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DIỄN ĐÀN TOÁN HỌC VIỆT NAMwww.maths.vn

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Tuyển tập

Bất đẳng thứ cVolume 1

Biên t ậ p: Võ Quốc Bá Cẩn Tác gi ả các bài toán: Trần Quốc LuậtThành viên tham gia gi ải bài:

1. Võ Quốc Bá Cẩn (nothing )2. NgôĐứ c Lộc (Honey_suck )

3. Trần Quốc Anh (nhocnhoc )4. Seasky

5. Materazzi


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Tran Quoc Luat's Inequalities

Vo Quoc Ba Can - Pham Thi Hang

February 25, 2009


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ii

Copyright c 2008 by Vo Quoc Ba Can


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Preface

"Life is good for only two things, discovering mathematics and teaching mathematics."

S. Poisson

Bat dang thuc la mot trong linh vuc hay va kho. Hien nay, co kha nhieu nguoi quan tam den no boi no thucsu rat don gian, quyen ru va ban khong can phai "hoc vet" nhieu dinh ly de co the giai duoc chung. Khi hoc bat dang thuc, hai dieu cuon hut chung ta nhat chinh la sang tao va giai bat dang thuc. Nham muc dich kichthich su sang tao cua hoc sinh sinh vien nuoc nha, dien dan mathsvn da co mot so topic sang tao bat dangthuc danh rieng cho cac ca nhan tren dien dan. Tuy nhien, cac topic do con roi rac nen ta can mot su tonghop lai thong nhat hon de cho ban doc tien theo doi, do la li do ra doi cua quyen sach nay. Quyen sach duoctrinh bay trong phan chinh bang tieng Anh voi muc dich giup chung ta ren luyen them ngoai ngu va co thegioi thieu no den cac ban trong va ngoai nuoc. Mac du da co gang bien soan nhung sai sot la dieu khong thetranh khoi, rat mong nhan duoc su gop y cua ban doc gan xa. Moi su dong gop y kien xin duoc gui ve tacgia theo: [email protected]. Xin chan than cam on!

Quyen sach nay duoc thuc hien vi much dich giao duc, moi viec mua ban trao doi thuong mai tren quyensach nay deu bi cam neu nhu khong co su cho phep cua tac gia.

Vo Quoc Ba Can

iii


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iv Preface


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Chapter 1

Problems

"Each problem that I solved became a rule, which served afterwards to solve other problems."

R. Descartes

1. Given a triangle ABC with the perimeter is 2 p: Prove that the following inequality holds

a p a

+ b

p b +

c p c s b + c p a + r c + a p b + s a + b p c :

2. Let a; b; c be nonnegative real numbers such that a2 + b2 + c2 + abc = 4: Prove that the followinginequality holds

a2 + b2 + c2 a2b2 + b2c2 + c2a2 :

3. Show that for any positive real numbers a ; b; c; we have

a3 + b3 + c3 + 6abc 3p abc (a + b + c)2 :4. Let a ; b; c be nonnegative real numbers with sum 1 : Determine the maximum and minimum values of

P (a ; b; c) = ( 1 + ab )2 +( 1 + bc)2 +( 1 + ca )2 :

5. Let a ; b; c be nonnegative real numbers with sum 1 : Determine the maximum and minimum values of

P (a ; b; c) = ( 1 4ab )2 +( 1 4bc)2 + ( 1 4ca )2 :

6. Let a ; b; c be positive real numbers. Prove that

b + ca

+ c + a

b +

a + bc

2

4(ab + bc + ca ) 1a2

+ 1b2

+ 1c2

:

1


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2 Problems

7. Let a ; b; c be the side of a triangle. Show that

∑cyc

(a + b)(a + c)p b + c a 4(a + b + c)p (b + c a)(c + a b)(a + b c):8. Given a triangle with sides a ; b; c satisfying a2 + b2 + c2 = 3: Show that

a + bp a + b c +

b + cp b + c a +

c + ap c + a b 6:

9. Given a triangle with sides a ; b; c satisfying a2 + b2 + c2 = 3: Show that

ap b + c a +

bp c + a b +

cp a + b c 3:

10. Show that if a ; b; c are positive real numbers, then

a

a + b +

b

b + c +

c

c + a 1

+ s 2abc

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

11. Show that if a ; b; c are positive real numbers, then

aa + b

2

+ b

b + c

2

+ c

c + a

2 34

+ a2b + b2c + c2a 3abc

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


(a2 + b2)(b2 + c2)(c2 + a2)8a2b2c2

a2 + b2 + c2

ab + bc + ca

2

:

13. Let a ; b; c be positive real numbers. Prove the inequality

(b + c)2

a(b + c + 2a) +

(c + a)2

b(c + a + 2b) +

(a + b)2

c(a + b + 2c) 3:

14. Let a ; b; c be positive real numbers. Prove the inequality

(b + c)2

a(b + c + 2a) +

(c + a)2

b(c + a + 2b) +

(a + b)2

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

b + c + 2a +

c + ac + a + 2b

+ a + b

a + b + 2c:


(b + c)2

a(b + c + 2a) +

(c + a)2

b(c + a + 2b) +

(a + b)2

c(a + b + 2c) 2 a

b + c +

bc + a

+ ca + b

:


a3b3 + b3c3 + c3a3 (b + c a)(c + a b)(a + b c)(a3 + b3 + c3):


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3

17. If a ; b; c are positive real numbers such that abc = 1; show that we have the following inequality

a3 + b3 + c3 ab + c

+ bc + a

+ ca + b

+ 32

:

18. Given nonnegative real numbers a ; b; c such that ab + bc + ca + abc = 4: Prove that

a2 + b2 + c2 + 2(a + b + c) + 3abc 4(ab + bc + ca ):

19. Let a ; b; c be real numbers with min fa ; b; cg 34 and ab + bc + ca = 3: Prove thata3 + b3 + c3 + 9abc 12 :

20. Let a ; b; c be positive real numbers such that a2b2 + b2c2 + c2a2 = 1: Prove that

(a2 + b2 + c2)2 + abcq (a2 + b2 + c2)3 4:21. Show that if a ; b; c are positive real numbers, the following inequality holds

(a + b + c)2(ab + bc + ca )2 +( ab + bc + ca )3 4abc (a + b + c)3 :

22. Let a ; b; c be real numbers from the interval [3; 4]: Prove that

(a + b + c)abc

+ bca

+ cab

3(a2 + b2 + c2):

23. Given ABC is a triangle. Prove that

8cos 2 Acos 2 Bcos 2 C + cos2 Acos2 Bcos2 C 0:

24. Let a ; b; c be positive real numbers such that a + b + c = 3 and ab + bc + ca 2max fab ; bc ; cag: Provethat

a2 + b2 + c2 a2b2 + b2c2 + c2a2 :


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4 Problems


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Chapter 2

Solutions

"Don't just read it; ght it! Ask your own questions, look for your own examples, discover your

own proofs. Is the hypothesis necessary? Is the converse true? What happens in the classical special case? What about the degenerate cases? Where does the proof use the hypothesis?"

P. Halmos , I Want to be a Mathematician

Problem 2.1 Given a triangle ABC with the perimeter is 2 p: Prove that the following inequality holds

a p a

+ b

p b +

c p c s b + c p a + r c + a p b + s a + b p c :

Solution. Setting x = p a ; y = p b and z = p c; then a = y + z ; b = z + x and c = x + y: The originalinequality becomes

y+ z x

+ z + x

y +

x+ y z r 2 + y+ z x + r 2 + z + x y + r 2 + x + y z :

By AM-GM Inequality , we have

4r 2 + y+ z x 2 + y+ z x + 4 = y+ z x + 6:It follows that

4

r 2 +

y+ z x

+

r 2 +

z + x y

+

r 2 +

x+ y z ! y+ z x + z + x y + x + y z + 18 :

We have to prove

4 y+ z

x +

z + x y

+ x + y

z y+ z

x +

z + x y

+ x+ y

z + 18 ;

5


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6 Solutions

or equivalently, y+ z

x +

z + x y

+ x+ y

z 6;which is obviously true by AM-GM Inequality .

Equality holds if and only if a = b = c:

Problem 2.2 Let a; b; c be nonnegative real numbers such that a 2 + b2 + c2 + abc = 4: Prove that the fol-lowing inequality holds

a2 + b2 + c2 a2b2 + b2c2 + c2a2 :

Solution. From the given condition, we see that there exist x; y; z 0 such that

a = 2 x

p ( x+ y)( x+ z ); b =

2 y

p ( y+ z )( y+ x); c =

2 z

p ( z + x)( z + y):

Using this substitution, we may write our inequality as

∑cyc

x2

( x+ y)( x+ z ) ∑cyc

4 y2 z 2

( y+ z )2( x+ y)( x+ z )

;

which is obviously true because

∑cyc

4 y2 z 2

( y+ z )2( x+ y)( x+ z ) ∑cyc yz ( y+ z )2

( y+ z )2( x+ y)( x+ z ) = ∑

cyc

yz ( x+ y)( x+ z )

;

and

∑cyc

x2

( x+ y)( x+ z ) = ∑

cyc

yz ( x+ y)( x+ z )

:

Our proof is completed. Equality holds if and only if a = b = c = 1 or a = b = p 2; c = 0 and its cyclic permutations.

Problem 2.3 Show that for any positive real numbers a ; b; c; we havea3 + b3 + c3 + 6abc 3p abc (a + b + c)2 :

Solution 1. According to the AM-GM Inequality , we have the following estimation

6 3p abc (a + b + c)2 (a + b + c)3 + 9 3p a2b2c2(a + b + c);which leads us to prove the sharper inequality

6(a3 + b3 + c3 + 6abc ) (a + b + c)3 + 9 3p a2b2c2(a + b + c);or

5(a3 + b3 + c3) 3∑cyc

ab (a + b) + 30abc 9 3p a2b2c2(a + b + c);From Schur's Inequality for third degree, we have

3(a3 + b3 + c3) 3∑cyc

ab (a + b)+ 9abc 0;


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7

and we deduce our inequality to

2(a3 + b3 + c3) + 21abc 9 3p a2b2c2(a + b + c);Again, the Schur's Inequality for third degree shows that

4(a3 + b3 + c3) + 15abc (a + b + c)3 ;

and with this inequality, we nally come up with

(a + b + c)3 15abc + 42abc 18 3p a2b2c2(a + b + c);

(a + b + c)3 + 27abc 18 3p a2b2c2(a + b + c);By AM-GM Inequality , we have that

2(a + b + c)3 + 54abc = ( a + b + c)3 + (a + b + c)3 + 27abc + 27abc

(a + b + c)3 + 27 3p a2b2c2(a + b + c) 9

3p a2b2c2(a + b + c)+ 27 3p a2b2c2(a + b + c)= 36 3p a2b2c2(a + b + c):

It shows that(a + b + c)3 + 27abc 18

3p a2b2c2(a + b + c);which completes our proof. Equality holds if and only if a = b = c: Solution 2 (by Seasky). Since the inequality being homogeneous, we can suppose without loss of generalitythat abc = 1: In this case, the inequality can be rewitten in the form

P (a ; b; c) = a3 + b3 + c3 + 6 (a + b + c)2 0:

We will now use mixing variables method to solve this inequality. Assuming a b c; we claim that

P (a ; b; c) P (t ; t ; c);

where t = p ab 1:Indeed, we have

P (a ; b; c) P (t ; t ; c) =

= a3 + b3 2abp ab (a2 + b2 2ab ) 2c a + b 2p ab= ( a b)2(a + b) + ab p a p b 2 (a b)2 2c p a p b 2

= p a p b2

p a + p b2

(a + b 1) + ab 2c 0;

becausep a + p b 2 (a + b 1) + ab 2c 4(2 1) + 1 2 = 3 > 0:


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8 Solutions

So, the above statement holds and all we have to do is to prove that P (t ; t ; c) 0; which is equivalent to eachof the following inequalities

2t 3 + c3 + 6 (2t + c)2 ;

2t 3 + 1t 6

+ 6 2t + 1t 2

2;

2t 9 + 6t 6 + 1 t 2(2t 3 + 1)2 ;

2t 9 + 6t 6 + 1 4t 8 + 4t 5 + t 2 ;

2t 9 4t 8 + 6t 6 4t 5 t 2 + 1 0;

(t 1)2 2t 7 2t 5 + 2t 4 + 2t 3 + 2t 2 + 2t + 1 0;

which is obivously true because t 1:

Problem 2.4 Let a; b; c be nonnegative real numbers with sum 1: Determine the maximum and minimumvalues of

P (a ; b; c) = ( 1 + ab )2 +( 1 + bc)2 +( 1 + ca )2 :

Solution. It is clear that min P = 3 with equality attains when a = 1; b = c = 0 and its cyclic permutions. Now, let us nd max P : We claim that max P = 10027 attains when a = b = c =

13 ; or

(1 + ab )2 +( 1 + bc)2 +( 1 + ca )2 100

27 ;

a2b2 + b2c2 + c2a2 + 2(ab + bc + ca ) 1927

;

(ab + bc + ca )2 + 2(ab + bc + ca ) 2abc 1927

;

(ab + bc + ca + 1)2

2abc 4627 :

According to AM-GM Inequality , we have

(ab + bc + ca + 1)2 43

(ab + bc + ca + 1):

And we deduce the inequality to

43

(ab + bc + ca + 1) 2abc 4627

;

4(ab + bc + ca ) 6abc 10

9 ;

which is obviously true by Schur's Inequality for third degree,

4(ab + bc + ca ) 6abc (1 + 9abc ) 6abc = 1 + 3abc 1 + 3 127

= 10

9 :

With the above solution, we have the conclusion for the requirement is min P = 3 and max P = 10027 :


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9

Problem 2.5 Let a; b; c be nonnegative real numbers with sum 1: Determine the maximum and minimumvalues of

P (a ; b; c) = ( 1 4ab )2 +( 1 4bc)2 +( 1 4ca )2 :

Solution (by Honey_suck). Notice that

1 1 4ab 1 (a + b)2 1 (a + b + c)2 = 0:

Hence(1 4ab )2 1;

and it follows that P (a ; b; c) 1 + 1 + 1 = 3;

which equality holds when a = 1; b = c = 0 and its cyclic permutions. And we conclude that max P = 3:Moreover,applying Cauchy Schwarz Inequality and AM-GM Inequality , we have

(1 4ab )2 + ( 1 4bc)2 +( 1 4ca )2 1

3(1 4ab + 1 4bc + 1 4ca )2

= 1

3 [3 4(ab + bc + ca )]2

13

3 413

2

= 2527

;

with equality holds when a = b = c = 13 : And we conclude that min P = 2527 :

This completes the proof.

Problem 2.6 Let a; b; c be positive real numbers. Prove that

b + ca

+ c + a

b +

a + bc

2

4(ab + bc + ca ) 1a2

+ 1b2

+ 1c2

:

Solution 1. We can rewite the inequality as

"∑cycab (a + b)#2

4(ab + bc + ca )(a2b2 + b2c2 + c2a2):

Assuming that a b c; then using AM-GM Inequality , we have that

4(ab + bc + ca )(a2b2 + b2c2 + c2a2) =

= 16(a + b)2(ab + bc + ca )(a2b2 + b2c2 + c2a2)

4(a + b)2

(a + b)2(ab + bc + ca ) + 4(a2b2 + b2c2 + c2a2) 2

4(a + b)2 :

It suf ces to show that

2ab (a + b) + 2bc(b + c) + 2ca (c + a) (a + b)2(ab + bc + ca ) + 4(a2b2 + b2c2 + c2a2)

a + b ;


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10 Solutions

2ab (a + b) + 2c2(a + b)+ 2c(a2 + b2) (a + b)(ab + bc + ca ) + 4(a2b2 + b2c2 + c2a2)

a + b ;

ab (a + b) + 2c2

(a + b) + c(a b)2 4a

2b2

a + b + 4c2(a2 + b2)

a + b ;

ab a + b 4aba + b

+ c(a b)2 2c22(a2 + b2)

a + b a b ;ab (a b)2

a + b + c(a b)2

2c2(a b)2

a + b ;

aba + b

+ c 2c2

a + b;

which is obviously true because2c2

a + b 2c2

c + c = c c +

aba + b

:

The proof is completed and we have equality iff a = b = c: Solution 2. Similar to solution 1, we need to prove

"∑cycab (a + b)#2

4(ab + bc + ca )(a2b2 + b2c2 + c2a2):

For all real numbers m; n; p; x; y; z ; we have the following interesing identity of Lagrange

( x2 + y2 + z 2)(m2 + n2 + p2) (mx + ny + pz )2 = ( my nx)2 +( nz py)2 +( px mz )2 :

Now, applying this identity with x = p ab ; y = p bc ; z = p ca ; m = ( a + b)p ab ; n = ( b + c)p bc ; and p =(c + a)p ca ; we obtain

(ab + bc + ca )"∑cycab (a + b)2

#"∑cycab (a + b)#2

= abc ∑cycc(a b)2 :

Moreover,∑cyc

ab (a + b)2 4(a2b2 + b2c2 + c2a2) = ∑cyc

ab (a b)2 ;

So, we may rewrite our inequality as

(ab + bc + ca )"∑cycab (a + b)2 4∑cyca2b2# (ab + bc + ca )"∑cycab (a + b)2#"∑cycab (a + b)#

2

;

(ab + bc + ca )∑cyc

ab (a b)2 abc ∑cyc

c(a b)2 ;

∑cyc

a2b2(a b)2 + abc ∑cyc

(a + b c)(a b)2 0;


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11

∑cyc

a2b2(a b)2 + 2abc ∑cyc

a(a b)(a c) 0;

which is obviously true by Schur's Inequality for third degree.

Problem 2.7 Let a; b; c be the side of a triangle. Show that

∑cyc

(a + b)(a + c)p b + c a 4(a + b + c)p (b + c a)(c + a b)(a + b c): Solution. The inequality can be rewritten in the form

∑cyc

(a + b)(a + c)

p (c + a b)(a + b c)4(a + b + c);


∑cyc(a + b)(a + c)

p (c + a b)(a + b c) = ∑cyc(a + b)(a + c)

p a2 (b c)2

∑cyc

(a + b)(a + c)a

= 3(a + b + c) +bca

+ cab

+ ab

c 4(a + b + c);

where the last inequality is valid because

bca

+ cab

+ ab

c = a

b2c

+ c2b

+ b c2a

+ a2c

+ c a2b

+ b2a

a + b + c:Equality holds iff a = b = c:

Problem 2.8 Given a triangle with sides a ; b; c satisfying a 2 + b2 + c2 = 3: Show that

a + bp a + b c +

b + cp b + c a +

c+ ap c + a b 6:

Solution. Firstly, to prove the original inequality, we will show that 1

ab + bc + ca 4a3(b + c a)

(b + c)2 +

4b3(c + a b)(c + a)2

+ 4c3(a + b c)

(a + b)2 ;

∑cyc

a2 4a3(b + c a)(b + c)2

a2 + b2 + c2 ab bc ca ;

1We may prove this statement easily by using tangent line technique, the readers can try it! In here, we present a nonstandard proof for it, this proof seems to be complicated but it is nice about its idea.


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12 Solutions

a2(2a b c)2

(b + c)2 +

b2(2b c a)2

(c + a)2 +

c2(2c a b)2

(a + b)2 a2 + b2 + c2 ab bc ca ;

Without loss of generality, we may assume that a b c; then

a2

(b + c)2 b2

(c + a)2 and (2a b c)2 (2b c a)2 :

Thus, using Chebyshev's Inequality , we have

a2(2a b c)2

(b + c)2 +

b2(2b c a)2

(c + a)2 12

a2

(b + c)2 +

b2

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

Notice that

14

(2a b c)2 +( 2b c a)2 (a2 ab + b2) 18

(a + b 2c)214

(a + b)2 ; (2)

And

a2

(b + c)2 + b2

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

(a + b + 2c)2 12 ;which yieds that

12

a2

(b + c)2 +

b2

(c + a)212

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

12

2(a + b)2

(a + b + 2c)212

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

14

2(a + b)2

(a + b + 2c)212

(a + b 2c)2 : (3)

From (1); (2) and (3); we obtain

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

+ b2(2b c a)2

(c + a)2 (a2 ab + b2) (

a + b)2(a + b 2c)22(a + b + 2c)2

14

(a + b)2 :

Using this inequality, we have to prove

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

2(a + b + 2c)2 14

(a + b)2 + c2(a + b 2c)2

(a + b)2 c2 c(a + b);

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

2(a + b + 2c)2 +

c2(a + b 2c)2

(a + b)2 14

(a + b 2c)2 ;

(a + b)2

2(a + b + 2c)2 +

c2

(a + b)2 14

;

which can be easily checked. Thus, the above statement is proved. Now, turning back to our problem, using Holder Inequality , we have

∑cyc

b + cp b + c a!

2

"∑cyc a3(b + c a)

(b + c)2 # ∑cyca!3

:


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13

It follows that

∑cycb + c

p b + c a!2 ∑

cyca

3

∑cyc

a3 (b+ c a)(b+ c)2

4 ∑cyc

a3

∑cycab :

Moreover, by AM-GM Inequality , we have

∑cyc

ab = v uut13 ∑cyc

ab! ∑cycab! ∑cyca2!v uuut

13 0@

∑cyc

ab + ∑cyc

ab + ∑cyc

a2

3 1A3

=∑cyc

a3

9 :

Hence

∑cyc

b + cp b + c a!

2 4 ∑cyc

a3

∑cyc

ab 36 :

Our proof is completed. Equality holds if and only if a = b = c = 1:

Problem 2.9 Given a triangle with sides a ; b; c satisfying a 2 + b2 + c2 = 3: Show that

ap b + c a +

bp c + a b +

cp a + b c 3:

Solution (by Materazzi). Applying Holder Inequality , we obtain

∑cyc

ap b + c a!

2

"∑cyca(b + c a)# (a + b + c)3 :And we deduce our inequality to show that

(a + b + c)3 9∑cyc

a(b + c a);

(a + b + c)3 9 [2(ab + bc + ca ) 3]:

Setting p = a + b + c; then it p 3 p 3 and 2 (ab + bc + ca ) = p2 3: Thus, we can rewrite the aboveinequality as

p3 9( p2 6);

p3 9 p2 + 54 0;

(3 p)(18 + 6 p p2) 0;

which is obviously true. Equality holds if and only if a = b = c = 1:


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14 Solutions

Problem 2.10 Show that if a ; b; c are positive real numbers, then

a

a + b +

b

b + c +

c

c + a 1 +

s 2abc

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

Solution. The given inequality is equivalent to

∑cyc

a2

(a + b)2 + 2∑

cyc

ab(a + b)(b + c) 1 + 2s 2abc(a + b)(b + c)(c + a) + 2abc(a + b)(b + c)(c + a) ;

Using the known inequality 2

∑cyc

a2

(a + b)2 1 2abc

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

we can deduce it to

∑cyc

ab(a + b)(b + c) s 2abc(a + b)(b + c)(c + a) + 2abc(a + b)(b + c)(c + a) ;

a2b + b2c + c2a + abc p 2abc (a + b)(b + c)(c + a): Now, we assume that c = min fa ; b; cg; applying AM-GM Inequality , we havea2b + b2c + c2a + abc = a(ab + c2) + bc(a + b)

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

2 + bc(a + b) +

a(a c)(b c)2

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

+ bc(a + b)

2r a(a + c)(b + c)2 bc(a + b)= p 2abc (a + b)(b + c)(c + a):Our proof is completed. Equality holds if and only if a = b = c: Problem 2.11 Show that if a ; b; c are positive real numbers, then

aa + b

2

+ bb + c

2

+ c

c + a

2 34

+ a2b + b2c + c2a 3abc

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

Solution. We havea2

(a + b)2 =

aa + b

ab(a + b)2

;

anda

a + b +

bb + c

+ cc + a

= 1 + a2b + b2c + c2a + abc(a + b)(b + c)(c + a)

:

2The proof will be left to the readers


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15

Hence, the given inequality can be rewritten as

14

+ 4abc

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

(a + b)2 +

bc(b + c)2

+ ca

(c + a)2 ;

a ba + b

2

+b cb + c

2

+c ac + a

2

2 16abc

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

Now, notice that

2 16abc

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

= 2[(a + b)(b + c)(c + a) 8abc ]

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

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

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

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

The above inequality is equivalent to

a ba + b

2

+b cb + c

2

+c ac + a

2

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

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

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

;

a ba + b

+ b cb + c

+ c ac + a

2

0;

which is obviously true. Equality holds if and only if a = b or b = c or c = a :


(a2 + b2)(b2 + c2)(c2 + a2)8a2b2c2

a2 + b2 + c2

ab + bc + ca

2

:

Solution 1. Without loss of generality, we may assume that a b c; then we have that

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

4

2

= (b c)2(8a2 b2 c2 6bc)

16 0;

b2 + c2 (b + c)2

2 :

It suf ces to prove that4a2 + ( b + c)2 (b + c)

16abc a2 + b2 + c2

ab + bc + ca:

This inequality is homogeneous, we may assume that b + c = 1; putting x = bc; then a 12 ; and 14 x 0:The above inequality becomes

4a2 + 116ax

a2 + 1 2 xa + x

;


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16 Solutions

(4a2 + 1)(a + x) 16ax(a2 + 1 2 x);

32ax2 16a3 4a2 + 16a 1 x+ a 4a2 + 1 0;

2a(4 x 1)2

+ 2a(8 x 1) 16a3

4a2

+ 16a 1 x+ a 4a2

+ 1 0;

2a(4 x 1)2 + ( 1 + 4a2 16a3) x+ a(4a2 1) 0;

2a(4 x 1)2 (2a 1)(8a2 + 2a + 1) x+ a(4a2 1) 0;

which is true because

4(2a 1)(8a2 + 2a + 1) x+ 4a(4a2 1) 4a(4a2 1) (2a 1)(8a2 + 2a + 1)= ( 2a 1)2 0:

Our proof is completed. Equality holds if and only if a = b = c: Solution 2. Put a = 1 x ; b =

1 y ; c =

1 z ; then the above inequality becomes

( x+ y+ z )2( x

2+ y

2)( y

2+ z

2)( z

2+ x

2) 8( x

2 y

2+ y

2 z

2+ z

2 x

2)

2:

Notice that( x2 + y2)( y2 + z 2)( z 2 + x2) = ( x2 + y2 + z 2)( x2 y2 + y2 z 2 + z 2 x2) x2 y2 z 2 :

Thus, we can rewrite the above inequality as

( x2 y2 + y2 z 2 + z 2 x2) ( x+ y+ z )2( x2 + y2 + z 2) 8( x2 y2 + y2 z 2 + z 2 x2) x2 y2 z 2( x+ y+ z )2 :

Now, we see that

( x+ y+ z )2( x2 + y2 + z 2) 8( x2 y2 + y2 z 2 + z 2 x2)= ∑

cyc x4 + 2∑

cyc xy( x2 + y2)+ 2 xyz ∑

cyc x 6∑

cyc x2 y2

= ∑cyc

x2( x y)( x z ) + 3∑cyc

xy( x y)2 + xyz ∑cyc

x xyz ∑cyc

x:

It suf ces to prove that

xyz ( x2 y2 + y2 z 2 + z 2 x2)( x+ y+ z ) x2 y2 z 2( x+ y+ z )2 ;

x2 y2 + y2 z 2 + z 2 x2 xyz ( x+ y+ z );

which is obviously true by AM-GM Inequality . Solution 3. Similar to solution 2 , we need to prove that

( x2 + y2)( y2 + z 2)( z 2 + x2)( x+ y+ z )2 8( x2 y2 + y2 z 2 + z 2 x2)2 :

By AM-GM Inequality , we have

( x+ y+ z )2 = x2 + y2 + z 2 + 2 xy+ 2 yz + 2 zx

x2 + y2 + z 2 + 4 x2 y2

x2 + y2 +

4 y2 z 2

y2 + z 2 +

4 z 2 x2

z 2 + x2 :


22/36

17

Hence, it suf ces to prove that

(m + n)(n + p)( p + m) m + n + p + 4mnm

+n

+ 4npn

+ p

+ 4 pm p

+m

8(mn + np + pm)2 ;

where m = a2 ; n = b2 ; p = c2 :This inequality is equivalent with

mn(m n)2 + np (n p)2 + pm( p m)2 0;

which is obviously true. Solution 4. 3Again, we will give the solution to the inequality

( x2 + y2)( y2 + z 2)( z 2 + x2) ( x+ y+ z )2 8( x2 y2 + y2 z 2 + z 2 x2)2 :

Assuming that x y z ; then by Cauchy Schwarz Inequality , we have

( x2 + z 2)( y2 + z 2) ( xy+ z 2)2 :

Hence, it suf ces to prove that

( x2 + y2)( xy+ z 2)2( x+ y+ z )2 8( x2 y2 + y2 z 2 + z 2 x2)2 ;

q 2( x2 + y2)( xy+ z 2)( x+ y+ z ) 4( x2 y2 + y2 z 2 + z 2 x2);We have

q 2( x2 + y2) = x+ y+ ( x y)2

x+ y+ p 2( x2 + y2) x+ y+

( x y)2

x+ y+ p 2( x+ y)

= x+ y+p 2 1 ( x y)2

x+ y x+ y+

3( x y)2

8( x+ y) :

Therefore

q 2( x2 + y2)( xy+ z 2)( x+ y+ z )= xy( x+ y)q 2( x2 + y2) + z ( x+ z )( y+ z )q 2( x2 + y2)

xy( x+ y)q 2( x2 + y2) + z ( x+ y)( x+ z )( y+ z ) + 3 z 2( x y)2

8 +

3( x y)2 xyz 8( x+ y)

:


xy( x+ y)

q 2( x2 + y2)+ z ( x+ y)( x+ z )( y+ z ) +

3 z 2( x y)2

8 +

3( x y)2 xyz 8( x+ y) 4( x

2 y2 + y2 z 2 + z 2 x2);

xy ( x+ y)q 2( x2 + y2) 4 xy + xyz x+ y+ 3( x y)2

8( x+ y)21 x2 10 xy+ 21 y2 z 2

8 + z 3( x+ y) 0:

3This proof seems to be the most complicated but the idea is very interesting.


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18 Solutions

We have

xy ( x+ y)q 2( x2 + y2) 4 xy xz ( x+ y)q 2( x2 + y2) 4 xy :Hence, it suf ces to prove that

f ( z ) = x ( x+ y)q 2( x2 + y2) 4 xy + xy x+ y+ 3( x y)2

8( x+ y)21 x2 10 xy+ 21 y2 z

8 + z 2( x+ y) 0:

Also, we have

f ( z ) f ( y) = 18

( y z ) 21 x2 + 13 y2 18 xy 8 xz 8 yz 0:

Therefore

f ( z ) f ( y) = x ( x+ y)q 2( x2 + y2) 4 xy y(10 x+ 13 y)( x y)2

8( x+ y)

= 2 x( x y)2( x2 + y2 + 4 xy)

( x+ y)p 2( x2 + y2) + 4 xy

y(10 x+ 13 y)( x y)2

8( x+ y) 0;

since

2 x( x2 + y2 + 4 xy)( x+ y)p 2( x2 + y2) + 4 xy

y(10 x+ 13 y)8( x+ y)

2 x( x2 + y2 + 4 xy)2( x2 + y2) + 4 xy

y(10 x+ 13 y)8( x+ y)

= 8 x3 + 22 x2 y 15 xy2 13 y3

8(a + b)2 0:Thus, our inequality is proved. Solution 5 (by Gabriel Dospinescu). We rewrite the inequality in the form

(a2 + b2)(b2 + c2)(c2 + a2)1

a +

1

b +

1

c

2

8(a2 + b2 + c2)2 :

Setting a2 + b2 = 2 x; b2 + c2 = 2 y and c2 + a2 = 2 z ; then x; y; z are the side lengths of a triangle and we mayrewrite the inequality as

1p y+ z x +

1p z + x y +

1p x+ y z

x+ y+ z p xyz :

By Holder Inequality , we have

∑cyc

1p y+ z x!

2

"∑cyc x3( y+ z x)# ( x+ y+ z )3 :It suf ces to show that xyz ( x+ y+ z ) ∑

cyc x3( y+ z x);

which is just Schur's Inequality for fourth degree.This completes the proof.


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19

Problem 2.13 Let a; b; c be positive real numbers. Prove the inequality

(b + c)2

a(b + c + 2a) +

(c + a)2

b(c + a + 2b) +

(a + b)2

c(a + b + 2c) 3:

Solution 1. After using AM-GM Inequality , it suf ces to prove that

(a + b)2(b + c)2(c + a)2 abc (a + b + 2c)(b + c + 2a)(c + a + 2b):

Now, applying Cauchy Schwarz Inequality and AM-GM Inequality , we obtain

(b + c)2(a + b)(a + c) (b + c)2 a + p bc2

= bca(b + c)p bc + b + c

2

bc(2a + b + c)2

:

Similarly, we have

(a + b)2(c + a)(c + b) ab(a + b + 2c)2 ;(c + a)2(b + c)(b + a) ca(c + a + 2b)2 :

Multiplying these inequalities and taking the square root, we get the result. Equality holds if and only if a = b = c: Solution 2. We need to prove the inequality

(a + b)2(b + c)2(c + a)2 abc (a + b + 2c)(b + c + 2a)(c + a + 2b):

According to the AM-GM Inequality , we have

abc (a + b + 2c)(b + c + 2a)(c + a + 2b) 64

27abc (a + b + c)3

6481

(a + b + c)2(ab + bc + ca )2 :

And we deduce the problem to

(a + b)2(b + c)2(c + a)2 6481

(a + b + c)2(ab + bc + ca )2 ;

9(a + b)(b + c)(c + a) 8(a + b + c)(ab + bc + ca );

ab (a + b)+ bc(b + c) + ca (c + a) 6abc ;

which is obviously true by AM-GM Inequality .


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20 Solutions

Solution 3 (by Materazzi). We will try to write the inequality as the sum of squares, which shows that theoriginal inequality is valid. Indeed, we have

∑cyc(b + c)2

a(2a + b + c) 3 = ∑cyc(b + c)2 a(2a + b + c)

a(2a + b + c)

= ( a + b + c)∑cyc

b + c 2aa(2a + b + c)

= ( a + b + c)∑cyc

c aa(2a + b + c)

a ba(2a + b + c)

= ( a + b + c)∑cyc

a bb(2b + c + a)

a ba(2a + b + c)

= ( a + b + c)∑cyc

(a b)2(2a + 2b + c)ab (2a + b + c)(2b + c + a)

;

which is obviously nonnegative. Solution 4. We have the following identity

4(b + c)2

a(2a + b + c) +

27aa + b + c

13 = (7a + 4b + 4c)(2a b c)2

a(a + b + c)(2a + b + c) 0;which yields that

(b + c)2

a(2a + b + c) 13

4 274

aa + b + c

:

It follows that

∑cyc

(b + c)2

a(2a + b + c) 39

4 274

aa + b + c

+ b

a + b + c +

ca + b + c

= 3:

Problem 2.14 Let a; b; c be positive real numbers. Prove the inequality

(b + c)2

a(b + c + 2a) +

(c + a)2

b(c + a + 2b) +

(a + b)2

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

b + c + 2a +

c + ac + a + 2b

+ a + b

a + b + 2c:

Solution 1. We may write our inequality as

∑cyc

(b + c)2

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

0;

∑cyc(b + c 2a) b + c

a(2a + b + c) 0:Assuming without loss of generality that a b c; then we can easily check that

b + c 2a c + a 2b a + b 2c;


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21

andb + c

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

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

c(2c + a + b):

Hence, we may apply the Chebyshev's Inequality as follow

∑cyc

(b + c 2a) b + c

a(2a + b + c) 1

3 "∑cyc(b + c 2a)#"∑cyc b + ca(2a + b + c)#= 0:

This completes the proof. Equality holds if and only if a = b = c: Solution 2 (by Honey_suck). We have a notice that

∑cyc

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

= ∑cyc

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

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

= ∑cyc

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

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

= ∑cyc

(a b)2(2a2 + 2b2 + c2 + 3ab + 3bc + 3ca )ab (2a + b + c)(2b + c + a)

;

which is obviously nonnegative. Solution 3. Again, we notice that

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

3(b + c 2a)2(a + b + c)

= (3a + 2b + 2c)(2a b c)2

2a(a + b + c)(2a + b + c) 0:It follows that

∑cyc

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

32 ∑cyc

b + c 2aa + b + c

= 0:


(b + c)2

a(b + c + 2a) +

(c + a)2

b(c + a + 2b) +

(a + b)2

c(a + b + 2c) 2 ab + c

+ bc + a

+ ca + b

:

Solution. We see that(b + c)2

a(b + c + 2a) =

b+ ca

+ 4a

2a + b + c2:

Hence, we may write the inequality in the form

∑cyc

b + ca

+ 4∑cyc

a2a + b + c 2∑cyc

ab + c

+ 6:

Now, using Cauchy Schwarz Inequality , we have

∑cyc

b + ca

= ∑cyc

ab

+ ac ∑cyc

4ab + c

:


27/36

22 Solutions

It suf ces to show that2∑

cyc

ab + c

+ 4∑cyc

a2a + b + c 6;

∑cyc

ab + c

+ ∑cyc

2a2a + b + c 3;


∑cyc

ab + c

+ ∑cyc

2a2a + b + c

= ∑cyc

a 1b + c

+ 1

2a+ b+ c2 !

∑cyc

4a

b + c + 2a+ b+ c2

= 8∑cyc

a2a + 3b + 3c

8(a + b + c)2

∑cyc

a(2a + 3b + 3c)

= 4(a + b + c)2

a2 + b2 + c2 + 3(ab + bc + ca );

and4(a + b + c)2 3(a2 + b2 + c2) 9(ab + bc + ca ) = a2 + b2 + c2 ab bc ca 0:

Equality holds if and only if a = b = c:


a3b3 + b3c3 + c3a3 (b + c a)(c + a b)(a + b c)(a3 + b3 + c3):

Solution 1. By Schur's Inequality for third degree, we have

a3b3 + b3c3 + c3a3 abc "∑cycab (a + b) 3abc#:It suf ces to prove that

abc "∑cycab (a + b) 3abc# (b + c a)(c + a b)(a + b c)(a3 + b3 + c3);which is equivalent to each of the following inequalities

∑cyc

ab (a + b) 3abc

a3 + b3 + c3 (b + c a)(c + a b)(a + b c)abc ;

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

abc 1∑cyc

ab (a + b) 3abc

a3 + b3 + c3 ;


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23

∑cyc

a(a b)(a c)

abc ∑cyc

a(a b)(a c)

a3 + b3 + c3 ;

(a3

+ b3

+ c3

abc )∑cyca(a b)(a c) 0;which is obviously true by AM-GM Inequality and Schur's Inequality for third degree. Solution 2. Without loss of generality, we may assume that a b c; then we have 2 casesCase 1. If b + c a ; then the inequality is trivial since

(b + c a)(c + a b)(a + b c) 0:

Case 2. If b + c a ; then we have

(c + a b)(a + b c) = a2 (b c)2 a2 ;b3c3 a2bc(b + c a)2 = bc [bc + a(b + c a)](a b)(a c) 0:


a3(b3 + c3)+ a2bc(b + c a)2 a2(b + c a)(a3 + b3 + c3);

a(b3 + c3) + bc(b + c a)2 (b + c a)(a3 + b3 + c3):

Now, since this inequality is homogeneous, we can assume that b + c = 1 and put x = bc ; then 1 a 12and 14 x a(1 a): The above inequality becomes

a(1 3 x) + x(1 a)2 (1 a)(a3 + 1 3 x);

f ( x) = ( 4 8a + a2) x+ a4 a3 + 2a 1 0:

We see that f ( x) is a linear function of x; thus

f ( x) min f 14 ; f (a(1 a)) :

But, we have

f 14

= a2(2a 1)2

4 0; and f (a(1 a)) = ( 2a 1)3 0:

Thus, our proof is completed. Equality holds if and only if a = b = c: Solution 3 (by nhocnhoc). By expanding, we see that the given inequality is equivalent to

∑cyc

a6 + 3∑cyc

a3b3 + 2abc ∑cyc

a3 ∑cyc

ab (a4 + b4) + ∑cyc

a2b2(a2 + b2) + abc ∑cyc

ab (a + b);

∑cyc

a6 + b6

2 + 3a3b3 ab (a4 + b4) a2b2(a2 + b2) + abc ∑

cyc

a3 + b3 ab (a + b) 0;

12 ∑cyc

(a4 + b4 3a2b2)(a b)2 + abc ∑cyc

(a b)2(a + b) 0;

S a (b c)2 + S b(c a)2 + S c(a b)2 0;


29/36

24 Solutions

where

S a = b4 + c4 3b2c2 + 2abc (b + c);

S b = c4

+ a4

3c2a

2

+ 2abc (c + a);S c = a4 + b4 3a2b2 + 2abc (a + b):

Without loss of generality, we may assume that a b c; then it is easy to check that S a ; S b 0: Moreover,we have

S b + S c = 2a4 + b4 3a2b2 + c4 + 2abc (2a + b + c) 3a2c2

2a4 + b4 3a2b2 + c4 + 2ac 2(2a + b + c) 3a2c2

= 2a4 + b4 3a2b2 + c4 + ac 2(a + 2b + 2c)> 2a4 + b4 3a2b2 = ( a2 b2)(2a2 b2) 0:

It follows that

S a (b c)2 + S b(c a)2 + S c(a b)2 S b(c a)2 + S c(a b)2

S b(c a)2 S b(a b)2

= S b(b c)(2a b c) 0:


Problem 2.17 If a; b; c are positive real numbers such that abc = 1; show that we have the following in-equality

a3 + b3 + c3 ab + c

+ bc + a

+ ca + b

+ 32

:

Solution. By GM-HM Inequality , we have that

4ab + c

+ 4bc + a

+ 4ca + b

+ 6 a 1b + 1c + b 1c + 1a + c 1a + 1b + 6= a2(b + c)+ b2(c + a) + c2(a + b) + 6abc

2a2(b + c) + 2b2(c + a)+ 2c2(a + b)= 2ab (a + b) + 2bc(b + c) + 2ca (c + a)

2(a3 + b3)+ 2(b3 + c3) + 2(c3 + a3)= 4(a3 + b3 + c3):

Hence

a3 + b3 + c3 ab + c

+ bc + a

+ ca + b

+ 32

:

Our proof is completed. Equality holds if and only if a = b = c = 1:

Remark 1 We can prove that the stronger inequality holds

a3 + b3 + c3 3 6 + 4p 2 ab + c

+ bc + a

+ ca + b

32

:


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25

Indeed, this inequality is equivalent to each of the following

a3 + b3 + c3

abc 3 6 + 4p 2 a

b + c +

b

c + a +

c

a + b

3

2;

(a + b + c) ∑cyc

(a b)(a c)

abc 3 + 2p 2 ∑cyc(2a + b + c)(a b)(a c)

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

X (a b)(a c)+ Y (b c)(b a) + Z (c a)(c b) 0;

where

X = (a + b + c)(a + b)(b + c)(c + a)

abc 3 + 2p 2 (2a + b + c)=

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

a(a + b + c)bc

+ 1 3 + 2p 2 (2a + b + c)

(a

+b

+c)(

b+

c)a

4a(a

+b

+c)(b + c)2 + 1 3 + 2p 2 (2a + b + c)

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

a(b + c) 3 2p 2 0;andY ; Z are similar. Now, assume that a b c; then we see that Z Y X ; thus

X (a b)(a c) + Y (b c)(b a)+ Z (c a)(c b) Y (b c)(b a) + Z (c a)(c b) Y (b c)(b a) + Y (c a)(c b)

= Y (b c)2 0:

Problem 2.18 Given nonnegative real numbers a ; b; c such that ab + bc + ca + abc = 4: Prove that

a2 + b2 + c2 + 2(a + b + c) + 3abc 4(ab + bc + ca ):

Solution. Setting p = a + b + c; q = ab + bc + ca and r = abc : The given condition gives us q + r = 4; andwe have to prove

p2 2q + 2 p + 3(4 q) 4q;

p2 + 2 p + 12 9q:

If p 4; then it is trivial because

p2 + 2 p + 12 16 + 8 + 12 = 36 9q:

If 4 p 3;4 applying Schur's Inequality for third degree, we have

p3 4 pq + 9r 0;

4In here, we have p 3 because p2

3 + p327 q + r = 4; which gives us p 3:


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26 Solutions

which yields that p3 4 pq + 9(4 q) 0;

or

q p3 + 36

4 p+ 9 :

It follows that

p2 + 2 p + 12 9q p2 + 2 p + 12 9 p3 + 364 p + 9

= (5 p + 18)(4 p)( p 3)

4 p + 9 0:This completes our proof. Equality holds if and only if a = b = c = 1 or a = b = 2; c = 0 and its cyclic permutations.

Problem 2.19 Let a; b; c be real numbers with min fa ; b; cg 34 and ab + bc + ca = 3: Prove that

a3 + b3 + c3 + 9abc 12 :

Solution. Notice that from the given condition, we have

a2 916

= 316

(ab + bc + ca ) 316

a(b + c) >18

a(b + c):

It follows that 8 a > b + c: Similarly, we have 8 b > c + a and 8c > a + b: Now, using AM-GM Inequality , we obtain

36 = 4(ab + bc + ca )

p 3(ab + bc + ca )

2(ab + bc + ca ) a + b + c + 3(ab + bc + ca )

a + b + c :

And we deduce the inequality to

3∑cyc

a3 + 27abc 2 ∑cyc

a! ∑cycab!+ 6(ab + bc + ca )2

a + b + c ;

3∑cyc

a3 + 27abc 2 ∑cyc

a! ∑cycab! 6(ab + bc + ca )2

a + b + c 2 ∑cyca! ∑cycab!;We have

3∑cyc

a3 + 27abc 4∑cyc

a

! ∑cycab

!=

∑cyc(3a b c)(a b)(a c);

and6(ab + bc + ca )2

a + b + c 2 ∑cyca! ∑cycab!= 2(ab + bc + ca )a + b + c ∑cyc(a b)(a c):


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Hence, we may write the above inequality as

X (a b)(a c)+ Y (b c)(b a) + Z (c a)(c b) 0;

where

X = 4(ab + bc + ca )

a + b + c + 6a 2b 2c +

(b c)2

a + b + c

= 4a(b + c) + ( b + c)2

a + b + c + 6a 2b 2c

= 6a2 +( b + c)(8a b c)

a + b + c > 0;

and Y ; Z are similar.We will now assume that a b c; then

X Y = 4a(b + c) + ( b + c)2 4b(c + a) (c + a)2

a + b + c + 8(a b)

= 8(a b)(a b)(a + b 2c)

a + b + c =

(a b)(7a + 7b + 10c)a + b + c 0:

It follows that

X (a b)(a c)+ Y (b c)(b a) + Z (c a)(c b) X (a b)(a c)+ Y (b c)(b a) Y (a b)(a c) + Y (b c)(b a)

= Y (a b)2 0:

This completes our proof. Equality holds if and only if a = b = c = 1:

Problem 2.20 Let a; b; c be positive real numbers such that a 2b2 + b2c2 + c2a2 = 1: Prove that

(a2 + b2 + c2)2 + abcq (a2 + b2 + c2)3 4: Solution. According to AM-GM Inequality and Cauchy Schwarz Inequality , we have

q (a2 + b2 + c2)3 q 3(a2 + b2 + c2)(a2b2 + b2c2 + c2a2)= q 3(a2 + b2 + c2) a + b + c:

It suf ces to show that(a2 + b2 + c2)2 + abc (a + b + c) 4;

a4 + b4 + c4 + abc (a + b + c) 2(a2b2 + b2c2 + c2a2);

∑cyc

a2(a b)(a c) + ∑cyc

ab (a b)2 0;

which is obviously true by Schur's Inequality for fourth degree. Equality holdsif and only if a = b = c = 14p 3or a = b = 1; c = 0 and its cyclic permutations.


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28 Solutions

Problem 2.21 Show that if a ; b; c are positive real numbers, the following inequality holds

(a + b + c)2(ab + bc + ca )2 +( ab + bc + ca )3 4abc (a + b + c)3 :

Solution. Setting x = 1a ; y = 1b ; z = 1c ; then we may write the above inequality in the form

( x+ y+ z )2( xy+ yz + zx)2 + xyz ( x+ y+ z )3 4( xy+ yz + zx)3 :

Using AM-GM Inequality , we have

xyz ( x+ y+ z )3 3 xyz ( x+ y+ z )( xy+ yz + zx);

and we can deduce the inequality to

( x+ y+ z )2( xy+ yz + zx) + 3 xyz ( x+ y+ z ) 4( xy+ yz + zx)2 ;

which can be easily simplied to

xy( x y)2 + yz ( y z )2 + zx( z x)2 0;

which is obviously true and this completes our proof. Equality holds if and only if a = b = c:

Problem 2.22 Let a; b; c be real numbers from the interval [3; 4]: Prove that

(a + b + c)abc

+ bca

+ cab

3(a2 + b2 + c2):

Solution 1. The given condition shows that a; b; c are the side lengths of a triangle, hence we may puta = y+ z ; b = z + x and c = x+ y where x; y; z > 0: The above inequality becomes

( x+ y+ z )

"∑cyc

( x+ y)( x+ z ) y+ z

#3( x2 + y2 + z 2 + xy+ yz + zx):

We have

( x+ y+ z )"∑cyc ( x+ y)( x+ z ) y+ z # = ∑cyc x( x+ y)( x+ z ) y+ z + ∑cyc( x+ y)( x+ z )= ∑

cyc

x( x2 + yz ) y+ z

+ 2∑cyc

x2 + 3∑cyc

xy:

From this, we may rewrite our inequality as

x( x2 + yz ) y+ z

+ y( y2 + zx)

z + x +

z ( z 2 + xy) x+ y x

2 + y2 + z 2 ;

x( x y)( x z ) y+ z

+ y( y z )( y x)

z + x +

z ( z x)( z y) x+ y 0;

which is obviously true by Vornicu Schur Inequality.Equality holds if and only if a = b = c:


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Solution 2 (by nhocnhoc). We will rewite the inequality as

a2b2 + b2c2 + c2a2 abc (a + b + c)

abc

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

a + b + c ;

abc

2a + b + c

(b c)2 + bca

2a + b + c

(c a)2 + cab

2a + b + c

(a b)2 0;

S a (b c)2 + S b(c a)2 + S c(a b)2 0;

where

S a = abc

2a + b + c

; S b = bca

2a + b + c

; S c = cab

2a + b + c

:

Without loss of generality, we may assume that a b c: It is easy to see that S a S b S c : Moreover, wehave

S b + S c = bca

+ cab

4a + b + c

2a

4a + b + c

= 2(b + c a)

a + b + c > 0

since b + c > 4 a :It follows that S a S b 0: And we obtain

S a (b c)2 + S b(c a)2 + S c(a b)2 S b(c a)2 + S c(a b)2

S b(c a)2 S b(a b)2

= S b(b c)(2a b c) 0:


Problem 2.23 Given ABC is a triangle. Prove that

8cos 2 Acos 2 Bcos 2 C + cos2 Acos2 Bcos2 C 0:

Solution (by Honey_suck). Since cos 2 A+ cos 2 B+ cos 2 C + 2cos Acos Bcos C = 1 and cos Acos Bcos C 18 ;it follows that

cos 2 A+ cos 2 B+ cos 2 C 34

;

and(1 cos 2 A cos 2 B cos 2 C )2 = 4cos 2 Acos 2 Bcos 2 C :

Setting a = cos2 A; b = cos 2 B and c = cos2 C ; then a + b + c 34 and (a + b + c 1)2 = 4abc : We need to

prove that8abc +( 2a 1)(2b 1)(2c 1) 0;

16abc + 2(a + b + c) 4(ab + bc + ca ) + 1;

4(a + b + c 1)2 + 2(a + b + c) 4(ab + bc + ca ) + 1:

By Schur's Inequality for third degree, we have

4(ab + bc + ca ) (a + b + c)3 + 9abc

a + b + c

= 4(a + b + c)3 + 9(a + b + c 1)2

4(a + b + c) :


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30 Solutions

It suf ces to show that

4(a + b + c 1)2 + 2(a + b + c) 4(a + b + c)3 + 9(a + b + c 1)2

4(a + b + c) + 1;

4( p 1)2 + 2 p 4 p3 + 9( p 1)2

4 p + 1; ( p = a + b + c)

3(4 p 3)( p 1)2

4 p 0;which is obviously true since p 34 :

Problem 2.24 Let a; b; c be positive real numbers such that a + b + c = 3 and ab + bc + ca 2max fab ; bc ; cag: Prove that a2 + b2 + c2 a2b2 + b2c2 + c2a2 : Solution. The desired inequality is equivalent to each of the following inequalities

(a + b + c)2(a2 + b2 + c2) 9(a2b2 + b2c2 + c2a2);

(a + b + c)2 9(a2b2 + b2c2 + c2a2)

a2 + b2 + c2 ;

3(a2 + b2 + c2)9(a2b2 + b2c2 + c2a2)

a2 + b2 + c2 3(a2 + b2 + c2) (a + b + c)2 ;

3 ∑cyc

(a2 b2)(a2 c2)

a2 + b2 + c2 2∑cyc(a b)(a c); X (a b)(a c) + Y (b c)(b a)+ Z (c a)(c b) 0;

where

X = 3(a + b)(a + c) 2(a2 + b2 + c2) + 2(b c)2

= a2 + 3a(b + c) bc > ab + bc + ca 2bc ab + bc + ca 2max fab ; bc ; cag 0;

and Y ; Z are similar. Now, we assume that a b c; then

X Y = ( a b)(a + b + 4c) 0:

It follows that

X (a b)(a c) + Y (b c)(b a)+ Z (c a)(c b)

X (a b)(a c) + Y (b c)(b a) Y (a b)(a c) + Y (b c)(b a)

= Y (a b)2 0:

This completes our proof. Equality holds if and only if a = b = c:


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