<s>[INST] For all triangle prove that $\sum\limits_{cyc}\frac{a}{(a+b)^2}\geq\frac{9}{4(a+b+c)}$ 
Let $a$, $b$ and $c$ be a sides-lengths of the triangle. Prove that:
  $$\frac{a}{(a+b)^2}+\frac{b}{(b+c)^2}+\frac{c}{(c+a)^2}\geq\frac{9}{4(a+b+c)}$$

The Buffalo way kills it, but I am looking for a nice proof for this nice inequality.
SOS (sums of squares) gives $\sum\limits_{cyc}\frac{(a-b)^2(ab+b^2-ac)}{(a+b)^2}\geq0$ and I don't see what comes next.
 [/INST] Have: $\Leftrightarrow \left(\sum _{cyc}a\right)\left(\sum _{cyc}\frac{a}{\left(b+c\right)^2}\right)\ge \frac{9}{4}$
By C-S and Nesbitt : $LHS\ge \left(\frac{a}{b+c}+\frac{b}{b+c}+\frac{c}{c+a}\right)^2=\frac{9}{4}$
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