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 $.

Solution 1:

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

Solution 2:

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.