<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] After using Ravi subtitution, we need to prove
$$\sum \frac{x+y}{(2y+z+x)^2} \geqslant \frac{9}{8(x+y+z)}.$$
By the Cauchy-Schwarz inequality we get
$$\sum \frac{x+y}{(2y+z+x)^2} \geqslant \frac{\left[\displaystyle \sum (x+y)(42x+3y+55z) \right]^2}{\displaystyle \sum (2y+z+x)^2(x+y)(42x+3y+55z)^2}.$$
Therefore, we will show that
$$8(x+y+z)\left[\displaystyle \sum (x+y)(42x+3y+55z) \right]^2 \geqslant 9\sum (2y+z+x)^2(x+y)(42x+3y+55z)^2,$$
or
$$11700 \sum xy^2(y-z)^2 + \sum \left(1954xyz+6723x^2y+\frac{3104xy^2}{3}+\frac{5053yz^2}{3}\right)(z+x-2y)^2$$
$$\geqslant 0.$$
Done.
</s>