Dung So Phuc Chung Minh BDT

USING COMPLEX NUMBER TO PROVE INEQUALITIES

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I.Theorem

Let a b a b, , ', ' be real numbers

Let complex numbers z = a+bi and z’ = a'+b i' ( 2

1

i = − )

We have z + z' ≥ +z z'

II.Application

Example 1: Let x y, be real numbers Prove that

2 2

x − x+ + x + x+ ≥

x − x+ + x + x+ ≥

(x 1) 1 (x 1) 16 29

⇔ − + + + + ≥

Let complex numbers z= − +x 1 i , z'= − − +x 1 4i and z''= − +2 5i

We have: z+ =z' z''

using inequality z + z' ≥ +z z', we have 2 2

(x−1) + +1 (x+1) +16≥ 29

Example 2: Let a a b b1, 2, ,1 2 be real numbers Prove that:

(a +a ) +(b +b ) ≤ a +b + a +b

Solution

Let complex numbers z= +a1 b i1 and z'= +a2 b i2

We have z+ =z' (a1+a2) (+ b1+b i2)

using inequality z + z' ≥ +z z',we have:

(a +a ) +(b +b ) ≤ a +b + a +b

Example 3: Let a b c, , be real numbers Prove that:

a +ab b+ + a +ac c+ ≥ b + +bc c

Solution

We have

a +ab b+ + a +ac c+ ≥ b + +bc c

⇔  +  +  +  +  +  ≥  −  + + 

b b

z= + +a i , ' 3

c c

z = − − +a i

We have ' 3 3

+ = − + + 

using inequality z + z' ≥ +z z', we have:

+ +  + + +  ≥ − + + 

Example 4: Let x, y, z be positive real numbers such that x + y + z = 3 Prove

3 3

x +xy+y + y +yz+z + x + +xz z ≥

Solution

S= x +xy+y + y +yz+z + x + +xz z

We have

= + +  + + +  + + + 

= + + 

3 '

= + + 

3 ''

= + + 

z+ + =z z x+ + +y z x+ +y z i

using inequality z + z' + z''≥ + +z z' z'' , we have:

+ +  + + +  + + +  ≥ + + + + +

Thus

Example5: Let a, b, c be positive real numbers such that ab + bc +ca = abc

Prove that:

3

Solution

Let x 1,y 1,z 1

= = = , we have: x, y , z > 0 and x + y + z =1

LHS =

x + y + y + z + z + x

Let complex numbers z= +x 2yi , 'z = +y 2zi, ''z = +z 2xi

We have z+ +z' z''= + + +(x y z) 2(x+ +y z i)

using inequality z + z'+ z'' ≥ + +z z' z'' , we have:

x + y + y + z + z + x ≥ x+ +y z + x+ +y z

Thus

3

Example6: Let a, b, c be positive real numbers such that x + y + z ≤1 Prove

82

Solution

Let complex numbers z x 1i

x

= + , z' y 1i

y

= + , z'' z 1i

z

= +

We have z z' z'' (x y z) (1 1 1)i

x y z

+ + = + + + + +

using inenquality z + z'+ z'' ≥ + +z z' z'', we have:

On the other hand, we have

+ + + + + = + + + + + − + +

We will use the AM-GM inequality, we have

thus

(x y z) ( ) 81(x y z) ( ) 80(x y z) 18.9 80.1 82

then LHS x2 12 y2 12 z2 12 82

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