Prove the inequality $\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$ with the constraint $abc=1$ If $a,b,c$ are positive reals such that $abc=1$, then prove that $$\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$$ I tried substituting $x/y,y/z,z/x$, but it didn't help(I got the reverse inequality). Need some stronger inequality. Thanks.

First let
$$ x = \sqrt{\frac{a}{a+8}}, \,\,  y = \sqrt{\frac{b}{b+8}}, \,\,  z = \sqrt{\frac{c}{c+8}} \,\, $$
Then $1 > x,y,z > 0$ and 
$$ a = \frac{8x^2}{1 - x^2}, \,\,   b = \frac{8y^2}{1 - y^2}, \,\,  c = \frac{8z^2}{1 - z^2},\,\, $$
So the question transforms to this:
Given that $1 > x,y,z > 0, \, \, \frac{512x^2y^2z^2}{(1 - x^2)(1 - y^2)(1 - z^2)} = 1$, prove that $x + y + z \geqslant 1$.
Prove this by contradiction. Suppose on the contrary that $x + y + z < 1$, then 
$$ \begin{align}
(1 - x^2)(1 - y^2)(1 - z^2) &= (1 - x)(1 + x)(1 - y)(1 + y)(1 - z)(1 + z) \\
&>(x + x + y + z)(y + z)(x + y + y + z)(x + z)(z + x + y + z)(x + y) \\
&\geqslant 4x^{\frac12}y^{\frac14}z^{\frac14}\cdot 2y^{\frac12}z^{\frac12} \cdot 4y^{\frac12}x^{\frac14}z^{\frac14}\cdot 2x^{\frac12}z^{\frac12} \cdot 4z^{\frac12}y^{\frac14}x^{\frac14}\cdot 2y^{\frac12}x^{\frac12}\\
&=512 x^{\frac12 + \frac14 + \frac12 + \frac14 + \frac12}y^{\frac14 + \frac12 + \frac12 + \frac14 + \frac12}z^{\frac14 + \frac12 +\frac14 + \frac12 + \frac12} \\
&= 512x^2y^2z^2
\end{align}$$
And this is contradictory to the condition.
