<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] Replace $(a,b,c) \to \left(\frac{a^2}{bc},\frac{b^2}{ca},\frac{c^2}{ab}\right)$ the inequality become
$$ \sum \frac{a}{\sqrt{a^2+8bc}} \geqslant 1.$$
We have
$$\left[5(a^2+b^2+c^2)+4(ab+bc+ca)\right]^2-(a^2+8bc)(5a+2b+2c)^2$$
$$=[5a(b+c)+17bc](2a-b-c)^2+[82a^2+25(b^2+c^2)+35a(b+c)+41bc](b-c)^2 \geqslant 0.$$
Therefore
$${\frac {a}{\sqrt {a^{2} + 8bc}}}\geq {\frac {a(5a + 2b + 2c)}{5\left(a^2 + b^2 + c^2\right) + 4(bc + ca + ab)}}.$$
Note
$$\sum{\frac {a(5a + 2b + 2c)}{5\left(a^2 + b^2 + c^2\right) + 4(ab+bc+ca)}} = 1,$$
The proof is completed.
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