<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] Let$$f(a,b,c)=\frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}}$$
$$g(a,b,c)=abc-1=0$$
Using Lagrange Multiplier
$$\large\frac{\frac{\partial f}{\partial a}}{\frac{\partial g}{\partial a}}=
\frac{\frac{\partial f}{\partial b}}{\frac{\partial g}{\partial b}}=
\frac{\frac{\partial f}{\partial c}}{\frac{\partial g}{\partial c}}=k
$$
We get $$\frac{a+2}{2bc(a+1)^{3/2}}=\frac{b+2}{2ac(b+1)^{3/2}}=\frac{c+2}{2ab(c+1)^{3/2}}=k$$
by solving this for $a,b,c$ we get $$a=b=c$$ and from constraint $g(a,b,c)=0$ we get $$a=b=c=1$$
$$f_{min}=\frac{1}{\sqrt{1+1}}+\frac{1}{\sqrt{1+1}}+\frac{1}{\sqrt{1+1}}=\frac{3}{\sqrt2}=\frac{3\sqrt{2}}{2}$$
$$f(a,b,c)=\frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}}\ge f_{min}$$
$$f(a,b,c)=\frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}}\ge \frac{3\sqrt{2}}{2}$$
</s>