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

• maybe $\sqrt{ab} + \sqrt{ac} + \sqrt{bc} \le a+b+c\leq 3$more straightforward. – chenbai Mar 8 '15 at 11:59
• @chenbai But $a+b+c\geq 3$. – WimC Mar 8 '15 at 12:47
• Looks OK now, +1. Maybe the first inequality could use some extra explanation. – WimC Mar 8 '15 at 13:26
• @Dr.MV The intention is not to spell out every detail, however you may note that the "last" inequality follows from simplification (cross multiply, expand, cancel) of the "first" inequality. – Macavity Mar 8 '15 at 16:20
• Macavity. Yes, I understand. Actually, my request was meant only as a suggestion to assist other potential readers that might not have your level of mathematical maturity. I might have shown, for example, that Cauchy-Schwartz, as it is usually expressed (i.e., $||x||^2||y||^2\ge\left(<x,y>\right)^2$), can be modified easily by letting $x=\sqrt(\frac{u}{v})$ and $y=\sqrt(v)$. – Mark Viola Mar 9 '15 at 3:43

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.

• If $\alpha$, $\beta$, and $gamma$ are real numbers, then have you not restricted the range of values for $a$, $b$, and $c$? – Mark Viola Mar 8 '15 at 8:19
• it must be $\cos(\alpha),\cos(\beta),\cos(\gamma)\geq 0$ – Dr. Sonnhard Graubner Mar 8 '15 at 8:49
• To Dr. MV. The condition gives $4=ab+ac+bc+abc$. Id est, $\sqrt{ab}\leq2$, $\sqrt{ac}\leq2$ and $\sqrt{bc}\leq2$. – Michael Rozenberg Mar 8 '15 at 14:56
• Michael Rozenberg. How did you arrive at your conclusions? The approach is not obvious to all. – Mark Viola Mar 8 '15 at 15:31
• Dr.MV. Let $\sqrt{ab}=2\cos\gamma$, $\sqrt{ac}=2\cos\beta$, where $\{\beta,\gamma\}\subset[0,\frac{\pi}{2}]$ and $\sqrt{bc}=2x$. Hence, $x^2+2\cos\beta\cos\gamma x+\cos^2\beta+\cos^2\gamma-1=0$. Thus, $x=-\cos\beta\cos\gamma+\sqrt{\cos^2\beta\cos^2\gamma-\cos^2\beta-\cos^2\gamma+1}$ or $x=-\cos\beta\cos\gamma+\sin\beta\sin\gamma$ or $x=\cos(\pi-\beta-\gamma)$. Let $\pi-\beta-\gamma=\alpha$ and we are ready to use $\cos\alpha+\cos\beta+\cos\gamma\leq\frac{3}{2}$. – Michael Rozenberg Mar 8 '15 at 19:09