<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] Let $a+b+c=3$.
Hence, we need to prove that:
$$\sum_{cyc}\frac{a}{(3-a)^2}\geq\frac{3}{4}$$ or
$$\sum_{cyc}\left(\frac{a}{(3-a)^2}-\frac{1}{4}\right)\geq0$$ or
$$\sum_{cyc}\frac{(a-1)(9-a)}{(3-a)^2}\geq0$$ or
$$\sum_{cyc}\left(\frac{(a-1)(9-a)}{(3-a)^2}-2(a-1)\right)\geq0$$ or
$$\sum_{cyc}\frac{(a-1)^2(9-2a)}{(3-a)^2}\geq0$$
and since $\{a,b,c\}\subset(0,3)$, we are done! 
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