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Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=3$. Prove that: $$\frac{a}{b\sqrt{c^2+3}}+\frac{b}{c\sqrt{a^2+3}}+\frac{c}{a\sqrt{b^2+3}}\geq\frac{a^2+b^2+c^2}{2}$$ I tried C-S, AM-GM, Holder and more, but without success.

Thank you!

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  • $\begingroup$ If it's of any use, I proved that the inequality is true in the case ab+ac+bc<=1 and I can post the proof here, again, if it is of any interest. $\endgroup$ – Raizen Nov 8 '16 at 13:53
  • $\begingroup$ Dear @Raizen Do not do it, please. I also have a very many similar "proofs", but I want to prove this inequality. If I'll find a full prove I'll post it here. Thank you! $\endgroup$ – Michael Rozenberg Nov 8 '16 at 21:24
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Your inequality is equivalent to :

$$\sum_{cyc}\frac{a}{b\sqrt{c^2+\frac{(a+b+c)^2}{3}}}\geq\frac{a^2+b^2+c^2}{(a+b+c)^3}\frac{27}{2}$$

We begin with a first substitution we put :

$\frac{b}{2}-a=-u$

$\frac{b}{2}+a=v$

$c+a=w$

The initial inequality become :

$$\sum_{cyc}\frac{\frac{v+u}{2}}{(v-u)\sqrt{(\frac{2w-u-v}{2})^2+\frac{(v-u+w)^2}{3}}}\geq \frac{(\frac{v+u}{2})^2+(v-u)^2+(\frac{2w-u-v}{2})^2}{(v-u+w)^3}\frac{27}{2}$$

So we start here with a second substitution we put :

$-u=\frac{p}{\sqrt{-r^2+q^2-p^2}}$

$v=\frac{r}{\sqrt{-r^2+q^2-p^2}}$

$w=\frac{q}{\sqrt{-r^2+q^2-p^2}}$

We get this :

$p+r+q=3\sqrt{-r^2+q^2-p^2}$

The initial inequality become

$$\sum_{cyc}\frac{\frac{-p+r}{2}}{(r+p)\sqrt{(\frac{2q+p-r}{2})^2+\frac{(r+p+q)^2}{3}}}\geq \frac{(\frac{-p+r}{2})^2+(r+p)^2+(\frac{2q-r+p}{2})^2}{(r+p+q)^3}\frac{27}{2}$$

We make a last substitution :

$r=R$

$p=RP$

$q=RQ$

So the initial condition become :

$P+1+Q=3\sqrt{-1+Q^2-P^2}$

Wich is equivalent to :

$Q=\frac{3}{8}\sqrt{(9P^2+2P+9)}+\frac{(1+P)}{8}$

The initial inequality become :

$$\sum_{cyc}\frac{\frac{-P+1}{2}}{(1+P)\sqrt{(\frac{2Q-1+P}{2})^2+\frac{(1+P+Q)^2}{3}}}\geq \frac{(\frac{-P+1}{2})^2+(1+P)^2+(\frac{2Q-1+P}{2})^2}{(1+P+Q)^3}\frac{27}{2}$$

So if you replace the value of $Q$ the inequality is just with the variable $P$

So you have many ways to treat it .

Edit :

We treat the case where $b>2a$ and $a\leq c$

It's easy to see that we have :

$$\sum_{cyc}\frac{a}{b\sqrt{c^2+3}}\geq 3(0.75-a)+0.5\frac{1}{\sqrt{(3-3a)^2+3}}+\frac{2a}{(3-3a)\sqrt{a^2+3}}+\frac{3-3a}{a\sqrt{4a^2+3}}\geq 3(0.75-a)+\frac{a^2+4a^2+(3-3a)^2}{2}\geq\frac{a^2+b^2+c^2}{2}$$

So it's conclude the proof .

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  • $\begingroup$ I believe that $u$ and $v $ are positives. What happens if $\frac{b}{2}>a$? $\endgroup$ – Michael Rozenberg Jul 13 '17 at 15:16
  • $\begingroup$ I edit my proof. $\endgroup$ – max8128 Jul 13 '17 at 16:48
  • $\begingroup$ How you are proving the inequality with $P$ and $Q$? $\endgroup$ – Michael Rozenberg Jul 13 '17 at 17:37

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