<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] Consider the function for positive $x$:
$$f(x) = \frac{x}{\sqrt{1+x}}-\frac1{\sqrt 2}-\frac3{4\sqrt 2}\log x$$
Note that $f(x) \ge 0 \implies f(a)+f(b)+f(c) \ge 0 \implies $ the given inequality.  Now
$$f'(x) = \frac{4x^2-3\sqrt2 (x+1)^{3/2}+8x}{8x(x+1)^{3/2}}$$  
We need to check the sign of the numerator, $4(x+1)^2-3\sqrt2(x+1)^{3/2}-4$.  Using $y = \sqrt{x+1}$, we get the numerator as
$$4y^4-3\sqrt2y^3-4 = (y-\sqrt2)(4y^3+\sqrt2y^2+2y+2\sqrt2)$$
As the second factor is positive, the numerator's sign is given by $y-\sqrt2$ which has the same sign as $x-1$, so $f'(x)< 0$ for $x < 1$ and $f'(x)> 0$ for $x> 1$.  Hence $f(x)\ge f(1)=0$. 
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