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.

Let $a=\frac{x^2}{yz},\, b=\frac{y^2}{zx},\,c=\frac{z^2}{xy}.$ Thus$,$ we need to prove$:$
$$\frac{x}{\sqrt{x^2+8yz}}+\frac{y}{\sqrt{y^2+8zx}} +\frac{z}{\sqrt{z^2+8xy}} \geqq 1$$
By AM-GM$:$ \begin{align*} \text{LHS} &=\sum\limits_{cyc} \frac{x}{\sqrt{x^2+8yz}} =\sum\limits_{cyc} \frac{x(x+y+z)}{\sqrt{(x^2+8yz)(x+y+z)^2}}\\&\geqq 2\sum\limits_{cyc} \frac{x(x+y+z)}{(x^2+8yz)+(x+y+z)^2} \geqq 1 \end{align*}
Where the last inequality equivalent to $$\frac{1}{2} \sum\limits_{cyc} \left( 8\,{x}^{3}y+31\,{x}^{2}{y}^{2}+8\,x{y}^{3}+202\,x{y}^{2}z+262
\,xy{z}^{2}+202\,x{z}^{3}+79\,{z}^{4} \right)  \left( x-y \right) ^{2}
 \geqq 0$$
Done.
