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one of my friends asked me if I could solve him a mathematics problem. It looks like this: $$\frac1{a^2 +2} + \frac1{b^2 +2} + \frac1{c^2 +2} \le \frac{\sqrt2}{2}\frac{\sqrt a+\sqrt b+\sqrt c}{\sqrt{abc}}$$ As I think it looks like an inequality between means, hope it helps. And sorry for my bad english by the way. :)

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  • $\begingroup$ Please check to make sure I wrote your formula correctly. $\endgroup$ – MJD Nov 20 '13 at 21:51
  • $\begingroup$ Do you want to prove this inequality or try to find when this inequality will be OK? if it is first case,then it is not true. if it is second case, it is complex as there are 3 varies. $\endgroup$ – chenbai Nov 21 '13 at 6:34
  • $\begingroup$ a necessary condition is $c\le (\dfrac{a(a^2+2)^2}{3})^{\frac{1}{3}}$ if $a\le b \le c$ $\endgroup$ – chenbai Nov 21 '13 at 6:48
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Without loss of generality, we assume that $0 < a \leq b \leq c.$ Then, by AM-GM we have, $$ \frac{a^2 + (\sqrt{2})^2}{2} \geq a\sqrt{2}, $$ and so $$ \frac{1}{a^2 + 2} + \frac{1}{b^2+2} + \frac{1}{c^2+2} \leq \frac{1}{2\sqrt{2}}\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right). $$ Now, $ \frac{1}{\sqrt{c}} \leq \frac{1}{\sqrt{b}} \leq \frac{1}{\sqrt{a}}, $ so using the rearrangement inequalities, and adding, we obtain the desired result.

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    $\begingroup$ there is problem.your RHS is not always < RHS(original),take $a=1,b=c=25$,your RHS=$\dfrac{27\sqrt{2}}{100}>\dfrac{22\sqrt{2}}{100} $ of original. $\endgroup$ – chenbai Nov 21 '13 at 6:21
  • $\begingroup$ Yes, the problem is that the rearrangement inequality does not give exactly what is needed. $\endgroup$ – Mark Fischler Apr 14 '17 at 19:03
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The starting inequality is obviously wrong!. Try $a=0$ and $b=c=1$.

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  • $\begingroup$ There's a missing condition here that $a,b,c > 0$. $\endgroup$ – AlexanderJ93 Jan 21 '17 at 20:16
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    $\begingroup$ Aren't $a,b,c$ implicitly taken to be positive, since the denominator of the right-hand side would otherwise be undefined and (real) square roots can only exist for nonnegative numbers? $\endgroup$ – A.Sh Jan 21 '17 at 20:51
  • $\begingroup$ @A.Sh It's exactly, which I say. We need to prove something, which wrong for $a=0$. $\endgroup$ – Michael Rozenberg Jan 21 '17 at 20:59
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@michael Rozenberg: Let's bury this one for good by presenting a point with all positive $a,b,c$ such that the inequality is violated.

Take $$ a = 25\\ b=25\\ c= \frac{59}{81} \\ \frac{1}{a^2+2}+\frac{1}{b^2+2}+\frac{1}{c^2+2} = \frac{4146953}{10410081} > 0.3983 \\ \frac{\sqrt{2}}{2}\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{abc}}= \frac{59+90\sqrt{59}}{2950}\sqrt{2} < 0.3597 \\\mbox{LHS } > 0.39 > 0.36 > \mbox{ RHS} $$ which violates the inequality.

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