Prove that: $\sum\limits_{cyc} \frac{b+c}{a^{2}+bc}\leq \sum\limits_{cyc} \frac{1}{a}$

Question

If $a$, $b$ and $c$ are positive then $$\sum\limits_{cyc} \frac{b+c}{a^{2}+bc}\leq \sum\limits_{cyc} \frac{1}{a}$$
The things I have done so far:
$$\frac{b+c}{a^2+bc}\leq \frac{b+c}{2a\sqrt{bc}}=\frac{b}{2a\sqrt{bc}}+\frac{c}{2a\sqrt{bc}}\leq \sqrt{2\left ( \frac{b^2}{4a^2bc}+\frac{c^2}{4a^2bc} \right )}= \sqrt{2\left ( \frac{b^2+c^2}{4a^2bc}\right )}$$
I just make it looks more difficult!

Answer 1

Let $a\geq b\geq c$.

Thus, $$\sum_{cyc}\frac{1}{a}-\sum_{cyc}\frac{b+c}{a^2+bc}=\sum_{cyc}\frac{(a-b)(a-c)}{a^3+abc}\geq$$ $$\geq\frac{(b-a)(b-c)}{b^3+abc}+\frac{(c-a)(c-b)}{c^3+abc}=(b-c)\left(\frac{a-c}{c^3+abc}-\frac{a-b}{b^3+abc}\right)\geq0.$$