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!

$a=x+2y,b=y+2z,c=z+2x \implies x=\dfrac{a-2b+4c}{9},y=\dfrac{b-2c+4a}{9},z=\dfrac{c-2a+4b}{9}$
the inequality become:
$\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2} \ge \dfrac{27}{5(ab+bc+ac)-2(a^2+b^2+c^2)}$
now use UVW method:
$3u=a+b+c,3v^2=ab+bc+ac,w^3=abc \implies u\ge v\ge w$
$\iff \dfrac{(3v^2)^2-6uw^3}{w^6}\ge \dfrac{3}{3v^2-2u^2} \iff w^6+2u(3v^2-2u^2)w^3-3v^4(3v^2-u^2) \le 0$
let $w^3=x,f(x)=x^2+2u(3v^2-2u^2)x-3v^4(3v^2-2u^2)$
$2u(3v^2-2u^2) \ge 0 $
$f_{max}(x)=f(w^3|w=v)=f(v^3)$, when $w=v \implies u=v=w \implies f(v^3)=0 \implies f(x) \le 0 $
when $u=v=w \implies a=b=c \implies x=y=z$
QED.
