1
$\begingroup$

How to prove the following inequality : $$ \sqrt{\frac{2a}{b+a}} + \sqrt{\frac{2b}{c+b}} + \sqrt{\frac{2c}{a+c}} \leq 3 $$ with $a>0,\ b>0$ and $c>0$.

$\endgroup$
  • $\begingroup$ @Gaffney, then we would get $1 + 0 + \sqrt{2} \le 3$, which seems correct. $\endgroup$ – zhoraster Oct 15 '15 at 11:39
  • 5
    $\begingroup$ Looks familiar. $\endgroup$ – A.Γ. Oct 15 '15 at 12:11
4
$\begingroup$

since $$9(a+b)(b+c)(a+c)\ge 8(ab+bc+ac)(a+b+c)$$ By Cauchy-Schwarz inequality we have $$\sum_{cyc}\sqrt{\dfrac{2a}{a+b}}\le\sqrt{\left[\sum(c+a)\right]\left[\sum_{cyc}\dfrac{2a}{(a+b)(c+a)}\right]} =\sqrt{\dfrac{8(a+b+c)(ab+bc+ac)}{(a+b)(b+c)(a+c)}}\le 3$$

$\endgroup$

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