Prove that $\sum \limits_{cyc}\frac {a}{(b+c)^2} \geq \frac {9}{4(a+b+c)}$

Since the inequality is homogeneous, WLOG assume that $a+b+c = 1$.

Then, the inequality becomes $\dfrac{a}{(1-a)^2}+\dfrac{b}{(1-b)^2}+\dfrac{c}{(1-c)^2} \ge \dfrac{9}{4}$.

Since the function $f(x) = \dfrac{x}{(1-x)^2}$ is concave up for $x > 0$, by Jensen's Inequality, we have:

$f(a)+f(b)+f(c) \ge 3f\left(\dfrac{a+b+c}{3}\right) = 3f\left(\dfrac{1}{3}\right) = \dfrac{9}{4}$, as desired.

Well I found a different approach for solving this problem. We could rewrite the inequality from the question as:

$$(a+b+c)\left(\sum \limits_{cyc}\frac {a}{(b+c)^2}\right) \ge \frac{9}{4}$$

By the Cauchy-Schwarz inequality:

$$(a+b+c)\left(\sum \limits_{cyc}\frac {a}{(b+c)^2}\right) \ge \left(\sum \limits_{cyc}\frac {a}{b+c}\right)^2$$

And by Nesbitt's inequality:

$$\left(\sum \limits_{cyc}\frac {a}{b+c}\right)\ge \frac{3}{2}$$

So

$$\left(\sum \limits_{cyc}\frac {a}{b+c}\right)^2 \ge \frac{9}{4}$$

Let $x=b+c,y=c+a,z=a+b$, then $a=\frac{y+z-x}{2},\dots$. Your inequality becomes $$(x+y+z)\left(\frac{y+z-x}{x^2}+\cdots\right)\geq 9.$$

Write $y+z-x=(x+y+z)-2x,\dots$ we need to show $$(x+y+z)^2\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right) -2(x+y+z)\left(\frac1x+\frac1y+\frac1z\right)\geq 9.$$

Use $3\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{x^2}\right)\geq \left(\frac1x+\frac1y+\frac1z\right)^2$, we only need to show

$$\frac13(x+y+z)^2\left(\frac1x+\frac1y+\frac1z\right)^2-2(x+y+z)\left(\frac1x+\frac1y+\frac1z\right)\geq 9.$$ The last inequality is correct because $(x+y+z)\left(\frac1x+\frac1y+\frac1z\right)\geq 9$.