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: $a=x+2y,b=y+2z,c=z+2x \implies x=\dfrac{a-2b+4c}{9},y=\dfrac{b-2c+4a}{9},z=\dfrac{c-2a+4b}{9}$
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.
A: 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}$$
which is equivalent to $(x+y)(y+z)(z+x) \geqslant 8xyz$
A: Also we can make the following.
From my previous post we need to prove that:
$$\sum_{cyc}(4a^5b+4a^4c-12a^4b^2+12a^4c^2+5a^3b^3+8a^4bc-19a^3b^2c+5a^3c^2b-7a^2b^2c^2)\geq0$$ or
$$6\sum_{cyc}ab(a^2-b^2-2ab+2ac)^2+\sum_{sym}(2a^5b-a^3b^3-4a^4bc+10a^3b^2c-7a^2b^2c^2)\geq0,$$
which is obviously true.
A: Alternative proof without full expanding or uvw:
Let
$$u = \frac{1}{x + 2y}, \quad
v = \frac{1}{y + 2z}, \quad w = \frac{1}{z + 2x}.$$
We have
\begin{align*}
 &\mathrm{LHS} - \mathrm{RHS} \\
 =\,& u^2 + v^2 + w^2 - \frac{1}{xy + yz + zx}\\
 =\,& \frac{(u-v)^2 + (v - w)^2 + (w-u)^2}{2} + uv + vw + wu - \frac{1}{xy + yz + zx}\\
 =\,& \frac{(u-v)^2 + (v - w)^2 + (w-u)^2}{2} + \frac{3(x + y + z)}{(x + 2y)(y + 2z)(z + 2x)} - \frac{1}{xy + yz + zx}\\
 =\,& \frac{(u-v)^2 + (v - w)^2 + (w-u)^2}{2}
 - \frac{(y-x)(y-z)(x-z)}{(x+2y)(y+2z)(z+2x)(xy+yz+zx)}. \tag{1}
\end{align*}
WLOG, assume that $z = \min(x, y, z)$.
From (1), we only need to prove the case $y \ge x \ge z$.
If $2x \ge y + z$, from (1),
we have
\begin{align*}
 \mathrm{LHS} - \mathrm{RHS}
 &\ge \frac{(v - u + v - w + w - u)^2}{6} - \frac{(y-x)\cdot (y - z + x - z)^2/4}{(x+2y)(y+2z)(y+2z)xy} \tag{2}\\
 &= \frac{2(x + y - 2z)^2}{3(x + 2y)^2(y + 2z)^2} - \frac{(y-x) (x + y - 2z)^2}{4(x+2y)(y+2z)^2xy}\\
 &= \frac{(x + y - 2z)^2}{(x+2y)(y+2z)^2}
 \left(\frac{2}{3(x + 2y)} - \frac{y-x}{4xy}\right)\\
 &\ge 0
\end{align*}
where in (2) we use $(u-v)^2 + (v - w)^2 + (w-u)^2
\ge (v-u + v - w + w - u)^2 /3$, and
$(y - z)(x - z) \le (y - z + x - z)^2/4$, and $z + 2x \ge y + 2z$,
and $xy + yz + zx \ge xy$;
and the last inequality follows from
$\frac{2}{3(x + 2y)} - \frac{y-x}{4xy}
\ge \frac{2}{3(x + 4x)} - \frac{y-x}{4xy} = \frac{15x  -7y}{60xy} \ge 0$.
If $2x < y + z$, from (1), we have
\begin{align*}
 \mathrm{LHS} - \mathrm{RHS}
 &\ge \frac{(v - u + w - v + w - u)^2}{6} - \frac{(x-z)(2y - x - z)^2/4}{(x+2y)(z+2x)(z+2x)xy} \tag{3}\\
 &= \frac{2(2y - x - z)^2}{(x+2y)^2(z+2x)^2} - \frac{(x-z)(2y - x - z)^2}{4(x+2y)(z+2x)^2xy}\\
 &= \frac{(2y - x - z)^2}{(x + 2y)(z + 2x)^2}
 \left(\frac{2}{3(x + 2y)} - \frac{x - z}{4xy}\right)\\
 &\ge 0
\end{align*}
where in (3) we use
$(u-v)^2 + (v-w)^2 + (w-u)^2 \ge (v - u + w - v + w - u)^2/3$, and $(y - x)(y - z) \le (y - x + y - z)^2/4$,
and $y+2z \ge z + 2x$, and $xy + yz + zx \ge xy$;
and the last inequality follows from
$\frac{2}{3(x + 2y)} - \frac{x - z}{4xy} = \frac{2xy + 3zx + 6yz - 3x^2}{12(x+2y)xy} \ge 0
$ (using $2xy + 2zx = 2x(y + z) \ge 2x \cdot 2x = 4x^2$).
We are done.
