USA tst 2010 day 1 problem 2

[SANZEED]

Let $a,b,c$ be positive real numbers such that $abc=1$. Show that
$\frac{1}{a^{5}(b+2c)^{2}}+\frac{1}{b^{5}(c+2a)^{2}}+\frac{1}{c^{5}(a+2b)^{2}}\geq \frac{1}{3}$.

[SANZEED]

Let $x=\frac{1}{a}, y=\frac{1}{b}, z=\frac{1}{c}$ then by calculation,
$\sum_{cyclic}\frac{1}{a^{5}(b+2c)^{2}}=\sum_{cyclic}\frac{x^{3}}{(2y+z)^{2}}$.
But by Holder, $(\sum_{cyclic}\frac{x^{3}}{(2y+z)^{2}})(\sum_{cyclic}2y+z)(\sum_{cyclic}2y+z)\geq (x+y+z)^{3}$.
Now notice that $(\sum_{cyclic}2y+z)^{2}=9(x+y+z)^{2}$ and $x+y+z\geq 3$ (by AM-GM). The result follows from these.