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:
$$\frac32-1 = \sum_{cyc} \left(\frac12-\frac1{2+a} \right)$$
$$\implies 1 = \sum_{cyc} \frac{a}{2+a} \ge \frac{(\sqrt a+\sqrt b+\sqrt c)^2}{a+b+c+6} \quad \text{by Cauchy-Schwarz inequality}$$
$$\implies 3 \ge \sqrt{ab}+\sqrt{bc}+\sqrt{ca}$$
