TITLE: Prove $\sum{\frac{1}{(x+2y)^2}} \geq\frac{1}{xy+yz+zx}$
QUESTION [8 upvotes]: 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!

REPLY [2 votes]: 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.