Prove hard inequality Let $a,b,c>0$, prove that: $$\frac{1}{(2a+b)^2}+\frac{1}{(2b+c)^2}+\frac{1}{(2c+a)^2}\geq\frac{1}{ab+bc+ca}$$
I tried to use the inequality $\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\geq\frac{(a+b+c)^2}{x+y+z} \forall x,y,z>0$ but that's all I can do. I think this inequality is too tight to use AM-GM or Cauchy Schwarz alone.
 A: After full expanding we need to prove that:
$$\sum_{cyc}(4a^{5}b+4a^{5}c-12a^{4}b^{2}+12a^{4}c^{2}+5a^{3}b^{3}+8a^{4}bc-19a^{3}b^{2}c+5a^{3}c^{2}b-7a^{2}b^{2}c^{2})\geq0$$ or
$$6\sum_{cyc}ab(a^{2}-b^{2}-2ab+2ac)^{2}+
\sum_{sym}(2a^{5}b-a^{3}b^{3}-4a^{4}bc+10a^{3}b^{2}c-7a^{2}b^{2}c^{2})\geq0.$$
Thus, it's enough to prove that:
$$\sum_{cyc}(a^5b+a^5c-a^3b^3-4a^4bc+5a^3b^2c+5a^3c^2b-7a^2b^2c^2)\geq0.$$
Can you end it now?
A: Let $x=2a+b,y=2b+c,z=2c+a$, then the inequality becomes:$$\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\ge\dfrac{27}{5x^2+5y^2+5z^2-2xy-2yz-2zx}$$ By using $a^2+b^2+c^2\ge ab+bc+ca$,
$$\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}=\dfrac{\left(xy\right)^2+\left(yz\right)^2+\left(zx\right)^2}{x^2y^2z^2}\ge\dfrac{x^2yz+xy^2z+xyz^2}{x^2y^2z^2}=\dfrac{xyz\left(x+y+z\right)}{x^2y^2z^2}=\dfrac{x+y+z}{xyz}$$ Then, by using $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}$, $$\dfrac{x+y+z}{xyz}=\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\ge\dfrac{9}{xy+yz+zx}$$ Lastly, use $a^2+b^2+c^2\ge ab+bc+ca$ again, $$\dfrac{27}{5x^2+5y^2+5z^2-2xy-2yz-2zx}\le\dfrac{27}{3xy+3yz+3zx}=\dfrac{9}{xy+yz+zx}$$ Therefore, $\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\ge\dfrac{27}{5x^2+5y^2+5z^2-2xy-2yz-2zx}$ is correct. Can you get it?
