Prove$:$ $\sum\limits_{cyc} (\frac{a}{b+c}-\frac{1}{2}) \geqq (\sum\limits_{cyc} ab)\Big[\sum\limits_{cyc} \frac{1}{(a+b)^2}\Big]-\frac{9}{4}$ For $a,b,c$ are reals and $a+b+c>0, ab+bc+ca>0, (a+b)(b+c)(c+a)>0.$ Prove$:$
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} -\frac{3}{2} \geqq (\sum\limits_{cyc} ab)\Big[\sum\limits_{cyc} \frac{1}{(a+b)^2}\Big]-\frac{9}{4}$$
My proof$:$
$$4(a+b+c) \prod (a+b)^2 (\text{LHS}-\text{RHS})$$
$$=\prod (a+b) \Big[\sum\limits_{cyc} (ab+bc-2ca)^2+(ab+bc+ca)\sum\limits_{cyc} (a-b)^2 \Big]$$
$$+(a+b+c)(a-b)^2 (b-c)^2 (c-a)^2 \geqq 0$$

From
$$ \sum \frac{a}{b+c} -\frac{3}{2} - \left(\sum \frac{ab+bc+ca}{(a+b)^2} -\frac{9}{4}\right) = \frac14 \sum \frac{(a-b)^2}{(a+b)^2} \geqslant 0.$$
We can see, the inequality also true for all $a,\,b,\,c$ are real numbers.
