﻿ If x,y,z >0, prove the equality \sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}> 2\sqrt{\dfrac{(x+y)(y+z)(z+x)}{xy+yz+zx}}

#### Question

If x,y,z >0, prove the equality

\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}> 2\sqrt{\dfrac{(x+y)(y+z)(z+x)}{xy+yz+zx}}

Collected in the board: Inequality

Steven Zheng posted 3 months ago

In an acute triangle, there exist x,y, z such that

\sqrt{x+y}=c, \sqrt{y+z}=a, \sqrt{z+x}=b, in which a, b,c are the three sides of the triangle,

Then

x+y = c^2, y+z=a^2, z+x=b^2

Solve the equation system, we get

x= \dfrac{-a^2+b^2+c^2}{2}

y = \dfrac{a^2+b^2-c^2}{2}

x=\dfrac{a^2-b^2+c^2}{2}

Substitue to RHS

2\sqrt{\dfrac{(x+y)(y+z)(z+x)}{xy+yz+zx}}

=\dfrac{2abc}{\sqrt{\dfrac{1}{4}( -a^4-b^4-c^4+2a^2b^2+2b^2c^2+2a^2c^2)}}

=\dfrac{4abc}{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}

Use Heron's formula

A = \dfrac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}

and

Area formula of a triabgle

A = \dfrac{abc}{4R}

R = \dfrac{abc}{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)} }

And

The equality is transformed to

a+b+c> 4R
(1)

According g to The Law of Sines,

\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R

The equality is further transformed to

\sin A+\sin B+\sin C> 2
(2)

Since \sin(x) is strictly concave on [0,\dfrac{\pi}{2} ] and the sequence (\dfrac{\pi}{2},\dfrac{\pi}{2},0) strictly majorizes the sequence (A,B,C) by Karamata's theorem,

\sin A+\sin B+\sin C>\sin \dfrac{\pi}{2}+\sin \dfrac{\pi}{2}+\sin 0=2

Therefore, we have proved the equality

\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}> 2\sqrt{\dfrac{(x+y)(y+z)(z+x)}{xy+yz+zx}}

is true.

Steven Zheng posted 3 months ago

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