By rearrangement or AM-GM $$x^2 + y^2 + 2z^2\ge xy+yz+zx+z^2=(x+z)(y+z)$$Again AM-GM and above gives $$2\sqrt {\frac {xy}{x^2 + y^2 + 2z^2}}\le 2\sqrt {\frac {xy}{(x+z)(y+z)}}\le\frac{x}{x+z}+\frac{y}{y+z}$$Similarly $$2\sqrt {\frac {yz}{y^2 + z^2 + 2x^2}}\le \frac{y}{x+y}+\frac{z}{x+z}$$$$2\sqrt {\frac {zx}{z^2 + x^2 + 2y^2}}\le \frac{z}{y+z}+\frac{x}{x+y}$$Add these and write "QED".

nice sol.

By $CSB$ : $(x^2+y^2)(1+1)\ge ( x+y)^2$ we get that $\sum \sqrt{\frac {xy}{x^2 + y^2 + 2z^2}} \leq \sum \sqrt {\frac {xy}{\frac{(x + y)^2 +(x +z)^2}{2}}}$ . By $AM-QM$ : $\sum \sqrt {\frac {xy}{\frac{(x + y)^2 +(x +z)^2}{2}}} \leq \sum \frac {\sqrt{xy}}{\frac{(x+y)+(x+z)}{2}} $ and now my $AM-GM$ we have $\sum \frac {\sqrt{xy}}{\frac{(x+y)+(x+z)}{2}} \leq \sum \sqrt{\frac{xy}{(x+y)(x+z)}}$ By $GM-AM$ we get: $\sqrt{\frac{xy}{(x+y)(x+z)}} \leq \frac{\sum (\frac{x}{x+z}+\frac{y}{x+y})}{2}=\frac{3}{2}$

Let $x$, $y$, and $z$ be positive real numbers. Prove that $$\sqrt{\frac{xy}{2z^2+xy}}+ \sqrt{\frac{yz}{2x^2+yz}}+ \sqrt{\frac{zx}{2y^2+zx}}\leq2$$