Let $a, b, c > 0$.Prove that: $\sum\limits_{cyc} \frac{a^2+2bc}{(b+c)^2}\geq \sum \frac{3}{2}\frac{a}{b+c}$
p/s: I tried to solve the problem by $S.O.S$. But I cannot solve it !! I have: The inequatily $\Leftrightarrow \sum\limits_{cyc} S_{c}(a-b)^2\geq 0$ with $S_c=\frac{(a+b)(a+b+2c)}{2(b+c)^2(a+c)^2}-\frac{1}{4(a+b)^2}-\frac{3}{4(a+b)(b+c)}$ However I cannot find a way....
There is a good reason why you couldn’t solve it : your inequality is false as claimed. Let $\varepsilon\in[-\frac{1}{2},\frac{1}{2}]$ be a variable tending to zero. Putting $a=1-2\varepsilon,b=1+\varepsilon,c=1+\varepsilon$, we have (using Landau’s $O$- notation)
$$ \begin{array}{lcl} \sum\limits_{cyc}\frac{a^2+2bc}{(b+c)^2} &=& \frac{9}{4}+\frac{27}{8}\varepsilon^2 +\frac{243}{16}\varepsilon^3+\frac{3}{32}\varepsilon^4+O(\varepsilon^5) \\ \sum\limits_{cyc}\frac{a}{b+c} &=& \frac{3}{2}+\frac{9}{4}\varepsilon^2 -\frac{9}{8}\varepsilon^3+\frac{27}{16}\varepsilon^4+O(\varepsilon^5) \\ \end{array} $$
Thus, when $\varepsilon$ is positive and very small, we have $$ \sum_{cyc}\frac{a^2+2bc}{(b+c)^2}-\frac{3}{2} \sum_{cyc}\frac{a}{b+c}=-\frac{81}{16}\varepsilon^3+\frac{81}{16}\varepsilon^4+O(\varepsilon^5) < 0. $$
Note that when $\varepsilon$ is negative and very small, the LHS above becomes positive, so that the question in the OP cannot be fixed by reversing the inequality.
