Prove the inequality $\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$ with the constraint $abc=1$

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.

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.

Let $a=\frac{x^2}{yz}$, $b=\frac{y^2}{xz}$ and $c=\frac{z^2}{xy}$, where $x$, $y$ and $z$ are positives.

Hence, by Holder and AM-GM we obtain: $$\sum_{cyc}\sqrt{\frac{a}{a+8}}=\sum_{cyc}\frac{x}{\sqrt{x^2+8yz}}=\sqrt{\frac{\left(\sum\limits_{cyc}\frac{x}{\sqrt{x^2+8yz}}\right)^2\sum\limits_{cyc}x(x^2+8yz)}{\sum\limits_{cyc}x(x^2+8yz)}}\geq$$ $$\geq\sqrt{\frac{(x+y+z)^3}{\sum\limits_{cyc}(x^3+8xyz)}}\geq\sqrt{\frac{(x+y+z)^3}{\sum\limits_{cyc}(x^3+3x^2y+3x^2z+2xyz)}}=1.$$ Done!