Given positives $a, b, c$, prove that $\frac{a}{(b + c)^2} + \frac{b}{(c + a)^2} + \frac{c}{(a + b)^2} \ge \frac{9}{4(a + b + c)}$. 
Given positives $a, b, c$, prove that $$\large \frac{a}{(b + c)^2} + \frac{b}{(c + a)^2} + \frac{c}{(a + b)^2} \ge \frac{9}{4(a + b + c)}$$

Let $x = \dfrac{b + c}{2}, y = \dfrac{c + a}{2}, z = \dfrac{a + b}{2}$
It needs to be sufficient to prove that $$\sum_{cyc}\frac{\dfrac{a + b}{2} + \dfrac{b + c}{2} - \dfrac{c + a}{2}}{\left(2 \cdot \dfrac{c + a}{2}\right)^2} \ge \frac{9}{\displaystyle 4 \cdot \sum_{cyc}\dfrac{c + a}{2}} \implies \sum_{cyc}\frac{z + x - y}{y^2} \ge \frac{9}{y + z + x}$$
According to the Cauchy-Schwarz inequality, we have that
$$(y + z + x) \cdot \sum_{cyc}\frac{z + x - y}{y^2} \ge \left(\sum_{cyc}\sqrt{\frac{z + x - y}{y}}\right)^2$$
We need to prove that $$\sum_{cyc}\sqrt{\frac{z + x - y}{y}} \ge 3$$
but I don't know how to.
Thanks to Isaac YIU Math Studio, we additionally have that $$(y + z + x) \cdot \sum_{cyc}\frac{z + x - y}{y^2} = \sum_{cyc}(z + x - y) \cdot \sum_{cyc}\frac{z + x - y}{y^2} \ge \left(\sum_{cyc}\frac{z + x - y}{y}\right)^2$$
We now need to prove that $$\sum_{cyc}\frac{z + x - y}{y} \ge 3$$, which could be followed from Nesbitt's inequality.
I would be greatly appreciated if there are any other solutions than this one.
 A: Hint : Using homogeneity, WLOG we may set $a+b+c=3$, then note $f(x) = \dfrac{x}{(3-x)^2}$ is convex and use Jensen’s inequality.
A: By Cauchy-Schwarz,
$$\left[\dfrac{a}{(b + c)^2} + \dfrac{b}{(c + a)^2} + \dfrac{c}{(a + b)^2}\right](a+b+c) \ge \left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)^2$$
Then by rearrangement inequality,
$$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{b}{b+c}+\dfrac{c}{c+a}+\dfrac{a}{a+b} \\ \dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{c}{b+c}+\dfrac{a}{c+a}+\dfrac{b}{a+b}$$ Sum up and get $$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{3}{2}$$
So from the first inequality, we get:
$$\left[\dfrac{a}{(b + c)^2} + \dfrac{b}{(c + a)^2} + \dfrac{c}{(a + b)^2}\right](a+b+c) \ge \dfrac{9}{4} \\ \dfrac{a}{(b + c)^2} + \dfrac{b}{(c + a)^2} + \dfrac{c}{(a + b)^2} \ge \dfrac{9}{4(a+b+c)}$$
A: Let $a+b+c=p$, then consider a function $f(a)=\dfrac{a}{(p-a)^2}$, then $f''(a)=\dfrac{2(a+p)}{(p-a)^4} >0$. So from the Jensen's in equality it follows that
$$\frac{f(a)+f(b)+f(c)}{3} \ge f\left[\frac{a+b+c}{3}\right]$$
So we get $$\frac{a}{(b+c)^2}+\frac{b}{(a+c)^2}+\frac{c}{(a+c)^2} \ge 3*\frac{p/3}{(p-p/3)^2}=\frac{9}{4(a+b+c)}$$
A: Recall Nesbitt's inequality
\begin{eqnarray*}
\frac{a}{b+c}+\frac{b}{c+a} + \frac{c}{a+b} \geq \frac{3}{2}.
\end{eqnarray*}
Using Cauchy-Schwartz and Nesbitt gives
\begin{eqnarray*}
&\left( 2 \cdot \frac{a}{b+c}+1 \right)^2 +\left( 2 \cdot \frac{b}{c+a}+1 \right)^2 + \left(2 \cdot \frac{c}{a+b} +1 \right)^2\\ &\geq 
\frac{1}{3}\left( 2 \cdot \frac{a}{b+c}+2 \cdot \frac{b}{c+a} + 2 \cdot \frac{c}{a+b} +3 \right)^2 \geq 12. 
\end{eqnarray*}
And this can be rearranged to give the inequality.
Edit: In light of Issac's answer ... By Cauchy-Schwartz,
\begin{eqnarray*}
\left(\dfrac{a}{(b + c)^2} + \dfrac{b}{(c + a)^2} + \dfrac{c}{(a + b)^2}\right)(a+b+c) \geq \left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)^2
\end{eqnarray*}
and the result now follows by Nesbitt's inequality.
A: Since our inequality is homogeneous, we can assume that $a+b+c=3.$
Thus, $$\sum_{cyc}\frac{a}{(b+c)^2}-\frac{9}{4(a+b+c)}=\sum_{cyc}\frac{a}{(3-a)^2}-\frac{3}{4}=\sum_{cyc}\left(\frac{a}{(3-a)^2}-\frac{1}{4}\right)=$$
$$=\frac{1}{4}\sum_{cyc}\frac{-a^2+10a-9}{(3-a)^2}=\frac{1}{4}\sum_{cyc}\frac{(a-1)(9-a)}{(3-a)^2}=$$
$$=\frac{1}{4}\sum_{cyc}\left(\frac{(a-1)(9-a)}{(3-a)^2}-2(a-1)\right)=\frac{1}{4}\sum_{cyc}\frac{(a-1)^2(9-2a)}{(3-a)^2}\geq0.$$
A: By Hölder's inequality:
$$ ..... \geq \frac{(a+b+c)^3}{4(ab+bc+ac)^2} \geq \frac{9(a+b+c)^3}{4(a+b+c)^4}$$
