Hoang Lan Huong

DIỄN ĐÀN TOÁN HỌC VIỆT NAM

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Tuyển tập

Bất đẳng thức Volume 1

Biên tập: Võ Quốc Bá Cẩn

Tác giả các bài toán: Trần Quốc Luật

Thành viên tham gia giải bài:

1. Võ Quốc Bá Cẩn (nothing)

2. Ngô Đức Lộc (Honey_suck)

3. Trần Quốc Anh (nhocnhoc)

4. Seasky

5. Materazzi

Tran Quoc Luat's Inequalities

Vo Quoc Ba Can - Pham Thi Hang

February 25, 2009

ii

Copyright c 2008 by Vo Quoc Ba Can

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 hocbat 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

iv Preface

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 2p: Prove that the following inequality holds

ap�a +

bp�b +

cp� c �

sb+ cp�a +

rc+ap�b +

sa+bp� 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� 3pabc(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+ab

+a+bc

�2� 4(ab+bc+ ca)

�1a2+1b2+1c2

�:

1

2 Problems

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

∑cyc(a+b)(a+ c)

pb+ 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+bpa+b� c

+b+ cpb+ c�a

+c+apc+a�b

� 6:

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

apb+ c�a

+bp

c+a�b+

cpa+b� c

� 3:

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

aa+b

+bb+ c

+cc+a

� 1+

s2abc

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

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

�2+

�bb+ c

�2+

�cc+a

�2� 34+a2b+b2c+ c2a�3abc(a+b)(b+ c)(c+a)

:

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

(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

�ab+ c

+bc+a

+c

a+b

�:


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

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

+c

a+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 minfa;b;cg � 34 and ab+bc+ ca= 3: Prove that

a3+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

8cos2Acos2Bcos2C+ cos2Acos2Bcos2C � 0:

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

a2+b2+ c2 � a2b2+b2c2+ c2a2:

4 Problems

Chapter 2

Solutions

"Don't just read it; �ght it! Ask your own questions, look for your own examples, discover yourown proofs. Is the hypothesis necessary? Is the converse true? What happens in the classicalspecial 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 2p: Prove that the following inequality holds

ap�a +

bp�b +

cp� c �

sb+ cp�a +

rc+ap�b +

sa+bp� 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+ zx+z+ xy+x+ yz

�r2+

y+ zx+

r2+

z+ xy+

r2+

x+ yz:

By AM-GM Inequality, we have

4r2+

y+ zx

� 2+ y+ zx+4=

y+ zx+6:

It follows that

4

r2+

y+ zx+

r2+

z+ xy+

r2+

x+ yz

!� y+ z

x+z+ xy+x+ yz+18:

We have to prove

4�y+ zx+z+ xy+x+ yz

�� y+ z

x+z+ xy+x+ yz+18;

5

6 Solutions

or equivalently,y+ zx+z+ xy+x+ yz

� 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 a2+ 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=2xp

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

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

; c=2zp

(z+ x)(z+ y):

Using this substitution, we may write our inequality as

∑cyc

x2

(x+ y)(x+ z)�∑cyc

4y2z2

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

which is obviously true because

∑cyc

4y2z2

(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 =p2;c = 0 and its cyclic

permutations. �

Problem 2.3 Show that for any positive real numbers a;b;c; we have

a3+b3+ c3+6abc� 3pabc(a+b+ c)2:

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

6 3pabc(a+b+ c)2 � (a+b+ c)3+9 3

pa2b2c2(a+b+ c);

which leads us to prove the sharper inequality

6(a3+b3+ c3+6abc)� (a+b+ c)3+9 3pa2b2c2(a+b+ c);

or5(a3+b3+ c3)�3∑

cycab(a+b)+30abc� 9 3

pa2b2c2(a+b+ c);

From Schur's Inequality for third degree, we have

3(a3+b3+ c3)�3∑cycab(a+b)+9abc� 0;

7

and we deduce our inequality to

2(a3+b3+ c3)+21abc� 9 3pa2b2c2(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 3pa2b2c2(a+b+ c);

(a+b+ c)3+27abc� 18 3pa2b2c2(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 3

pa2b2c2(a+b+ c)

� 9 3pa2b2c2(a+b+ c)+27 3

pa2b2c2(a+b+ c)

= 36 3pa2b2c2(a+b+ c):

It shows that(a+b+ c)3+27abc� 18 3

pa2b2c2(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 =pab� 1:

Indeed, we have

P(a;b;c)�P(t; t;c) =

=�a3+b3�2ab

pab�� (a2+b2�2ab)�2c

�a+b�2

pab�

= (a�b)2(a+b)+ab�pa�

pb�2� (a�b)2�2c

�pa�

pb�2

=�pa�

pb�2 ��p

a+pb�2(a+b�1)+ab�2c

�� 0;

because �pa+

pb�2(a+b�1)+ab�2c� 4(2�1)+1�2= 3> 0:

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

2t3+ c3+6� (2t+ c)2;

2t3+1t6+6�

�2t+

1t2

�2;

2t9+6t6+1� t2(2t3+1)2;

2t9+6t6+1� 4t8+4t5+ t2;

2t9�4t8+6t6�4t5� t2+1� 0;

(t�1)2�2t7�2t5+2t4+2t3+2t2+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 minP = 3 with equality attains when a = 1;b = c = 0 and its cyclic permutions.Now, let us �nd maxP:We claim that maxP= 100

27 attains when a= b= c=13 ; or

(1+ab)2+(1+bc)2+(1+ ca)2 � 10027;

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� 46

27;

4(ab+bc+ ca)�6abc� 109;

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

4(ab+bc+ ca)�6abc� (1+9abc)�6abc= 1+3abc� 1+3 � 127=109:

With the above solution, we have the conclusion for the requirement is minP= 3 and maxP= 10027 : �

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 thatP(a;b;c)� 1+1+1= 3;

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

(1�4ab)2+(1�4bc)2+(1�4ca)2 � 13(1�4ab+1�4bc+1�4ca)2

=13[3�4(ab+bc+ ca)]2

� 13

�3�4 � 1

3

�2=2527;

with equality holds when a= b= c= 13 : And we conclude that minP=

2527 :

This completes the proof. �

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

+c+ab

+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)

�24(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;

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 � 4a2b2

a+b+4c2(a2+b2)a+b

;

ab�a+b� 4ab

a+b

�+ c(a�b)2 � 2c2

�2(a2+b2)a+b

�a�b�;

ab(a�b)2a+b

+ c(a�b)2 � 2c2(a�b)2a+b

;

aba+b

+ c� 2c2

a+b;

which is obviously true because2c2

a+b� 2c2

c+ c= c� c+ ab

a+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+ z2)(m2+n2+ p2)� (mx+ny+ pz)2 = (my�nx)2+(nz� py)2+(px�mz)2:

Now, applying this identity with x =pab;y =

pbc;z =

pca;m = (a+ b)

pab;n = (b+ c)

pbc; and p =

(c+a)pca; we obtain

(ab+bc+ ca)

"∑cycab(a+b)2

#�"∑cycab(a+b)

#2= abc∑

cycc(a�b)2:

Moreover,∑cycab(a+b)2�4(a2b2+b2c2+ c2a2) =∑

cycab(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)∑cycab(a�b)2 � abc∑

cycc(a�b)2;

∑cyca2b2(a�b)2+abc∑

cyc(a+b� c)(a�b)2 � 0;

11

∑cyca2b2(a�b)2+2abc∑

cyca(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)

pb+ 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)pa2� (b� c)2

� ∑cyc

(a+b)(a+ c)a

= 3(a+b+ c)+�bca+cab+abc

�� 4(a+b+ c);

where the last inequality is valid because

bca+cab+abc

= 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 a2+b2+ c2 = 3: Show that

a+bpa+b� c

+b+ cpb+ c�a

+c+apc+a�b

� 6:

Solution. Firstly, to prove the original inequality, we will show that1

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� 4a

3(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 prooffor it, this proof seems to be complicated but it is nice about its idea.

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)2and (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 that14�(2a�b� c)2+(2b� c�a)2

�� (a2�ab+b2)� 1

8(a+b�2c)2� 1

4(a+b)2; (2)

Anda2

(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)2� 12

��(2a�b� c)2+(2b� c�a)2

��

� 12

�2(a+b)2

(a+b+2c)2� 12

��(2a�b� c)2+(2b� c�a)2

�� 1

4

�2(a+b)2

(a+b+2c)2� 12

�(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)2

2(a+b+2c)2� 14(a+b)2:

Using this inequality, we have to prove

(a+b)2(a+b�2c)22(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)22(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+ cpb+ c�a

!2"∑cyc

a3(b+ c�a)(b+ c)2

#�

∑cyca

!3:

13

It follows that ∑cyc

b+ cpb+ c�a

!2�

�∑cyca�3

∑cyc

a3(b+c�a)(b+c)2

�4�

∑cyca�3

∑cycab

:

Moreover, by AM-GM Inequality, we have

∑cycab =

vuut13

∑cycab

! ∑cycab

! ∑cyca2!

�

vuuut13

0@ ∑cycab+ ∑

cycab+ ∑

cyca2

3

1A3 =�

∑cyca�3

9:

Hence ∑cyc

b+ cpb+ c�a

!2�4�

∑cyca�3

∑cycab

� 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 a2+b2+ c2 = 3: Show that

apb+ c�a

+bp

c+a�b+

cpa+b� c

� 3:

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

apb+ c�a

!2"∑cyca(b+ c�a)

#� (a+b+ c)3:

And we deduce our inequality to show that

(a+b+ c)3 � 9∑cyca(b+ c�a);

(a+b+ c)3 � 9 [2(ab+bc+ ca)�3] :

Setting p = a+ b+ c; then itp3 � p � 3 and 2(ab+ bc+ ca) = p2� 3: Thus, we can rewrite the above

inequality asp3 � 9(p2�6);

p3�9p2+54� 0;

(3� p)(18+6p� p2)� 0;

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

14 Solutions

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

aa+b

+bb+ c

+cc+a

� 1+

s2abc

(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+2

s2abc

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

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

;

Using the known inequality2

∑cyc

a2

(a+b)2� 1� 2abc

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

we can deduce it to

∑cyc

ab(a+b)(b+ c)

�

s2abc

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


;

a2b+b2c+ c2a+abc�p2abc(a+b)(b+ c)(c+a):

Now, we assume that c=minfa;b;cg ; applying AM-GM Inequality, we have

a2b+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)

� 2ra(a+ c)(b+ c)

2�bc(a+b)

=p2abc(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+

�cc+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

;

andaa+b

+bb+ c

+cc+a

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

:

2The proof will be left to the readers

15

Hence, the given inequality can be rewritten as

14+


� 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: �

Problem 2.12 Let a;b;c be positive real numbers. Prove that

(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 that �4a2+(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 becomes4a2+116ax

� a2+1�2xa+ x

;

16 Solutions

(4a2+1)(a+ x)� 16ax(a2+1�2x);

32ax2��16a3�4a2+16a�1

�x+a

�4a2+1

�� 0;

2a(4x�1)2+2a(8x�1)��16a3�4a2+16a�1

�x+a

�4a2+1

�� 0;

2a(4x�1)2+(1+4a2�16a3)x+a(4a2�1)� 0;

2a(4x�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=1y ;c=

1z ; then the above inequality becomes

(x+ y+ z)2(x2+ y2)(y2+ z2)(z2+ x2)� 8(x2y2+ y2z2+ z2x2)2:

Notice that(x2+ y2)(y2+ z2)(z2+ x2) = (x2+ y2+ z2)(x2y2+ y2z2+ z2x2)� x2y2z2:

Thus, we can rewrite the above inequality as

(x2y2+ y2z2+ z2x2)�(x+ y+ z)2(x2+ y2+ z2)�8(x2y2+ y2z2+ z2x2)

�� x2y2z2(x+ y+ z)2:

Now, we see that

(x+ y+ z)2(x2+ y2+ z2)�8(x2y2+ y2z2+ z2x2)= ∑

cycx4+2∑

cycxy(x2+ y2)+2xyz∑

cycx�6∑

cycx2y2

= ∑cycx2(x� y)(x� z)+3∑

cycxy(x� y)2+ xyz∑

cycx� xyz∑

cycx:

It suf�ces to prove that

xyz(x2y2+ y2z2+ z2x2)(x+ y+ z)� x2y2z2(x+ y+ z)2;

x2y2+ y2z2+ z2x2 � 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+ z2)(z2+ x2)(x+ y+ z)2 � 8(x2y2+ y2z2+ z2x2)2:

By AM-GM Inequality, we have

(x+ y+ z)2 = x2+ y2+ z2+2xy+2yz+2zx

� x2+ y2+ z2+4x2y2

x2+ y2+4y2z2

y2+ z2+4z2x2

z2+ x2:

17

Hence, it suf�ces to prove that

(m+n)(n+ p)(p+m)�m+n+ p+

4mnm+n

+4npn+ p

+4pmp+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+ z2)(z2+ x2)(x+ y+ z)2 � 8(x2y2+ y2z2+ z2x2)2:

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

(x2+ z2)(y2+ z2)� (xy+ z2)2:


(x2+ y2)(xy+ z2)2(x+ y+ z)2 � 8(x2y2+ y2z2+ z2x2)2;q2(x2+ y2)(xy+ z2)(x+ y+ z)� 4(x2y2+ y2z2+ z2x2);

We have q2(x2+ y2) = x+ y+

(x� y)2

x+ y+p2(x2+ y2)

� x+ y+ (x� y)2

x+ y+p2(x+ y)

= x+ y+

�p2�1

�(x� y)2

x+ y� x+ y+ 3(x� y)

2

8(x+ y):

Therefore q2(x2+ y2)(xy+ z2)(x+ y+ z)

= xy(x+ y)q2(x2+ y2)+ z(x+ z)(y+ z)

q2(x2+ y2)

� xy(x+ y)q2(x2+ y2)+ z(x+ y)(x+ z)(y+ z)+

3z2(x� y)28

+3(x� y)2xyz8(x+ y)

:


xy(x+ y)q2(x2+ y2)+ z(x+ y)(x+ z)(y+ z)+

3z2(x� y)28

+3(x� y)2xyz8(x+ y)

� 4(x2y2+ y2z2+ z2x2);

xy�(x+ y)

q2(x2+ y2)�4xy

�+ xyz

�x+ y+

3(x� y)28(x+ y)

��21x2�10xy+21y2

�z2

8+ z3(x+ y)� 0:

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

18 Solutions

We havexy�(x+ y)

q2(x2+ y2)�4xy

�� xz

�(x+ y)

q2(x2+ y2)�4xy

�:


f (z) = x�(x+ y)

q2(x2+ y2)�4xy

�+ xy

�x+ y+

3(x� y)28(x+ y)

��21x2�10xy+21y2

�z

8+ z2(x+ y)� 0:

Also, we havef (z)� f (y) = 1

8(y� z)

�21x2+13y2�18xy�8xz�8yz

�� 0:

Therefore

f (z) � f (y) = x�(x+ y)

q2(x2+ y2)�4xy

�� y(10x+13y)(x� y)

2

8(x+ y)

=2x(x� y)2(x2+ y2+4xy)(x+ y)

p2(x2+ y2)+4xy

� y(10x+13y)(x� y)2

8(x+ y)� 0;

since

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

p2(x2+ y2)+4xy

� y(10x+13y)8(x+ y)

� 2x(x2+ y2+4xy)2(x2+ y2)+4xy

� y(10x+13y)8(x+ y)

=8x3+22x2y�15xy2�13y3

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)�1a+1b+1c

�2� 8(a2+b2+ c2)2:

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

1py+ z� x +

1pz+ x� y +

1px+ y� z �

x+ y+ zpxyz :

By Holder Inequality, we have ∑cyc

1py+ z� x

!2"∑cycx3(y+ z� x)

#� (x+ y+ z)3:

It suf�ces to show thatxyz(x+ y+ z)�∑

cycx3(y+ z� x);

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

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+

pbc�2

= bc�a(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 ifa= 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) � 6427abc(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. �

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�a

a(2a+b+ c)� a�ba(2a+b+ c)

�= (a+b+ c)∑

cyc

�a�b

b(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)� 134� 274� aa+b+ c

:

It follows that

∑cyc

(b+ c)2

a(2a+b+ c)� 394� 274

�a

a+b+ c+

ba+b+ c

+c

a+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;

21

andb+ c

a(2a+b+ c)� c+ab(2b+ c+a)

� a+bc(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)22a(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

+c

a+b

�:

Solution. We see that(b+ c)2

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

+4a

2a+b+ c�2:

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

:

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

= ∑cyca

1b+ c

+1

2a+b+c2

!

� ∑cyc

4ab+ c+ 2a+b+c

2

= 8∑cyc

a2a+3b+3c

� 8(a+b+ c)2

∑cyca(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

#:


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

∑cycab(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�∑cycab(a+b)�3abc

a3+b3+ c3;

23

∑cyca(a�b)(a� c)

abc�

∑cyca(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 � 1

2and 14 � x� a(1�a): The above inequality becomes

a(1�3x)+ x(1�a)2 � (1�a)(a3+1�3x);

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

∑cyca6+3∑

cyca3b3+2abc∑

cyca3 �∑

cycab(a4+b4)+∑

cyca2b2(a2+b2)+abc∑

cycab(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;

Sa(b� c)2+Sb(c�a)2+Sc(a�b)2 � 0;

24 Solutions

where

Sa = b4+ c4�3b2c2+2abc(b+ c);Sb = c4+a4�3c2a2+2abc(c+a);Sc = a4+b4�3a2b2+2abc(a+b):

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

Sb+Sc = 2a4+b4�3a2b2+ c4+2abc(2a+b+ c)�3a2c2

� 2a4+b4�3a2b2+ c4+2ac2(2a+b+ c)�3a2c2

= 2a4+b4�3a2b2+ c4+ac2(a+2b+2c)> 2a4+b4�3a2b2 = (a2�b2)(2a2�b2)� 0:

It follows that

Sa(b� c)2+Sb(c�a)2+Sc(a�b)2 � Sb(c�a)2+Sc(a�b)2

� Sb(c�a)2�Sb(a�b)2

= Sb(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

+c

a+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):

Hencea3+b3+ c3 � a

b+ c+

bc+a

+c

a+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+4

p2�� a

b+ c+

bc+a

+c

a+b� 32

�:

25

Indeed, this inequality is equivalent to each of the following

a3+b3+ c3

abc�3�

�6+4

p2�� a

b+ c+

bc+a

+c

a+b� 32

�;

(a+b+ c)∑cyc(a�b)(a� c)

abc��3+2

p2�∑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+2

p2�(2a+b+ c)

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

a

�a(a+b+ c)

bc+1��3+2

p2�(2a+b+ c)

� (a+b+ c)(b+ c)a

�4a(a+b+ c)(b+ c)2

+1��3+2

p2�(2a+b+ c)

= (2a+b+ c)�(a+b+ c)(2a+b+ c)

a(b+ c)�3�2

p2�� 0;

and Y;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+2p+3(4�q)� 4q;

p2+2p+12� 9q:

If p� 4; then it is trivial because

p2+2p+12� 16+8+12= 36� 9q:

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

p3�4pq+9r � 0;4In here, we have p� 3 because p

2

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

26 Solutions

which yields thatp3�4pq+9(4�q)� 0;

or

q� p3+364p+9

:

It follows that

p2+2p+12�9q � p2+2p+12�9 � p3+364p+9

=(5p+18)(4� p)(p�3)

4p+9� 0:

This completes our proof. Equality holds if and only if a = b = c = 1 or a = b = 2;c = 0 and its cyclicpermutations. �

Problem 2.19 Let a;b;c be real numbers with minfa;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)� 3

16a(b+ c)>

18a(b+ c):

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

36 = 4(ab+bc+ ca)p3(ab+bc+ ca)

� 2(ab+bc+ ca)�a+b+ c+

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

�:

And we deduce the inequality to

3∑cyca3+27abc� 2

∑cyca

! ∑cycab

!+6(ab+bc+ ca)2

a+b+ c;

3∑cyca3+27abc�2

∑cyca

! ∑cycab

!� 6(ab+bc+ ca)

2

a+b+ c�2

∑cyca

! ∑cycab

!;

We have

3∑cyca3+27abc�4

∑cyca

! ∑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):

27

Hence, we may write the above inequality as


where

X =4(ab+bc+ ca)a+b+ c

+6a�2b�2c+ (b� c)2a+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 a2b2+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 haveq(a2+b2+ c2)3 �

q3(a2+b2+ c2)(a2b2+b2c2+ c2a2)

=q3(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);

∑cyca2(a�b)(a� c)+∑

cycab(a�b)2 � 0;

which is obviously true by Schur's Inequality for fourth degree. Equality holds if and only if a= b= c= 14p3

or a= b= 1;c= 0 and its cyclic permutations. �

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 � 3xyz(x+ y+ z)(xy+ yz+ zx);

and we can deduce the inequality to

(x+ y+ z)2(xy+ yz+ zx)+3xyz(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+ z2+ 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∑cycx2+3∑

cycxy:

From this, we may rewrite our inequality as

x(x2+ yz)y+ z

+y(y2+ zx)z+ x

+z(z2+ xy)x+ y

� x2+ y2+ z2;

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: �

29

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;

Sa(b� c)2+Sb(c�a)2+Sc(a�b)2 � 0;where

Sa =abc� 2a+b+ c

; Sb =bca� 2a+b+ c

; Sc =cab� 2a+b+ c

:

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

Sb+Sc =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 Sa � Sb � 0: And we obtain

Sa(b� c)2+Sb(c�a)2+Sc(a�b)2 � Sb(c�a)2+Sc(a�b)2

� Sb(c�a)2�Sb(a�b)2

= Sb(b� c)(2a�b� c)� 0:


Problem 2.23 Given ABC is a triangle. Prove that

8cos2Acos2Bcos2C+ cos2Acos2Bcos2C � 0:

Solution (by Honey_suck). Since cos2A+cos2B+cos2C+2cosAcosBcosC= 1 and cosAcosBcosC� 18 ;

it follows thatcos2A+ cos2B+ cos2C � 3

4;

and(1� cos2A� cos2B� cos2C)2 = 4cos2Acos2Bcos2C:

Setting a = cos2A;b = cos2B and c = cos2C; then a+ b+ c � 34 and (a+ b+ c� 1)

2 = 4abc: We need toprove that

8abc+(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+9abca+b+ c

=4(a+b+ c)3+9(a+b+ c�1)2

4(a+b+ c):

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)24(a+b+ c)

+1;

4(p�1)2+2p� 4p3+9(p�1)2

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

3(4p�3)(p�1)24p

� 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 �2maxfab;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);


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�2maxfab;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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Hoang Lan Huong