+

z

3√xyz

)2

+

(y

z+

x

3√xyz

)2

+

(z

x+

y

3√xyz

)2]

≥1

9

(x

y+y

z+z

x+x+ y + z

3√xyz

)2

from which follows(x

y+

z

3√xyz

)2

+

(y

z+

x

3√xyz

)2

+

(z

x+

y

3√xyz

)2

≥1

3

(x

y+y

z+z

x+x+ y + z

3√xyz

)2

≥1

3

(3 3

√x

y·y

z·z

x+x+ y + z

3√xyz

)2

=1

3

(3 +

x+ y + z

3√xyz

)2

≥ 12

becausex+ y + z

3√xyz

≥ 3 on account of AM-GM inequality. Equality holds when x =

y = z and we are done. �

Solution 2. Setting a =x

3√xyz

, b =y

3√xyz

and c =z

3√xyz

into the statement yields

(a+

b

c

)2

+

(b+

c

a

)2

+

(c+

a

b

)2

≥ 12

To prove the preceding inequality we set ~u =

(a+

b

c, b+

c

a, c+

a

b

), ~v = (1, 1, 1)

into the CBS inequality and we obtain(a+

b

c

)2

+

(b+

c

a

)2

+

(c+

a

b

)2

8

=1

3(12 + 12 + 12)

[(a+

b

c

)2

+

(b+

c

a

)2

+

(c+

a

b

)2]

≥1

3

[(a+

b

c

)+

(b+

c

a

)+

(c+

a

b

)]2Now, taking into account that abc = 1 and applying the AM-GM inequality twice, weget (

a+b

c

)+

(b+

c

a

)+

(c+

a

b

)

≥ 3 3

√(a+

b

c

)(b+

c

a

)(c+

a

b

)≥ 3

3

√23

√ab

c

bc

a

ca

b= 6

Therefore, on account of the preceding, we have

1

3

[(a+

b

c

)+

(b+

c

a

)+

(c+

a

b

)]2≥ 12

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

x+

√y +√z = z +

√y +√x

Prove that 40xz < 1.

Solution. From x+√y +√z = z +

√y +√x we obtain

x− z =

√y +√x−

√y +√z

and

(x−z)(√

y +√x+

√y +√z

)=

(√y +√x+

√y +√z

)(√y +√x−

√y +√z

)=√x−√z

Since x 6= z, dividing by both sides of the preceding expression by√x−√z yields

(√x+√z) (√

y +√x+

√y +√z

)= 1

On the other hand, since x, y, and z are positive, then 4√x <

√y +√x and 4

√z <√

y +√z. Therefore,

(√x+√z) (

4√x+ 4√z)<(√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

4 8√

(xz)3 ≤(√x+√z) (

4√x+ 4√z)< 1

9

Thus, (xz)3/8 <1

4and

xz <

(1

4

)8/3

=1

3√65536

<1

3√64000

=1

40

from which follows 40xz < 1.�

3. Let a, b, and c be positive real numbers. Prove that(5a− bb+ c

)2

+

(5b− cc+ a

)2

+

(5c− aa+ b

)2

≥ 12

Solution. WLOG we can assume that a ≥ b ≥ c from which immediately follows

that a + b ≥ a + c ≥ b + c and1

b+ c≥

1

c+ a≥

1

a+ b. Since the first and the last

sequences are sorted in the same way, by applying rearrangement inequality, we get

2

(a

b+ c+

b

c+ a+

c

a+ b

)≥ 2

(a

a+ b+

b

b+ c+

c

c+ a

)2

(a

b+ c+

b

c+ a+

c

a+ b

)≥ 2

(b

a+ b+

c

b+ c+

a

c+ a

)a

b+ c+

b

c+ a+

c

a+ b≥

a

a+ b+

b

b+ c+

c

c+ a

Adding up the preceding inequalities, yields

5

(a

b+ c+

b

c+ a+

c

a+ b

)≥ 6 +

a

a+ b+

b

b+ c+

c

c+ a

from which we obtain

1

3

(5a− bb+ c

+5b− cc+ a

+5c− aa+ b

)≥ 2

Taking into account AM-QM inequality, we have√√√√1

3

[(5a− bb+ c

)2

+

(5b− cc+ a

)2

+

(5c− aa+ b

)2]≥

1

3

(5a− bb+ c

+5b− cc+ a

+5c− aa+ b

)≥ 2

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

bc4√3a2 + 4

+ca

4√3b2 + 4

+ab

4√3c2 + 4

≤2

3

4√3

Solution. Squaring both terms, we get(bc

4√3a2 + 4

+ca

4√3b2 + 4

+ab

4√3c2 + 4

)2

≤4

9

√3

10

Applying CBS to the vectors ~u =(√

bc,√ca,√ab)

and

~v =

(√bc

√3a2 + 4

,

√ca

√3b2 + 4

,

√ab

√3c2 + 4

)we obtain (

bc4√3a2 + 4

+ca

4√3b2 + 4

+ab

4√3c2 + 4

)2

≤ (ab+ bc+ ca)

(bc

√3a2 + 4

+ca

√3b2 + 4

+ab

√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,

√3

(bc

√3a2 + 4

)=

bc√a2 + 4/3

≤bc

√a2 + ab+ bc+ ca

=bc√

(a+ b)(a+ c)≤

1

2

(bc

a+ b+

bc

a+ c

)The last inequality holds in account of HM-GM inequality. Likewise,

√3

(ca

√3b2 + 4

)≤

1

2

(ca

b+ c+

ca

b+ a

)and

√3

(ab

√3c2 + 4

)≤

1

2

(ab

c+ a+

ab

c+ b

)Adding the preceding inequalities, we obtain

√3

(bc

√3a2 + 4

+ca

√3b2 + 4

+ab

√3c2 + 4

)

≤1

2

(ca

b+ c+

ca

b+ a+

bc

a+ b+

bc

a+ c+

ab

c+ a+

ab

c+ b

)=

1

2(a+ b+ c) = 1

Therefore,bc

√3a2 + 4

+ca

√3b2 + 4

+ab

√3c2 + 4

≤√3

3

Equality holds when a = b = c = 2/3. From the preceding, we have(bc

4√3a2 + 4

+ca

4√3b2 + 4

+ab

4√3c2 + 4

)2

≤4

3·√3

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

(a2 + b2 + c2

)4 (a2 3

√a

b+ c+ b2 3

√b

c+ a+ c2 3

√c

a+ b

)3

≥1

2

11

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 theinequality claimed by (a2 + b2 + c2)3 we have

(a2 + b2 + c2)7

a2 3

√ab+c

+ b2 3

√bc+a

+ c2 3

√ca+b

a2 + b2 + c2

3

≥ (a2 + b2 + c2)7

(a2 + b2 + c2

a2 b+ca

+ b2 c+ab

+ c2 a+bc

)

=(a2 + b2 + c2)8

2(ab+ bc+ ca)=

(a2 + b2 + c2)8

2≥

1

2

In the preceding we have used the inequality f(1/3) ≥ f(−1), where

f(α) =

(w1a

α + w2bα + w3c

α

w1 + w2 + w3

)1/α

is the mean powered inequality with weights w1 =a2

a2 + b2 + c2, w2 =

b2

a2 + b2 + c2,

and w3 =c2

a2 + b2 + c2respectively. Equality holds when a = b = c = 1/

√3, and we

are done. �

Solution 2. First, using the constrain, we write the inequality claimed in the mostconvenient form

a2 3

√a

b+ c+ b2 3

√b

c+ a+ c2 3

√c

a+ b≥

13√2

(ab+ bc+ ca

a2 + b2 + c2

)4/3

=13√2

(ab+ bc+ ca

a2 + b2 + c2

)4/3

(ab+ bc+ ca)

To prove the preceding inequality, we consider the function f : [0,+∞)→ R definedby f(t) = t7/3 which is convex, as can be easily checked. Applying Jensen’s inequali-

ty, with q1 =(b+ c)2

A, q2 =

(c+ a)2

A, q3 =

(a+ b)2

A, where A = (a+ b)2+(b+ c)2+

(c+ a)2; and x1 =a

b+ c, x2 =

b

c+ a, x3 =

c

a+ b, we have

q1f

(a

b+ c

)+ q2f

(b

c+ a

)+ q3f

(c

a+ b

)

≥ f(q1

a

b+ c+ q2

b

c+ a+ q3

c

a+ b

),

or equivalently,

(b+ c)2

A

(a

b+ c

)7/3

+(c+ a)2

A

(b

c+ a

)7/3

+(a+ b)2

A

(c

a+ b

)7/3

≥((a(b+ c) + b(c+ a) + c(a+ b)

A

)7/3

12

Rearranging terms, and after simplification, we obtain

a2 3

√a

b+ c+ b2 3

√b

c+ a+ c2 3

√c

a+ b≥

[2(ab+ bc+ ca)

]7/3[2(a2 + b2 + c2 + ab+ bc+ ca)

]4/3

≥

[2(ab+ bc+ ca)

]7/3[4(a2 + b2 + c2)

]4/3 =13√2

(ab+ bc+ ca

a2 + b2 + c2

)4/3

(ab+ bc+ ca)

on account of the well-known inequality a2 + b2 + c2 ≥ ab+ bc+ ca. Equality holdswhen 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

c3(a+ b)5

)1/5

≥3

2

Solution 1. To prove the inequality claimed we consider the function f : R → Rdefined by f(t) = t5. This function is convex in [0,+∞) as can be easily checked.Applying Jensen’s inequality, namely

3∑k=1

qkf(xk) ≥ f(

3∑k=1

qkxk

)

with

q1 =a2

a2 + b2 + c2, q2 =

b2

a2 + b2 + c2, q3 =

c2

a2 + b2 + c2

and

x1 =1

a(b+ c), x2 =

1

b(c+ a), x3 =

1

c(a+ b),

yields

1

a2 + b2 + c2

[a2

(1

a(b+ c)

)5

+ b2(

1

b(c+ a)

)5

+ c2(

1

c(a+ b)

)5]

≥

(ab+c

+ bc+a

+ ca+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

a+ b

)5

where we have used the constrain a2 + b2 + c2 = 1. Now, it is easy to see that

a

b+ c+

b

c+ a+

c

a+ b≥

3

2

13

Indeed, WLOG we can assume that a ≥ b ≥ c from which follows a+b ≥ a+c ≥ b+cand 1

b+c≥ 1

c+a≥ 1

a+b. Now, applying rearrangement inequality, we get

a

b+ c+

b

c+ a+

c

a+ b≥

a

a+ b+

b

b+ c+

c

c+ aa

b+ c+

b

c+ a+

c

a+ b≥

b

a+ b+

c

b+ c+

a

c+ a

Adding the preceding expressions, we obtain

a

b+ c+

b

c+ a+

c

a+ b≥

3

2(Nesbitt’s Inequality)

Equality holds when a = b = c.

Finally, substituting this result in the preceding the statement follows. Equalityholds when a = b = c =

√3/3, and we are done. �

Solution 2. We recall that Holder’s inequality can be stated as follows: Let a1, a2, ..., anand b1, b2, ..., bn, (n ≥ 2), be sequences of positive real numbers and let p, q ∈ R∗+ suchthat 1

p+ 1

q= 1, then the following inequality holds(

n∑i=1

api

) 1p(

n∑i=1

bqi

) 1q

≥n∑i=1

aibi

Particularizing the preceding result for n = 3 with p = 5/4, q = 5 for which (1/p +1/q = 1), we get (

3∑i=1

a5/4i

) 45(

3∑i=1

b5i

) 15

≥3∑i=1

aibi

Putting in the above result a1 = a8/5, a2 = b8/5, a3 = c8/5, b1 =1

a3/5(b+ c), b2 =

1

b3/5(c+ a), b3 =

1

c3/5(a+ b)yields

[(a8/5

)5/4+(b8/5

)5/4+(c8/5

)5/4]4/5 [( 1

a3/5(b + c)

)5

+

(1

b3/5(c + a)

)5

+

(1

c3/5(a + b)

)5]1/5

≥a

b+ c+

b

c+ a+

c

a+ b

or equivalently,

(a2+b2+c2)4/5(

1

a3(b+ c)5+

1

b3(c+ a)5+

1

c3(a+ b)5

)1/5

≥a

b+ c+

b

c+ a+

c

a+ b

from which the statement follows on account of the constrain and Nesbitt’s inequal-ity. Equality holds when when a = b = c =

√3/3, and we are done.

�

Solution 3 by Emilio Fernandez Moral. On account of mean inequalities we have(1

a3(b+ c)5+

1

b3(c+ a)5+

1

c3(a+ b)5

)1/5

≥(

33√a3b3c3(a+ b)5(b+ c)5(c+ a)5

)1/5

14

=5

√3

abc

13√(a+ b)(b+ c)(c+ a)

≥31/5+1/2

2 (abc)1/5≥

3

2

because 3√(a+ b)(b+ c)(c+ a) ≤

(a+ b) + (b+ c) + (c+ a)

3=

2(a+ b+ c)

3≤

2

√1

3and (abc)1/5 = (a2b2c2)1/10 =

(3√a2b2c2

)3/10≤(a2 + b2 + c2

3

)3/10

=

(1

3

)3/10

and 31/5+1/2+3/10 = 3. Equality holds when a = b = c =√3/3. �

15

Jose Luis Dıaz-BarreroRSME Olympic CommitteeUPC–BARCELONA TECHjose.luis.diaz@upc.edu

 

Some ResultsHereafter, 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) =

(aα1 + aα2 + . . .+ aαn

n

)1/α

is nondecreasing and the following limits hold

limα→−∞

f(α) = min1≤i≤n

{ai

}, lim

α→0f(α) = n

√a1a2 . . . an, lim

α→+∞f(α) = max

1≤i≤n

{ai

}

Proof. Let 0 < a < b. Consider the function g : [0,+∞)→ R defined by g(α) = αb/a.Since

g′′(α) =b

a

(b

a− 1

)αb/a−2 ≥ 0, (α ≥ 0),

then g is convex in [0,+∞). Applying Jensen’s inequality, we get

g

1

n

n∑j=1

αaj

≤ 1

n

n∑j=1

g(αaj )

or 1

n

n∑j=1

αaj

b/a ≤ 1

n

n∑j=1

αbj

from which follows f(a) ≤ f(b). For negative values of α consider the functionh(α) = (−α)b/a which is convex in (−∞, 0).

Let L1 = limα→−∞

f(α). Then, lnL1 = limα→−∞

ln

(aα1 + aα2 + . . .+ aαn

n

)1/α

. That is,

lnL1 = limα→−∞

[1

αln

(aα1 + aα2 + . . .+ aαn

n

)]= lim

α→−∞

aα1 ln a1 + aα2 ln a2 + . . .+ aαn ln an

aα1 + aα2 + . . .+ aαn

= ln

(min

1≤i≤n

{ai

})

16

from which follows L1 = min1≤i≤n

{ai

}. Likewise, we obtain that L2 = lim

α→+∞f(α) =

max1≤i≤n

{ai

}. Finally, denoting by L = lim

α→0f(α), we have

lnL = limα→0

[1

αln

(aα1 + aα2 + . . .+ aαn

n

)]= lim

α→0

aα1 ln a1 + aα2 ln a2 + . . .+ aαn ln an

aα1 + aα2 + . . .+ aαn

=ln a1 + ln a2 + . . .+ ln an

n= ln n

√a1a2 . . . an

from which immediately follows limα→0

f(α) = n√a1a2 . . . an and the proof is complete.

�In 1821 Cauchy published his famous inequality as the second of the two noteson the theory of inequalities that formed the final part of his book Cours d’AnalyseAlgebrique. Namely,

Theorem 2 Let a1, a2, ..., an, b1, b2, ..., bn be any real numbers. Then, the followinginequality holds: (

n∑k=1

akbk

)2

≤(

n∑k=1

a2k

)(n∑k=1

b2k

)

Proof. Consider the quadratic polynomial A(x) =

n∑k=1

(akx− bk)2. Then,

A(x) = x2n∑k=1

a2k − 2x

n∑k=1

akbk +

n∑k=1

b2k ≥ 0

Since the preceding equation is nonnegative its discriminant must be less or equalthan zero. That is, (

n∑k=1

akbk

)2

−(

n∑k=1

a2k

)(n∑k=1

b2k

)≤ 0

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 Holder. It is statedand proved in the following theorem.

Theorem 3 Let a1, a2, ..., an and b1, b2, ..., bn, (n ≥ 2) be sequences of positive realnumbers and let p, q ∈ R∗+ such that 1

p+ 1

q= 1, then the following inequality holds(

n∑i=1

api

) 1p(

n∑i=1

bqi

) 1q

≥n∑i=1

aibi.

Proof 1. We will argue by induction. Assume that the inequality holds for n and wehave to prove it for n+ 1. In fact,

n+1∑i=1

aibi =

n∑i=1

aibi + an+1bn+1 ≤(

n∑i=1

api

) 1p(

n∑i=1

bqi

) 1q

+ an+1bn+1

17

≤(

n∑i=1

api + apn+1

) 1p(

n∑i=1

bqi + bqn+1

) 1q

=

(n+1∑i=1

api

) 1p(n+1∑i=1

bqi

) 1q

.

To complete the inductive process, we must verify that the case when n = 2 alsoholds. Indeed, the inequality

a1b1 + a2b2 ≤ (ap1 + ap2)1p (bq1 + bq2)

1q ,

immediate follows from the fact that the function

f(x) = (ap1 + ap2)1p (bq1 + xq)

1q − a1b1 − a2x,

is strictly positive for all x > 0. This completes the proof.

Proof 2. First, we write the inequality in the most convenient form∑ni=1 aibi(∑n

i=1 api

) 1p(∑n

i=1 bqi

) 1q

≤ 1

or equivalently,n∑i=1

(api∑ni=1 a

pi

) 1p(

bqi∑ni=1 b

qi

) 1q

≤ 1.

Now, using the power mean inequality,

x1αy

1β ≤

1

αx+

1

βy,

we have

n∑i=1

(api∑ni=1 a

pi

) 1p(

bqi∑ni=1 b

qi

) 1q

≤n∑i=1

[1

p

api∑ni=1 a

pi

+1

q

bqi∑ni=1 b

qi

]

=1

p

∑ni=1 a

pi∑n

i=1 api

+1

q

∑ni=1 b

qi∑n

i=1 bqi

=1

p+

1

q= 1,

as desired. Equality holds if and only if the n-tuples (ap1, ap2, ..., a

pn) and (bq1, b

q2, ..., b

qn)

are proportional. �Notice that for p = q = 2, we get the inequality

(a1b1 + a2b2 + ...+ anbn)2 ≤ (a2

1 + a22 + ...+ a2

n)(b21 + b22 + ...+ b2n)

This is the Cauchy-Bunyakowski-Schwarz inequality.

Using Holder’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 realnumbers and let p > 1, be a real number, then(

n∑i=1

(ai + bi)p

) 1p

≤(

n∑i=1

api

) 1p

+

(n∑i=1

bpi

) 1p

When p < 1 inequality reverses.

18

Proof. Let be q = pp−1

, then 1p+ 1

q= 1, and applying Holder’s inequality, we obtain

n∑i=1

(ai + bi)p =

n∑i=1

ai(ai + bi)p−1 +

n∑i=1

bi(ai + bi)p−1

≤(

n∑i=1

api

) 1p(

n∑i=1

(ai + bi)(p−1)q

) 1q

+

(n∑i=1

bpi

) 1p(

n∑i=1

(ai + bi)(p−1)q

) 1q

=

( n∑i=1

api

) 1p

+

(n∑i=1

bpi

) 1p

( n∑i=1

(ai + bi)p

)p−1p

Multiplying the preceding inequality by(∑n

i=1(ai + bi)p)−p−1

p , we get(n∑i=1

(ai + bi)p

) 1p

≤(

n∑i=1

api

) 1p

+

(n∑i=1

bpi

) 1p

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 ofpositive real numbers and let c1, c2, · · · , cn be a permutation of b1, b2, · · · , bn. Thesum S = a1b1 + a2b2 + · · ·+ anbn is maximal if the two sequences a1, a2, · · · , an andb1, b2, · · · , bn are sorted in the same way and minimal if the two sequences are sortedoppositely, one increasing and the other decreasing.

Proof. Let ai > aj. Consider the sums

S1 = a1c1 + · · ·+ aici + · · ·+ ajcj + · · ·+ ancn

S2 = a1c1 + · · ·+ aicj + · · ·+ ajci + · · ·+ ancn

We have obtained S2 from S1 by switching the positions of ci and cj. Then

S1 − S2 = aici + ajcj − aicj − ajci = (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 ine-quality 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 nonnegativereal numbers such that q1 + q2 + . . .+ qn = 1. Then for all ak ∈ I (1 ≤ k ≤ n) holds

f

(n∑k=1

qkak

)≤

n∑k=1

qkf(ak)

Equality holds when a1 = a2 = . . . = an. If f is concave the preceding inequalityreverses.

Proof. We will argue by mathematical induction. For n = 2, we have

f(q1a1 + q2a2) ≤ q1f(a1) + q2f(a2)

The preceding inequality holds because f is convex and q1 + q2 = 1. Assume that

f

(n∑k=1

qkak

)≤

n∑k=1

qkf(ak)

and we have to see that

f

(n+1∑k=1

qkak

)≤

n+1∑k=1

qkf(ak)

when q1 + q2 + . . . + qn + qn+1 = 1. Let q =∑nk=1 qk. Then, we have

n∑k=1

qk

q= 1,

n∑k=1

(qk

q

)ak ∈ I and q + qn+1 = 1. Now, taking into account the case when n = 2

and the inductive hypothesis, yields

f

(n+1∑k=1

qkak

)= f

[q

(n∑k=1

(qk

q

)ak

)+ qn+1an+1

]

≤ qf

(n∑k=1

(qk

q

)ak

)+ qn+1f(an+1)

≤ q

n∑k=1

(qk

q

)f(ak) + qn+1f(an+1)

=

n+1∑k=1

qkf(ak)

and by the principle of mathematical induction (PMI) the statement is proven. Equa-lity holds when a1 = a2 = . . . = an.

�

20

Think about

“Pulchra dicuntur quae visa placent– beauty is that which, be-ing seen, pleases”. This definition, applies well to Mathematicalbeauty in which lack of understanding is so often responsible oflack of pleasure.

Thomas Aquina’s definition of Beauty