3-variable inequality

jhsinutopia

$x,y,z>0$
Prove that
$\sum_{cyc} x\sqrt{\dfrac{x}{x+y}}\le \sqrt{\dfrac{3(x^2+y^2+z^2)}{2}}$

kuing

由 CS 有
\begin{align*}
\sum x\sqrt{\frac x{x+y}}&\leqslant\sqrt{\sum x^2(y+z)\sum\frac x{(x+y)(y+z)}}\\
&=\sqrt{\frac{\sum x^2(y+z)\sum x(z+x)}{(x+y)(y+z)(z+x)}},
\end{align*}
则只需证
\[2\sum x^2(y+z)\sum x(z+x)\leqslant3(x^2+y^2+z^2)(x+y)(y+z)(z+x),\]
展开可整理为
\[(x+y+z)\sum xy(x-y)^2\geqslant0,\]
显然成立。