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$:$
$$4(a+b+c) \prod (a+b)^2 (\text{LHS}-\text{RHS})$$
$$=\prod (a+b) \Big[\sum\limits_{cyc} (ab+bc-2ca)^2+(ab+bc+ca)\sum\limits_{cyc} (a-b)^2 \Big]$$
$$+(a+b+c)(a-b)^2 (b-c)^2 (c-a)^2 \geqq 0$$
 A: 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.
A: Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that
$$\frac{\sum\limits_{cyc}(a^3+a^2b+a^2c+abc)}{\prod\limits_{cyc}(a+b)}+\frac{3}{4}\geq\frac{(ab+ac+bc)\sum\limits_{cyc}(a+b)^2(a+c)^2}{\prod\limits_{cyc}(a+b)^2}$$ or
$$\frac{27u^3-27uv^2+3w^3+9uv^2-3w^3+3w^3}{9uv^2-w^3}+\frac{3}{4}\geq\frac{3v^2\sum\limits_{cyc}(a^2+3v^2)^2}{(9uv^2-w^3)^2}$$ or
$$4(9u^3-6uv^2+w^3)(9uv^2-w^3)+(9uv^2-w^3)^2\geq$$
$$\geq4v^2((9u^2-6v^2)^2-2(9v^4-6uw^3)+6(9u^2-6v^2)v^2+27v^4)$$ and since $$-4+1<0,$$ we see that our inequality it's $f(w^3)\geq0,$ where $$f(w^3)=-w^6+A(u,v^2)w^3+B(u,v^2),$$ which says that $f$ is a concave function.
Thus, by $uvw$ (see here https://artofproblemsolving.com/community/c6h278791 ) it's enough to prove our inequality in two following cases:


*

*$w^3\rightarrow0^+$;

*Two variables are equal (in this case it's enough to assume $b=c=1$). 
But in the last case we obtain: 
$$(a-1)^2\geq0,$$ which says that our inequality is true even for any reals $a$, $b$ and $c$ such that $\prod\limits_{cyc}(a+b)\neq0.$
