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.
 A: Let $a=\frac{x}{y}$, $b=\frac{y}{z}$ and $c=\frac{z}{x}$, where $x$, $y$ and $z$ are positives.
Hence, we need to prove that
$$\sum_{cyc}\frac{\frac{x}{y}}{\sqrt{1+\frac{x}{y}}}\geq\frac{3}{\sqrt2}$$ or
$$\sum_{cyc}\frac{x}{\sqrt{y(x+y)}}\geq\frac{3}{\sqrt2}.$$
Now, by Holder
$$\left(\sum_{cyc}\frac{x}{\sqrt{y(x+y)}}\right)^2\sum_{cyc}xy(x+y)(2z+x+y)^3\geq\left(\sum_{cyc}x(2z+x+y)\right)^3.$$
Thus, it remains to prove that
$$2\left(\sum_{cyc}(x^2+3xy)\right)^3\geq9\sum_{cyc}xy(x+y)(2z+x+y)^3$$ or
$$\sum_{sym}(x^6+9x^5y+24x^4y^2+18x^3y^3+9x^4yz-36x^3y^2z-25x^2y^2z^2)\geq0,$$
which is obviously true.
Done!
A: 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 $f(x)\ge f(1)=0$. 
A: By the same way as Mr. Mike$,$ it's enough to prove $$2\left(\sum_{cyc}(x^2+3yz)\right)^3\geqslant 9\sum xy(x+y)(2z+x+y)^3$$
Or
$$\frac18\sum \left( 16\,{x}^{4}+100\,{z}^{4}+104\,{x}^{3}y+243\,{y}^{2}{z}^{2
}+330\,{z}^{3}x+416\,{y}^{2}zx+342\,{z}^{2}xy \right)  \left( x-y
 \right) ^{2}+$$
$$+\frac18\sum x{y}^{2} \left( 18\,y+41\,x \right)  \left( z+x-2\,y \right) ^{2}\geqslant 0$$
A: 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.
A: 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 $$a=b=c=1$$
$$f_{min}=\frac{1}{\sqrt{1+1}}+\frac{1}{\sqrt{1+1}}+\frac{1}{\sqrt{1+1}}=\frac{3}{\sqrt2}=\frac{3\sqrt{2}}{2}$$
$$f(a,b,c)=\frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}}\ge f_{min}$$
$$f(a,b,c)=\frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}}\ge \frac{3\sqrt{2}}{2}$$
A: 
Thus, it remains to prove that $$2\left(\sum_{cyc}(x^2+3xy)\right)^3\geq9\sum_{cyc}xy(x+y)(2z+x+y)^3$$

We have \begin{align*}LHS-RHS&=(x^3+y^3+z^3)\sum x(x-y)(x-z)+(x^2+y^2+z^2-xy-yz-zx)^3\\ &+7(xy+yz+zx)\sum x^2(x-y)(x-z)\\ &+[x^2+y^2+z^2+3(xy+yz+zx)][x^2y^2+y^2z^2+z^2x^2-xyz(x+y+z)]\\ &+ (x+y+z)[4(x^2+y^2+z^2)+9(xy+yz+zx)]\sum z(x-y)^2\\ &+ (x+y+z)(xy+yz+zx)\sum (x+y+7z)(x-y)^2 \ge 0\end{align*}
A: Some observations:
$$A=\frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \tag{1}$$
$$A=\sqrt{1+a}+\sqrt{1+b}+\sqrt{1+c} +\frac{1}{\sqrt{1+a}}+\frac{1}{\sqrt{1+b}}+\frac{1}{\sqrt{1+c}} -\left(\frac{2}{\sqrt{1+a}}+\frac{2}{\sqrt{1+b}}+\frac{2}{\sqrt{1+c}} \right)$$
$$A\ge(2+2+2)-\left(\frac{2}{\sqrt{1+a}}+\frac{2}{\sqrt{1+b}}+\frac{2}{\sqrt{1+c}} \right)\tag{2}$$
Comparing with the original definition:
$$A=\frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2},\tag{3}$$
We conclude that we need to prove:
$$\frac{1}{\sqrt{1+a}}+\frac{1}{\sqrt{1+b}}+\frac{1}{\sqrt{1+c}} \le 3-\frac{3\sqrt{2}}{4}\tag{4}$$
