<s>[INST] 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.
 [/INST] 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!
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