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.

The required inequality is trivialized by the claim below.  The equality case is when $a=b=c=1$.

Claim: If $a,b,c>0$ are such that $abc=1$, then $\displaystyle\sqrt{\frac{a}{a+8}}\geq \frac{a^{4/9}}{a^{4/9}+b^{4/9}+c^{4/9}}$.  The equality holds if and only if $a=b=c=1$.

Proof: Note that the required inequality is equivalent to
$$\left(a^{4/9}+b^{4/9}+c^{4/9}\right)^2 \geq a^{-1/9}(a+8)\,,$$
which is also equivalent to
$$\left(b^{4/9}+c^{4/9}\right)\left(a^{4/9}+a^{4/9}+b^{4/9}+c^{4/9}\right) \geq 8a^{-1/9}\,.$$
To prove the previous inequality, we invoke the AM-GM Inequality twice:
$$b^{4/9}+c^{4/9}\geq 2b^{2/9}c^{2/9}$$
and
$$a^{4/9}+a^{4/9}+b^{4/9}+c^{4/9}\geq 4a^{1/9}a^{1/9}b^{1/9}c^{1/9}=4a^{2/9}b^{1/9}c^{1/9}\,.$$
Thus, 
$$
\begin{align}
\left(b^{4/9}+c^{4/9}\right)\left(a^{4/9}+a^{4/9}+b^{4/9}+c^{4/9}\right)  &\geq \left(2b^{2/9}c^{2/9}\right)\left(4a^{2/9}b^{1/9}c^{1/9}\right)
\\
&=8a^{2/9}b^{1/3}c^{1/3}=8a^{-1/9}\left(abc\right)^{1/3}=8a^{-1/9}\,,
\end{align}$$
which is what we want.  By the equality condition of the AM-GM Inequality, the equality happens iff $a=b=c=1$.
P.S.:  I just realized why this inequality looks so familiar.  It is equivalent to IMO'2001#2 (http://imo.wolfram.com/problemset/IMO2001_solution2.html).  Substitute $a$, $b$, and $c$ by $x^3$, $y^3$, and $z^3$, then homogenize the required inequality via the condition $xyz=1$, and you will see what I'm talking about.
