TITLE: Proving $\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab} \le\frac{1}{2}(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac})$
QUESTION [7 upvotes]: 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: 
$$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{a+b+c}{abc}\le2(a+b+c)$$

REPLY [10 votes]: Hint: $$\frac{1}{a^2+bc} \le \frac{1}{2a\sqrt{bc}} = \frac{1}{2\sqrt{ab}\sqrt{ac}}.$$
Now apply AM-GM inequality.

 $$\frac{1}{a^2+bc} + \frac{1}{b^2+ac} + \frac{1}{c^2+ab} \le \frac{1}{2\sqrt{ab}\sqrt{ac}} + \frac{1}{2\sqrt{ab}\sqrt{bc}} + \frac{1}{2\sqrt{ac}\sqrt{bc}} \le $$
 $$\le \frac14 \left(\frac{1}{ab}+\frac{1}{ac} + \frac{1}{ab}+\frac{1}{bc} + \frac{1}{ac}+\frac{1}{bc}\right)=\frac12\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right).$$