USING SIMPLE INEQUALITIES TO SOLVE SOME BEAUTIFUL PROBLEMS

[3] Vasile Cirtoaje, Vo Quoc Ba Can, Tran Quoc Anh, Inequalities with Beautiful Solutions , GIL Publishing house, 2009. [4] Pham Kim Hung, Secrets in Inequality , GIL Publishing house, 2[r]

USING SIMPLE INEQUALITIES TO SOLVE SOME

BEAUTIFUL PROBLEMS

Nguyen Dang Khoa - Hung Vuong High school for the gifted - Phu Tho

March 2019

1 Introduction

In the inequality world, there are many simple but effective inequalities, which are very useful for some cases Today, I bring the readers an inequality, and I think it is helpful for you in the future Although I wrote this article very carefully, limited knowledge and language description of the author can lead to some unavoidable mistakes Remember your feedbacks and recommendations are always welcomed, so don’t be hesitant!

Now, let’s start to find incredible things!

First, we have a basic inequality

• For any non-negative real numbers a, b, c, we obtain

a2

a2+ 2bc +

b2

b2+ 2ac +

c2

c2+ 2ab ≥ 1 (∗)

It is simple inequality and we know solution by applying Cauchy - Schwarz inequality

a2

a2+ 2bc +

b2

b2+ 2ac +

c2

c2+ 2ab ≥ (a + b + c)

2

a2+ 2bc + b2+ 2ac + c2+ 2ab =

(a + b + c)2 (a + b + c)2 = 1

We can rewrite that inequality as the following form

bc

a2+ 2bc +

ca

b2+ 2ac +

ab

c2+ 2ab ≤ 1 The equality holds if and only if a = b = c or one of these numbers a, b, c must be equal to 0

Now, We replace (a, b, c) −→ 1

a,

1

b,

1

c and we have the inequality (∗) becomes

• For any non-negative real numbers a, b, c, we have the inequality

bc 2a2+ bc +

ca 2b2+ ca +

ab 2c2+ ab ≥ 1 (∗∗)

or

a2

2a2+ bc+

b2

2b2+ ca +

c2

2c2+ ab ≤ 1 The inequality (∗∗) is also proved by Cauchy - Schwarz inequality, but the thing which author want to show you is relation between two inequalities (∗) and (∗∗)

From those inequalities, we will discover, extend and do some beautiful problems

2 Examples and problems

First, we can change the inequality (∗) is equivalent to

a2

a2+ 2bc +

b2

b2+ 2ac +

c2

c2+ 2ab ≥ 1 or a

3

a3+ 2abc+

b3

b3+ 2abc +

c3

c3+ 2abc ≥ 1

We see denominators of those fractions have 2abc, if we let abc = 1 then we have inequality is equivalent to

a3

a3+ 2 +

b3

b3 + 2 +

c3

c3+ 2 ≥ 1 or

1

a3+ 2 +

1

b3 + 2 +

1

c3+ 2 ≤ 1

So we have example 1

Example 1 Given positive real numbers a, b, c such that abc = 1 Prove that

1

a3+ 2 +

1

b3 + 2 +

1

c3+ 2 ≤ 1 Since abc = 1 then we also have inequality

1

a + 2+

1

b + 2+

1

c + 2 ≤ 1

We can prove this inequality by another way

Proof Sine abc = 1 then there exists x, y, z are positive such that a = x

2

yz, b =

y2

zx, c =

z2

xy.

So the inequality becomes

yz

x2+ 2yz +

zx

y2+ 2zx +

xy

z2+ 2xy ≤ 1

It is inequality (∗), so we are done

Example 2 Given positive real numbers a, b, c satisfying abc = 1 Show that

1 2a + 1 +

1 2b + 1 +

1 2c + 1 ≥ 1 Proof We will solve this example like example 1

Let a = x

2

yz, b =

y2

zx, c =

z2

xy (x, y, z > 0).

The inequality becomes

yz 2x2+ yz +

zx 2y2+ zx+

xy 2z2 + xy ≥ 1 This is inequality (∗∗), we are done

Now, combining example 1 and example 2, we have a new example

Example 3 Let a, b, c are positive real numbers satisfying abc = 1 Show that

1 2a + 1 +

1 2b + 1+

1 2c + 1 ≥ 1

a + 2+

1

b + 2+

1

c + 2 and

a

a + 2 +

b

b + 2 +

c

c + 2 ≥ a

2a + 1 +

b 2b + 1 +

c 2c + 1 Example 4 (Nguyen Dang Khoa) Given positive real numbers a, b, c such that a2+b2+c2 = 1 Prove the inequality

 a

a2+ 2bc

2

+

 b

b2+ 2ca

2

+

 c

c2+ 2ab

2

≥ 1

We can see this inequality has square of fraction so we always think use inequality

3(a2+ b2+ c2) ≥ (a + b + c)2 But if we use it, we have



a

a2+ 2bc

2

+

 b

b2+ 2ca

2

+

 c

c2 + 2ab

2

≥

 a

a2+ 2bc +

b

b2+ 2ca +

c

c2+ 2ab

2

3

So it is enough to show that a

a2+ 2bc +

b

b2+ 2ca +

c

c2+ 2ab ≥√3

But it is hard inequality, we can use S.O.S method to prove it Is that the author mean? Now, let consider a beautiful solution

Proof Applying Cauchy-Schwarz inequality and inequality (∗) we have

"



a

a2+ 2bc

2

+

 b

b2+ 2ca

2

+

 c

c2+ 2ab

2# (a2+ b2+ c2)

≥



a2

a2+ 2bc +

b2

b2+ 2ac +

c2

c2+ 2ab

2

≥ 1 Combining assumption a2+ b2+ c2 = 1, we get

 a

a2+ 2bc

2

+

 b

b2+ 2ca

2

+

 c

c2+ 2ab

2

≥ 1 With the same idea, we have another inequality

a2

(a2+ 2bc)3 + b

2

(b2+ 2ac)3 + c

2

(c2+ 2ab)3 ≥ 1

We can use Holder inequality to prove it

We have



a2

(a2+ 2bc)3 + b

2

(b2+ 2ac)3 + c

2

(c2 + 2ab)3

 (a2+ b2+ c2)(a2+ b2 + c2)

≥



a2

a2+ 2bc +

b2

b2+ 2ac +

c2

c2+ 2ab

3

≥ 1

It implies a

2

(a2+ 2bc)3 + b

2

(b2+ 2ac)3 + c

2

(c2+ 2ab)3 ≥ 1 From that examples, we have the general problem

General problem Given positive real numbers a, b, c such that a2 + b2 + c2 = 1 and k is positive integer Prove that

a2

(a2+ 2bc)k + b

2

(b2+ 2ca)k + c

2

(c2+ 2ab)k ≥ 1 Proof Note that a2+ 2bc ≤ a2+ b2+ c2 = 1, it implies (a2+ 2bc)k≤ a2 + 2bc

So we obtain

a2

(a2+ 2bc)k + b

2

(b2+ 2ca)k + c

2

(c2+ 2ab)k ≥ a

2

a2+ 2bc +

b2

b2+ 2ca +

c2

c2+ 2ab ≥ 1

This solution is very simple, it shows that for some cases general problem is not harder than original problem

Example 5 (Nguyen Dang Khoa) Given the positive real numbers a, b, c have the sum is 1 Show that

a 3bc + a+

b 3ca + b+

c 3ab + c ≥ 3

2 First, we have assumption is a + b + c = 1 so we have idea is multiple a + b + c or use directly Cauchy - Schwarz inequality But this method is not really useful So we need more skillful in some steps

Proof We rewrite the inequality in the form

bc 3bc + a+

ca 3ca + b+

ab 3ab + c ≤ 1

2 Here, we can use a + b + c = 1 and apply Cauchy - Schwarz inequality, we have

bc

3bc + a =

bc 3bc + a(a + b + c) =

bc (a2+ 2bc) + (ab + bc + ca) ≤ 1

4

 bc

a2+ 2bc +

bc

ab + bc + ca



Simillarly, we constitute the different inequalities and we get

bc 3bc + a+

ca 3ca + b+

ab 3ab + c ≤ 1

4



X ab 2ab + c2 +X ab

ab + bc + ca



≤ 1 2

So we are done The equality occurs if a = b = c = 1

3. Example 6 (Nguyen Dang Khoa) Prove this following inequality with a, b, c are non-negative real numbers

bc 2a2 + (b + c)2 + ca

2b2+ (c + a)2 + ab

2c2 + (a + b)2 ≤ 1

2 Proof Applying Cauchy - Schwarz inequality, we obtain

bc

2a2+ (b + c)2 = bc

2a2+ b2+ 2bc + c2 = bc

a2+ 2bc + (a2+ b2+ c2) ≤ 1

4

 bc

a2+ 2bc +

bc

a2+ b2+ c2



We constitute other inequalities, we have

X bc

2a2 + (b + c)2 ≤ 1

4

 bc

a2+ 2bc +

ca

b2+ 2ca +

ab

c2+ 2ab +

ab + bc + ca

a2+ b2+ c2



It is easy to see that ab + bc + ca

a2+ b2+ c2 ≤ 1 and we use the inequality (∗), i.e

bc 2a2 + (b + c)2 + ca

2b2+ (c + a)2 + ab

2c2 + (a + b)2 ≤ 1

2 The equality holds if and only if a = b = c or a = 0, b = c or b = 0, a = c or c = 0, a = b

Now, we change that inequality in the form

2a2+ b2+ c2 2a2+ (b + c)2 + 2b

2+ a2+ c2 2b2+ (b + c)2 + 2c

2+ a2+ b2 2c2+ (a + b)2 ≥ 2 Let a2+ b2+ c2 = 1 then we have a new inequality

Example 6.1 Given the positive real number a , b, c sastisfying a2+ b2+ c2 = 1 Prove that

a2+ 1

a2 + 2bc + 1+

b2+ 1

b2+ 2ca + 1 +

c2+ 1

c2+ 2ab + 1 ≥ 2 Example 7 (Nguyen Dang Khoa) Prove the following inequality with a, b, c are positive real numbers

a2 + bc

a2+ 2bc +

b2+ ac

b2+ 2ac +

c2+ ab

c2+ 2ab ≥ 2 Proof The original inequality is eqivalent to

2(a2+ bc)

a2+ 2bc +

2(b2+ ac)

b2+ 2ac +

2(c2+ ab)

c2+ 2ab ≥ 4 or

2(a2+ bc)

a2+ 2bc − 1 + 2(b

2+ ac)

b2+ 2ac − 1 + 2(c

2+ ab)

c2+ 2ab − 1 ≥ 1 or

a2

a2+ 2bc +

b2

b2+ 2ca +

c2

c2+ 2ab ≥ 1 This is inequality (∗), so that inequality has proved, as desire

Example 8 Prove that, for all non-negative real numbers a, b, c, we have an inequality

a2 + bc 2a2+ bc+

b2+ ca 2b2+ ca +

c2+ ab 2c2+ ab ≥ 2

Proof This inequality is equivalent to

2a2+ 2bc 2a2+ bc +

2b2+ 2ca 2b2+ ca +

2c2 + 2ab 2c2+ ab ≥ 4 or

2a2+ 2bc 2a2+ bc − 1 + 2b

2+ 2ca 2b2+ ca − 1 + 2c

2+ 2ab 2c2+ ab − 1 ≥ 1 or

bc 2a2+ bc+

ca 2b2+ ca +

ab 2c2+ ab ≥ 1 This is inequality (∗∗), so we are done

Example 9 (Nguyen Dang Khoa) Given non-negative numbers a, b, c Show that

a2 8a2 + (b + c)2 + b

2

8b2+ (c + a)2 + c

2

8c2 + (a + b)2 ≤ 1

4 Proof Applying AM - GM inequality and using inequality (∗∗), we have

a2

8a2+ (b + c)2 + b

2

8b2+ (c + a)2 + c

2

8c2 + (a + b)2 ≤X a

2

8a2+ 4bc =

1 4

X a2

2a2+ bc ≤ 1

4

So we are done

The equality occurs if and only if a = b = c or two numbers of a, b, c are equal to 0 Example 10 (Vasile Cirtoaje) Given three non-negative real numbers satisfying

ab + bc + ca = 3 Prove that

1

a2 + 1 +

1

b2+ 1 +

1

c2+ 1 ≥ 3

2

Proof Since 1

a2+ 1 = 1 −

a2

a2 + 1 the we can rewrite inequality in the form

a2

a2 + 1 +

b2

b2+ 1 +

c2

c2+ 1 ≤ 3

2 or

a2

3a2+ 3 +

b2

3b2+ 3 +

c2

3c2+ 3 ≤ 1

2 or

4a2 3a2 + ab + bc + ca +

4b2 3b2+ ab + bc + ca +

4c2 3c2+ ab + bc + ca ≤ 2 Using Cauchy - Schwarz inequality, we have

4a2

= (a + a)

2

≤ a

2

+ a

2

X 4a2

3a2+ ab + bc + ca ≤X a

a + b + c +

X a2

2a2 + bc ≤ 2 The equality holds if and only if a = b = c or a = b; c = 0 and cyclic

Example 11 (Tigran Sloyan) Given a, b, c are positive real numbers Prove that

a2 (2a + b)(2a + c)+

b2 (2b + c)(2b + a)+

c2 (2c + a)(2c + b) ≤ 1

3 Proof Applying Cauchy - Schwarz inequality, we obtain

9a2

(2a + b)(2a + c) =

(2a + a)2

2a(a + b + c) + (2a2+ bc) ≤ 2a

a + b + c +

a2

2a2+ bc

We write two similar inequalities and then add up all these relations We will find that

X 9a2

(2a + b)(2a + c) ≤X 2a

a + b + c +

X a2

2a2+ bc ≤ 3 Hence, we are done

Example 12 (Vo Quoc Ba Can) Given positive real numbers a, b, c Show that

a2

5a2 + (b + c)2 + b

2

5b2+ (a + c)2 + c

2

5c2 + (a + b)2 ≥ 1

3 Proof Using Cauchy - Schwarz inequality, we get

9a2

5a2+ (b + c)2 = (2a + a)

2

(a2+ b2+ c2) + 2(2a2+ bc) ≤ a

2

a2+ b2+ c2 + 2a

2

2a2+ bc Using that inequality and inequality (∗∗), we have

9X a

2

5a2+ (b + c)2 ≤X a

2

a2+ b2 + c2 + 2X a

2

2a2+ bc ≤ 3 Therefore, the initial inequality is proved

Example 13 (Nguyen Dang Khoa) Given non-negative real numbers a, b, c satisfying

a + b + c = 2 Prove that

a2

4a2 + 2 + bc+

b2

4b2+ 2 + ca +

c2

4c2 + 2 + ab ≤ 1

3

Proof The initial inequality is equivalent to

9a2 8a2+ 4 + 2bc +

9b2 8b2+ 4 + 2ca+

9c2 8c2+ 4 + 2ab ≤ 3

2 or

X 9a2 8a2+ (a + b + c)2+ 2bc ≤ 3

2 Note that, we have (a + b + c)2 ≥ max {4a(b + c), 4b(c + a), 4c(a + b)}

So we need to show that

X 9a2

8a2+ 4a(b + c) + 2bc ≤ 3

2 Now, by using Cauchy - Schwarz, we get

9a2 8a2 + 4a(b + c) + 2bc =

(2a + a)2 4a(a + b + c) + 2(2a2+ bc) ≤ a

a + b + c +

a2 2(a2+ 2bc)

It implies

X 9a2 8a2+ 4 + 2bc ≤X a

a + b + c +

1 2

X a2

a2+ 2bc ≤ 3

2 The equality holds if and only if b = c = 1, a = 0 and cyclic

Example 14 Given a, b, c are positive real numbers have the sum is 1 Prove that

ab 3ab + 2b + c+

bc 3bc + 2c + a+

ca 3ca + 2a + b ≤ 1

4 Proof Observe that

3ab + 2b + c = 3ab + c(a + b + c) + 2b = (ab + bc + ca) + (2ab + c2) + 2b Using Cauchy - Schwarz inequality, we have

(1 + 1 + 2)2

3ab + 2b + c ≤ 1

ab + bc + ca+

1 2ab + c2 + 2

b Therefore,

16X ab

3ab + 2b + c ≤X



ab

ab + bc + ca+

ab 2ab + c2 + 2a



= 3 +X ab

2ab + c2 ≤ 4

So we are done

2.2 Practice problems

Problem 1 (Nguyen Dang Khoa) Given positive real numbers a, b, c such that abc = 1 Prove that

(a + 1)2

a + 2 +

(b + 1)2

b + 2 +

(c + 1)2

c + 2 ≤ 4

3(a + b + c) Problem 2 Prove that, for all positive real numbers a, b, c such that a + b + c = 1, we have

ab 3ab + 2b + c+

bc 3bc + 2c + a+

ca 3ca + 2a + b ≤ 1

4 Problem 3 Given non-negative a, b, c satifying a + b + c = 1 Show that

1 6a2+ 1 +

1 6b2+ 1 +

1 6c2+ 1 ≥ 9

5 Problem 4 Prove if a, b, c > 0 and a + b + c = 3 then

a

a + bc+

b

b + ca +

c

c + ab ≥ 3

2 Problem 5 Given non-negative real number a, b, c such that a + b + c > 0 Prove that

a2

3a2 + (b + c)2 + b

2

3b2+ (c + a)2 + c

2

3c2 + (a + b)2 ≤ 1

2 Problem 6 Given a, b, c are positive real numbers Prove the following inequality

ab

a + 3b + 2c +

bc

b + 3c + 2a +

ca

c + 3a + 2b ≤ a + b + c

6

3 Reference

[1] Vasile Cirtoaje, Algebraic inequality: Old and new methods, GIL Publishing house, 2004 [2] Vo Quoc Ba Can, Tran Quoc Anh, Using Cauchy - Schwarz method to prove inequalities, NXBDPSP

[3] Vasile Cirtoaje, Vo Quoc Ba Can, Tran Quoc Anh, Inequalities with Beautiful Solutions, GIL Publishing house, 2009

[4] Pham Kim Hung, Secrets in Inequality, GIL Publishing house, 2007

[5] J.Michael Steele, The Cauchy - Schwarz Master Class, Cambridge University Press, 2004 [6] Mathematical Reflections

[7] AoPS Inequality forum