prove that $\frac{2}{b(a+b)}+\frac{2}{c(b+c)}+\frac{2}{a(c+a)} \ge \frac{27}{(a+b+c)^2}$ 
Prove that $$\frac{2}{b(a+b)}+\frac{2}{c(b+c)}+\frac{2}{a(c+a)} \ge \frac{27}{(a+b+c)^2},$$ where $a,b,c$ are positive reals.

After applying AM-GM I got
$$
\frac{2}{b(a+b)}+\frac{2}{c(b+c)}+\frac{2}{a(c+a)}
\ge \frac{36}{(a+b+c)^2+a^2+b^2+c^2},
$$
which is similar to original expression but not quite enough. Any hints??

By Holder $$\sum_{cyc}\frac{2}{b(a+b)}=\frac{1}{(a+b+c)^2}\sum_{cyc}b\sum_{cyc}(a+b)\sum_{cyc}\frac{1}{b(a+b)}\geq\frac{27}{(a+b+c)^2}.$$
