<s>[INST] Inequality with mean inequality 
If $a,b,c > 0$ and $a+b+c = 18 abc$, prove that
$$\sqrt{3 +\frac{1}{a^{2}}}+\sqrt{3 +\frac{1}{b^{2}}}+\sqrt{3 +\frac{1}{c^{2}}}\geq 9$$

I started writing the left member as $\frac{\sqrt{3a^{2}+ 1}}{a}+ \frac{\sqrt{3b^{2}+ 1}}{b} + \frac{\sqrt{3c^{2}+ 1}}{c}$ and I applied AM-QM inequality, but I obtain something with $\sqrt[4]{3}$.
 [/INST] Hints:
Put Let $a=\frac{x}{3\sqrt2},$ $b=\frac{y}{3\sqrt2}$ and $c=\frac{z}{3\sqrt2}.$
And $x=\tan(u)$,$y=\tan(v)$,$z=\tan(w)$ with $u+v+w=\pi$
Then the function $f(u)=\sqrt{3+\frac{18}{\tan(u)^2}}$ is convex on $(0,\pi)$ see here
Remains to apply Jensen's inequality .
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