<s>[INST] How to prove the inequality $ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$ How to prove the inequality $$ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$$
for $a,b,c>0$ and $abc=1$?
I have tried prove $\frac{a}{\sqrt{1+a}}\ge \frac{3a+1}{4\sqrt{2}}$
Indeed,$\frac{{{a}^{2}}}{1+a}\ge \frac{9{{a}^{2}}+6a+1}{32}$
$\Leftrightarrow 32{{a}^{2}}\ge 9{{a}^{2}}+6a+1+9{{a}^{3}}+6{{a}^{2}}+a$ 
$\Leftrightarrow 9{{a}^{3}}-17{{a}^{2}}+7a+1\le 0$ 
$\Leftrightarrow 9{{\left( a-1 \right)}^{2}}\left( a+\frac{1}{9} \right)\le 0$ (!)
It is wrong. Advice on solving this problem.
 [/INST] By Hölder's inequality
$$\left(\frac a{\sqrt{1+a}}+\frac b{\sqrt{1+b}}+\frac c{\sqrt{1+c}}\right)^2\Big(a(1+a)+b(1+b)+c(1+c)\Big)\ge(a+b+c)^3,$$
so we only need to prove
\begin{align*}(a+b+c)^3&\ge\frac92\Big(a(1+a)+b(1+b)+c(1+c)\Big)\\(a+b+c)(2(a+b+c)^2-9)&\ge9(a^2+b^2+c^2)\\\end{align*}
AM-GM tells us that $(a+b+c)^2=(a+b+c)^2+9-9\ge6(a+b+c)-9$, so it suffices to prove
\begin{align*}(a+b+c)(12(a+b+c)^2-27)&\ge9(a^2+b^2+c^2)\\4(a+b+c)^2&\ge3(a^2+b^2+c^2)+9(a+b+c)\\a^2+b^2+c^2+8(ab+bc+ca)&\ge9(a+b+c)\end{align*}
And I'll leave this last inequality up to you. Use the fact that $abc=1$, which haven't been used so far.
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