The power of substitution II
proving an inequality with three variables

Here's a problem by an anonymous author, communicated to me by Leo Giugiuc, along with a beautiful and instructive solution.

The power of  substitution II: proving an inequality with three variables

Solution

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Copyright © 1996-2017 Alexander Bogomolny

Let $a,b,c\,$ be non-negative numbers, no two zero. Prove that

$\displaystyle\frac{ab}{(a+b)^2}+\frac{bc}{(b+c)^2}+\frac{ca}{(c+a)^2}\le\frac{1}{4}+\frac{4abc}{(a+b)(b+c)(c+a)}.$

Solution

Define $x=\displaystyle\frac{a}{a+b},\,$ $y=\displaystyle\frac{b}{b+c},\,$ $z=\displaystyle\frac{c}{c+a}.\,$ From the definition, $1-x=\displaystyle\frac{b}{a+b},\,$ $1-y=\displaystyle\frac{c}{b+c},\,$ $1-z=\displaystyle\frac{a}{c+a}.\,$ Further, $x,y,z\in [0,1]\,$ and $xyz=(1-x)(1-y)(1-z).\,$ It follows that

$2xyz=1-(x+y+z)+(xy+yz+zx).$

The required inequality is rewritten as,

$4x(1-x)+4y(1-y)+4z(1-z)\le 1+16xyz.$

This is equivalent to

$4(x^2+y^2+z^2)-12(x+y+z)+8(xy+yz+zx)+9\ge 0,$

or

$4(x+y+z)^2-12(x+y+z)+9\ge 0,$

which is the same as

$[2(x+y+z)^2-3]^2\ge 0.$

That completes the proof.

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Copyright © 1996-2017 Alexander Bogomolny

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