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# Inequality 11(2009 Mediterranean Mathematical Olympiad)

Problem:

If $\displaystyle x,y,z$ are positive real numbers prove that

$\displaystyle \frac{xy}{xy+x^2+y^2}+\frac{yz}{yz+y^2+z^2}+\frac{zx}{zx+z^2+x^2}\leq \frac{x}{2x+z}+\frac{y}{2y+x}+\frac{z}{2z+y}$.

Solution (An idea by Vo Quoc Ba Can):

Using the Cauchy-Schwarz Inequality we have that

$\displaystyle \frac{x}{2x+z}+\frac{y}{2y+x}+\frac{z}{2z+y}=\frac{x^2}{2x^2+xz}+\frac{y^2}{2y^2+yx}+\frac{z^2}{2z^2+zy}\geq \frac{(x+y+z)^2}{2(x^2+y^2+z^2)+xy+yz+zx}$.

Thus it suffices to show that

$\displaystyle \sum_{cyc}\frac{xy}{x^2+xy+y^2}\leq \frac{(x+y+z)^2}{2(x^2+y^2+z^2)+xy+yz+zx}$,

which is equivalent to

$\displaystyle \sum_{cyc}\left(\frac{1}{3}-\frac{xy}{x^2+xy+y^2}\right)\geq 1-\frac{(x+y+z)^2}{2(x^2+y^2+z^2)+xy+yz+zx}$,

or

$\displaystyle \sum_{cyc}\frac{(x-y)^2}{3(x^2+xy+y^2)}\geq \frac{x^2+y^2+z^2-xy-yz-zx}{2(x^2+y^2+z^2)+xy+yz+zx}$.

Since $\displaystyle x^2+y^2+z^2-xy-yz-zx=\frac{(x-y)^2+(y-z)^2+(z-x)^2}{2}$ the above inequality can be rewritten as

$\displaystyle \sum_{cyc}(x-y)^2\left[\frac{1}{3(x^2+xy+y^2)}-\frac{1}{2(x^2+y^2+z^2)+xy+yz+zx}\right]\geq 0$,

which is obviously true, Q.E.D.

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