<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] $$\sqrt{3 +\frac{1}{a^{2}}}+\sqrt{3 +\frac{1}{b^{2}}}+\sqrt{3 +\frac{1}{b^{2}}}$$
$$
= \sqrt{
9 +
\frac{1}{a^{2}} + 
\frac{1}{b^{2}} +
\frac{1}{c^{2}} + 
2\sum_{cyc}
{\sqrt{
\left(3 +
\frac{1}{a^{2}}\right)
\left(3 +\frac{1}{b^{2}}\right)}}
}
$$
$$\tag{By C-S}\geqslant \sqrt{9 +
\frac{1}{a^{2}} + 
\frac{1}{b^{2}} +
\frac{1}{c^{2}} + 
2\sum_{cyc}{3 + \frac{1}{ab}}}$$
$$= \sqrt{9 +
\frac{1}{a^{2}} + 
\frac{1}{b^{2}} +
\frac{1}{c^{2}} + 
54}$$
$$\geqslant \sqrt{9 +18 + 54} =\sqrt{81} = 9$$
Which uses $\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} = 18$ and $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \geqslant \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}$.
</s>