chebyshev inequality voquocbacan

8/9/2019 Chebyshev Inequality VoQuocBaCan

1/12

Chebyshev inequality

Vo Quoc Ba CanStudent, Can Tho University, Vietnam

Chebyshev inequality is a useful instrument in solving inequalities where therearrangement of variables is possible. In this short note, we will give some tech-niques to use this nice inequality.

Theorem 1 (Chebyshev inequality). Consider two sequences ofn real numbers(a1, a2, . . . , an) and (b1, b2, . . . , bn). If the two sequences are sorted, that is a1 a2 an, and b1 b2 bn (or a1 a2 an, and b1 b2 bn) then

a1b1 + a2b2 + + anbn 1

n (a1 + a2 + + an)(b1 + b2 + + bn)

Equality occurs if and only if a1 = a2 = = an or b1 = b2 = = bn.Indeed, we have

n(a1b1 + a2b2 + + anbn) (a1 + a2 + + an)(b1 + b2 + + bn)

=n

i,j=1

(ai aj)(bi bj) 0

Since (a1, a2, . . . , an) and (b1, b2, . . . , bn) are sorted then (ai aj)(bi bj) 0.We have the following corollary

Corollary 1 (Corollary of Chebyshev inequality). Ifa1

a2

an

,

and b1 b2 bn (or a1 a2 an, and b1 b2 bn) then

a1b1 + a2b2 + + anbn 1n

(a1 + a2 + + an)(b1 + b2 + + bn)

Equality occurs if and only if a1 = a2 = = an or b1 = b2 = = bn.Example 1 (Cauchy Schwarz inequality). Givenn real numbers a1, a2, . . . , an,prove that

n(a21 + a22 + + a2n) (a1 + a2 + + an)2

Solution. This is direct consequence of Chebyshev inequality. We apply Chebyshevinequality for sequence (a1, a2, . . . , an) and itself.

Example 2. Given a,b,c,d be positive real numbers such that

a2 + b2 + c2 + d2 = 1,

prove thata2

b + c + d+

b2

c + d + a+

c2

d + a + b+

d2

a + b + c 2

3

1


2/12

Solution. Without loss of generality, we suppose that a b c d. We canestablish two sequences easily

a2 b2 c2 d2,1

b + c + d 1

c + d + a 1

d + a + b 1

a + b + c

Now applying Chebyshev inequality, we have

LHS 14

(a2 + b2 + c2 + d2)

1

a + b + c+

1

b + c + d+

1

c + d + a+

1

d + a + b

By AM - HM inequality,

1

a + b + c +

1

b + c + d +

1

c + d + a +

1

d + a + b 16

3(a + b + c + d)

By Cauchy Schwarz inequality,

a + b + c + d

4(a2 + b2 + c2 + d2) = 2

Adding up these inequalities, we have the result. Equality holds if and only ifa = b = c = d = 1

2.

Example 3. Let a1, a2, . . . , an be positive real numbers with sum 1, prove that

a1

2 a1 +a2

2 a2 + +an

2 an n

2n 1

Solution. Since this inequality is symmetric with a1, a2, . . . , an, without loss ofgenerality, we suppose that a1 a2 an, then

1

2 a1 1

2 a2 1

2 anNow, applying Chebyshev inequality, we have

LHS 1n

(a1 + a2 + + an)

1

2 a1 +1

2 a2 + +1

2 an

n2n (a1 + a2 + + an) =

n

2n 1

Equality holds if and only if a1 = a2 = = an =1

n .

Example 4. Let a1, a2, . . . , an be positive real numbers with sum 1, prove that

a11 a1

+a2

1 a2+ + an

1 an

a1 +

a2 + + an

n 1

2


3/12

Solution. Without loss of generality, assume a1 a2 an, then1

1 a1 1

1 a2 1

1 anNow, by Chebyshev inequality,

LH S 1n

(a1 + a2 + + an)

11 a1

+1

1 a2+ + 1

1 an

By AM - HM inequality,

11 a1

+1

1 a2+ + 1

1 an n

2

1 a1 +

1 a2 + +

1 an

By Cauchy Schwarz inequality,

1 a1 +

1 a2 + +

1 an

n(n a1 a2 an) =

n(n 1)

a1 +

a2 + + an

n(a1 + a2 + + an) =

n

Adding up these inequalities, the desired inequality is proved. Equality hold if andonly if a1 = a2 = = an = 1n .Example 5. Let a, b,c,d,e > 0 such that

1

4 + a+

1

4 + b+

1

4 + c+

1

4 + d+

1

4 + e= 1,

prove that

a4 + a2 +

b4 + b2 +

c4 + c2 +

d4 + d2 +

e4 + e2 1

Solution. The condition can be rewritten as

1 a4 + a

+1 b4 + b

+1 c4 + c

+1 d4 + d

+1 e4 + e

= 0

We have to prove cyc

1

4 + acyc

a

4 + a2

cyc

1 a(4 + a)(4 + a2)

0

Without loss of generality, suppose that a b c d e, then we can easilyestablish that

1 a4 + a

1 b4 + b

1 c4 + c

1 d4 + d

1 e4 + e

1

4 + a2 1

4 + b2 1

4 + c2 1

4 + d2 1

4 + e2

3


4/12

Applying Chebyshev inequality, we have

LHS cyc

1

5

1 a4 + a

cyc

1

4 + a2

= 0

Equality holds if and only if a = b = c = d = e = 1.

Example 6. Let a,b,c,d be positive real numbers such that a + b + c + d = 4,prove that

1

11 + a2+

1

11 + b2+

1

11 + c2+

1

11 + d2 1

3

Solution. We can rewrite the inequality as

cyc

112

1

11 + a2 0

cyc

a2 1a2 + 11

0

cyc

(a 1) a + 1a2 + 11

0

Without loss of generality, assume a b c d > 0, then it is easy to verify thata + 1

a2 + 11 b + 1

b2 + 11 c + 1

c2 + 11 d + 1

d2 + 11

By Chebyshev inequality, we have

LHS 14

cyc

(a 1)

cyc

a + 1

a2 + 11

= 0

Equality holds if and only if a = b = c = d = 1.

Example 7. Prove that for ifa , b, c are three sides of a triangle such that a+b+c =1, we have

x + y zz2 + xy

+

y + z x

x2 + yz+

z + x y

y2 + zx 2

Solution. Note that (m + n + p)2 m2 + n2 + p2 for any m,n,p 0, hence itsuffices to prove

cyc

x + y zz2 + xy

4

cyc

(x + y z)(x + y + z)z2 + xy

4

4


5/12

cyc (x + y)2 z2

z

2

+ xy

+ 1 7

cyc

(x + y)2

z2 + xy+cyc

xy

z2 + xy 7

We will show that cyc

xy

z2 + xy 1

and cyc

(x + y)2

z2 + xy 6

The first is trivial by Cauchy Schwarz inequality, indeed

cyc

xyz2 + xy

cyc

xy2z2 + xy

(xy + yz + zx)2cyc xy(2z

2 + xy)= 1

For the second, rewrite it as

cyc

x2 + y2 2z2z2 + xy

0

Since x, y, z are three sides of a triangle then the sequences

(y2 + z2 2x2, z2 + x2 2y2, x2 + y2 2z2)and

1x2 + yz , 1y2 + zx , 1z2 + xy

are sorted, hence by Chebyshev inequality,

LHS 13

cyc

(x2 + y2 2z2)

cyc

1

z2 + xy

= 0

Equality holds if and only if (x, y, z) =12

, 12

, 0

.

Now we are going to find more applications in which the use of Chebyshevinequality is really delicate.

Example 8. Let a,b,c,d be positive real numbers such that

a + b + c + d =1

a+

1

b+

1

c+

1

d

prove that

2(a + b + c + d)

a2 + 3 +

b2 + 3 +

c2 + 3 +

d2 + 3

5


6/12

Solution. We can rewrite the inequality ascyc

2a

a2 + 3

0

cyc

a2 12a +

a2 + 3

0

We cant use Chebyshev inequality now since the sequences

(a2 1, b2 1, c2 1, d2 1)

and

1

2a +

a2 + 3,

1

2b +

b2 + 3,

1

2c +

c2 + 3,

1

2d +

d2 + 3arent sorted, but if we rewrite it as

cyc

a21a

2 +

1 + 3a2

0

We can apply Chebyshev inequality now since the functions f(x) = x21x

andg(x) = 1

2+1+ 3

x2

increase on the interval (0, +), hence by Chebyshev inequality

cycf(a)g(a) 1

4

cyc

f(a)

cyc

g(a)

But cyc

f(a) =cyc

a2 1a

=cyc

a cyc

1

a= 0

Adding them up, we have the result. Equality hold if and only if a = b = c = d =1.

The special thing in above solution is we cant use Chebyshev inequality im-mediatly but after some changes, we can apply Chebyshev inequality to solve.The generalization is

To prove the inequality x1y1 + x2y2 + + xnyn 0, we can provex1

a1 (a1y1) +x2

a2 (a2y2) + +xn

an (anyn) 0where a1, a2, . . . , an are real numbers such that

x1

a1,

x2

a2, . . . ,

xn

an

6


7/12

and

(a1y1, a2y2, . . . , anyn)are sorted, then after using Chebyshev inequality, we can deduce the inequality toprove the more simple

x1

a1+

x2

a2+ + xn

an 0

anda1y1 + a2y2 + + anyn 0

Here are some examples

Example 9. Leta,b,c be positive real numbers such that a + b + c = 3, prove that

1

9

ab+

1

9

bc+

1

9

ca 3

8

Solution. Setting x = ab,y = bc,z = ca, we have to prove

cyc

1

9 x 3

8

cyc

1 x9 x 0

Now, we must choose ax, ay, az such that

(ax(1 x), ay(1 y), az(1 z)) and

1

(9 + x)ax,

1

(9 + y)ay,

1

(9 + y)ay

are sorted. Let ax = 6 + x, ay = 6 + y, az = 6 + z. Notes that max{y + z, z + x, x +y} 9

4(why?1), then if x y, we have

(1x)(6+x)(1y)(6+y) = (xy)(x+y+3) 0 (1x)(6+x) (1y)(6+y)1

(9 x)(6 + x) 1

(9 y)(6 + y) =(x y)(x + y 3)

(9 x)(9 y)(6 + x)(6 + y) 0

1(9 x)(6 + x)

1

(9 y)(6 + y)Now, since the inequality is symmetric, we can assume x y z and from theabove, we can establish

(1 x)(6 + x) (1 y)(6 + y) (1 z)(6 + z)1

(9 x)(6 + x) 1

(9 y)(6 + y) 1

(9 z)(6 + z)1y + z = a(b + c) 1

4(a + b + c)2 = 9

4, similarly z + x 9

4, x + y 9

4

7


8/12

Hence by Chebyshev inequality, we have

cyc

1 x9 x =

cyc

(1 x)(6 + x)(9 x)(6 + x)

1

3

cyc

(1 x)(6 + x)

cyc

1

(9 x)(6 + x)

Thus, it suffices to provecyc

(1 x)(6 + x) = 18 15(x + y + z) x2 y2 z2 0

a2b2 + b2c2 + c2a2 + 5(ab + bc + ca) 18 (ab + bc + ca)2 + 5(ab + bc + ca) 6(abc + 3)

By Schur inequality, we have 3(abc + 3)

4(ab + bc + ca), it suffices to prove

(ab + bc + ca)2 + 5(ab + bc + ca) 8(ab + bc + ca)

ab + bc + ca 3which is trivial since a + b + c = 3. The desired inequality is proved. Equalityholds if and only if a = b = c = 1.

Example 10 (Vasile Cirtoaje). Prove that for any a,b,c,d 0 such that a +b + c + d = 4, we have

1

5 abc +1

5 bcd +1

5 cda +1

5 dab 1

Solution. Setting x = abc, y = bcd, z = cda, t = dab, we have to provecyc

1

5 x 1

cyc

1 x5 x 0

cyc

(1 x)(2 + x)(5 x)(2 + x) 0

Note that x + y = bc(a + d) 127

(a + b + c + d)3 = 6427

< 3 then if x y, we have

(1x)(2+x)(1y)(2+y) = (xy)(x+y+1) 0 (1x)(2+x) (1y)(2+y)1

(5 x)(2 + x) 1

(5 y)(2 + y) =(x y)(x + y 5)

(5 x)(5 y)(x + 2)(y + 2) 0

1(5 x)(2 + x)

1

(5 y)(2 + y)

8


9/12

Hence the sequences

((1 x)(2 + x), (1 y)(2 + y), (1 z)(2 + z), (1 t)(2 + t))and

1

(5 x)(2 + x) ,1

(5 y)(2 + y) ,1

(5 z)(2 + z) ,1

(5 t)(2 + t)

are sorted. Applying Chebyshev inequality, we have

LHS 14

cyc

(1 x)(2 + x)

cyc

1

(5 x)(2 + x)

It suffice to prove

cyc

(1

x)(2 + x) = 8

x

y

z

t

x2

y2

z2

t2

0

a2b2c2 + b2c2d2 + c2d2a2 + d2a2b2 + abc + bcd + cda + dab 8This inequality can be proved easily by Mixing variables (do it!). The inequalityis proved. Equality holds if and only if a = b = c = d = 1.

Example 11. Leta , b, c 0 such that a + b + c = 3, prove that1

a2 a + 3 +1

b2 b + 3 +1

c2 c + 3 1

Solution. The inequality is equivalent to

cyc1

3 1

a2 a + 3 0cyc

a(a 1)a2 a + 3 0

cyc

a 1a 1 + 3

a

0

Assume that a b c, then a 1 b 1 c 1. Since a + b + c = 3 thenab, bc, ca 3, therefore

1

a 1 + 3a

1b 1 + 3

b

1ca 1 + 3

c

By Chebyshev inequality,

LHS 13

cyc

(a 1)

cyc

1

a 1 + 3a

= 0

Equality holds if and only if a = b = c = 1.

9


10/12

Example 12. Prove that for any a,b,c 0 and 2 k 0, we havea2 bc

b2 + c2 + ka2+

b2 cac2 + a2 + kb2

+c2 ab

a2 + b2 + kc2 0

Solution. Rewrite the inequality as

cyc

(a2 bc)(b + c)(ka2 + b2 + c2)(b + c)

0

Note that if a b then

(a2 bc)(b + c) (b2 ca)(c + a) = (a b)(c2 + ab + 2bc + 2ca) 0

(a2

bc)(b + c)

(b2

ca)(c + a)

(ka2+b2+c2)(b+c)(kb2+c2+a2)(c+a) = (ab)(a2+b2+c2(k1)(ab+bc+ca)) 0

1(ka2 + b2 + c2)(b + c)

1(kb2 + c2 + a2)(c + a)

Therefore the sequences

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

and 1

(ka2 + b2 + c2)(b + c),

1

(kb2 + c2 + a2)(c + a),

1

(kc2 + a2 + b2)(a + b)

are sorted. Hence applying Chebyshev inequality, we have

LHS 13

cyc

(a2 bc)(b + c)

cyc

1

(ka2 + b2 + c2)(b + c)

Moreover, we have cyc

(a2 bc)(b + c) = 0

We are done. Equality holds if and only if a = b = c for k < 2, for k = 2 equalityholds also for (a , b, c) (1, 1, 0).

Example 13. For any a , b, c

0, then

3(a + b + c)

a2 + 8bc +

b2 + 8ca +

c2 + 8ab

10


11/12

Solution. The inequality is equivalent tocyc

3a a2 + 8bc 0

a2 bc

3a +

a2 + 8bc 0

cyc

(a2 bc)(b + c)3a +

a2 + 8bc

(b + c)

0

Now, notes that if a b then(a2 bc)(b + c) (b2 ca)(c + a) = (a b)(c2 + ab + 2bc + 2ca) 0

(a2 bc)(b + c) (b2 ca)(c + a)We will show that

3a +

a2 + 8bc

(b + c)

3b +

b2 + 8ca

(c + a)

c(a b)

8a2 + 8b2 + 8c2 + 6ab + 15bc + 15ca

(b + c)

a2 + 8bc + (c + a)

b2 + 8ca 3

0

By AM - GM inequality,

8a2 + 8b2 + 8c2 + 6ab + 15bc + 15ca

(b + c)

a2 + 8bc + (c + a)

b2 + 8ca 3

2(8a2 + 8b2 + 8c2 + 6ab + 15bc + 15ca)

((b + c)2 + a2 + 8bc) + ((c + a)2 + b2 + 8ca) 3

= 5a2

+ 5b2

+ 5c2

+ 6aba2 + b2 + c2 + 5bc + 5ca

> 0

Therefore the sequences

((a2 bc)(b + c), (b2 ca)(c + a), (c2 ab)(a + b))and

13a +

a2 + 8bc

(b + c)

,1

3b +

b2 + 8ca

(c + a),

13c +

c2 + 8ab

(a + b)

are sorted. Applying Chebyshev inequality, we have

cyc

(a2 bc)(b + c)3a + a2 + 8bc (b + c)

13

cyc

(a2 bc)(b + c)

cyc

13a +

a2 + 8bc

(b + c)

= 0

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

11


12/12

Exercise

1. Prove that if a,b,c are three sides of a triangle then

a2 bc3a2 + b2 + c2

+b2 ca

3b2 + c2 + a2+

c2 ab3c2 + a2 + b2

0

2. If a, b,c,d > 0 such that

a2 + b2 + c2 + d2 = 4

then1

3 abc +1

3 bcd +1

3 cda +1

3 dab 2

3. If a1, a2, . . . , an are nonnegative real numbers such that

a1 + a2 + + an = 1a1

+1

a2+ + 1

an

prove that1

a21 + n 1+

1

a22 + n 1+ + 1

a2n + n 1 1

4. If a, b,c > 1 such that

1

a2 1 +1

b2 1 +1

c2 1 = 1

prove that

1a + 1

+ 1b + 1

+ 1d + 1

1

PS: We will back to these problems with another technique UndeterminedCoefficient Technique (UCT), this is also a useful tool to solve an inequality.

12

chebyshev inequality voquocbacan

Documents