<s>[INST] Prove that $\sum \limits_{cyc}\frac {a}{(b+c)^2} \geq \frac {9}{4(a+b+c)}$ 
Given $a$, $b$ and $c$ are positive real numbers. Prove that:$$\sum \limits_{cyc}\frac {a}{(b+c)^2} \geq \frac {9}{4(a+b+c)}$$

Additional info: We can't use induction. We should mostly use Cauchy inequality. Other inequalities can be used rarely.
Things I have done so far: The inequality look is similar to Nesbitt's inequality.
We could re-write it as:  $$\sum \limits_{cyc}\frac {a}{(b+c)^2}(2(a+b+c)) \geq \frac{9}{2}$$
Re-write it again:$$\sum \limits_{cyc}\frac {a}{(b+c)^2}\sum \limits_{cyc}(b+c) \geq \frac{9}{2}$$
Cauchy appears:
$$\sum \limits_{cyc}\frac {a}{(b+c)^2}\sum \limits_{cyc}(b+c) \geq \left(\sum \limits_{cyc}\sqrt\frac{a}{b+c}\right)^2$$
So, if I prove $\left(\sum \limits_{cyc}\sqrt\frac{a}{b+c}\right)^2 \geq \frac {9}{2}$ then problem is solved.
Re-write in semi expanded form:$$2\left(\sum \limits_{cyc}\frac{a}{b+c}+2\sum \limits_{cyc}\sqrt\frac{ab}{(b+c)(c+a)}\right) \geq 9$$
We know that $\sum \limits_{cyc}\frac{a}{b+c} \geq \frac {3}{2}$.So$$4\sum \limits_{cyc}\sqrt\frac{ab}{(b+c)(c+a)} \geq 6$$
So the problem simplifies to proving this $$\sum \limits_{cyc}\sqrt\frac{ab}{(b+c)(c+a)} \geq \frac{3}{2}$$
And I'm stuck here.
 [/INST] 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.
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