If $\frac {1}{2+a} + \frac {1}{2+b} + \frac {1}{2+c} = 1$, prove $\sqrt{ab} + \sqrt{ac} + \sqrt{bc} \leq 3$ Let $a,b,c$ be non-negative numbers such that $$\frac {1}{2+a} + \frac {1}{2+b} + \frac {1}{2+c} = 1.$$ 
Prove that $ \sqrt{ab} + \sqrt{ac} + \sqrt{bc} \leq 3 $.

The condition gives that there are $\alpha\geq0$, $\beta\geq0$ and $\gamma\geq0$ such that $\alpha+\beta+\gamma=\pi$ for which $\sqrt{ab}=2\cos\gamma$, $\sqrt{ac}=2\cos\beta$ and $\sqrt{bc}=2\cos\alpha$.
Hence, we need to prove that $\cos\alpha+\cos\beta+\cos\gamma\leq\frac{3}{2}$, which is obvious.
