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

  • $\begingroup$ maybe $\sqrt{ab} + \sqrt{ac} + \sqrt{bc} \le a+b+c\leq 3$more straightforward. $\endgroup$ – chenbai Mar 8 '15 at 11:59
  • 1
    $\begingroup$ @chenbai But $a+b+c\geq 3$. $\endgroup$ – WimC Mar 8 '15 at 12:47
  • $\begingroup$ Looks OK now, +1. Maybe the first inequality could use some extra explanation. $\endgroup$ – WimC Mar 8 '15 at 13:26
  • 1
    $\begingroup$ @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. $\endgroup$ – Macavity Mar 8 '15 at 16:20
  • $\begingroup$ 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 $y=\sqrt(v)$. $\endgroup$ – 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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.