Prove $$\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\le\frac{1}{2}\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right),$$ where $a,b,c > 0$ and $a,b,c \in \mathbb{R}$
Well, I've been trying for 3 good hours, nothing worked at all. I already applied HM < AM but I'm still stuck. It gave the following: $$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{a+b+c}{abc}\le2(a+b+c)$$
Hint: $$\frac{1}{a^2+bc} \le \frac{1}{2a\sqrt{bc}} = \frac{1}{2\sqrt{ab}\sqrt{ac}}.$$ Now apply AM-GM inequality.
$$\frac{1}{a^2+bc} + \frac{1}{b^2+ac} + \frac{1}{c^2+ab} \le \frac{1}{2\sqrt{ab}\sqrt{ac}} + \frac{1}{2\sqrt{ab}\sqrt{bc}} + \frac{1}{2\sqrt{ac}\sqrt{bc}} \le $$
$$\le \frac14 \left(\frac{1}{ab}+\frac{1}{ac} + \frac{1}{ab}+\frac{1}{bc} + \frac{1}{ac}+\frac{1}{bc}\right)=\frac12\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right).$$
