Problem 6: Hard inequality

by henderson, Feb 16, 2016, 7:26 PM

$\bf\color{red}Problem \ 6 \$ If $x,y,z>0$ and $x\sqrt{y}+y\sqrt{z}+z\sqrt{x}=3$ , prove that:
$\frac{x}{\sqrt{x+2y}}+\frac{y}{\sqrt{y+2z}}+\frac{z}{\sqrt{z+2x}}\ge\sqrt{xyz+2}.$ $\bf\color{red}My \ solution \$ Using $\text{Holder inequality},$ we have
$\left(\sum_{cyc}\frac{x}{\sqrt{x+2y}}\right)\left(\sum_{cyc}\frac{x}{\sqrt{x+2y}}\right)\left(\sum_{cyc}\frac{x(x+2y)}{3\sqrt{3}}\right)\geq \left(\frac{x}{\sqrt{3}}+\frac{y}{\sqrt{3}}+\frac{z}{\sqrt{3}}\right)^3$ $\implies$ $\left(\sum_{cyc}\frac{x}{\sqrt{x+2y}}\right)^2\geq (x+y+z)$ $\implies$
$\sum_{cyc}\frac{x}{\sqrt{x+2y}}\geq \sqrt{x+y+z}.$ $\left(\boxed{1}\right)$
On the other hand, from the given condition we get $3=x\sqrt{y}+y\sqrt{z}+z\sqrt{x}\geq 3\sqrt[3]{(xyz)^{\frac{3}{2}}},$ so $xyz\leq 1$ $\left(\boxed{2}\right)$
$\text{Cauchy-Schwarz inequality}$ implies $\frac{1}{3}\left(x+y+z\right)^3\geq (x+y+z)(xy+yz+zx)\geq (x\sqrt{y}+y\sqrt{z}+z\sqrt{x})^2=9,$ so $x+y+z\geq 3$ $\left(\boxed{3}\right)$

Finally, using $\boxed{2}$ and $\boxed{3}$ in $\boxed{1},$ we conclude
$\sum_{cyc}\frac{x}{\sqrt{x+2y}}\geq \sqrt{x+y+z}\geq \sqrt{3}\geq \sqrt{xyz+2},$ as desired.

This post has been edited 6 times. Last edited by henderson, Sep 12, 2016, 2:41 PM

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Problem 6: Hard inequality

by henderson, Feb 16, 2016, 7:26 PM

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