Prove that $\sum\limits_{\mathrm{cyc}}{\frac{1}{(x+2y)^2}} \geq\frac{1}{xy+yz+zx}$ for $x, y, z > 0$ 
Let $x,y,z>0$. Prove that
$$\frac{1}{(x+2y)^2}+\frac{1}{(y+2z)^2}+\frac{1}{(z+2x)^2} \geq\frac{1}{xy+yz+zx}.$$

I tried to apply Cauchy - Schwarz's inequality but I couldn't prove this inequality!

Assume $z=min\{x,y,z\}$, we prove two inequalities
$$$$1. $$\sum_{cyc} \frac{1}{(2x+y)^2} \geqslant \frac{1}{9xy}+\frac{8}{9(x+z)(y+z)}$$
which is true by expanding
$$$$2.$$\frac{1}{9xy}+\frac{8}{9(x+z)(y+z)} \geqslant \frac{1}{xy+yz+zx}$$
which is equivalent to $(x+y)(y+z)(z+x) \geqslant 8xyz$
