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 $\sqrt[4]{3}$.
 A: $$\sqrt{3 +\frac{1}{a^{2}}}+\sqrt{3 +\frac{1}{b^{2}}}+\sqrt{3 +\frac{1}{b^{2}}}$$
$$
= \sqrt{
9 +
\frac{1}{a^{2}} + 
\frac{1}{b^{2}} +
\frac{1}{c^{2}} + 
2\sum_{cyc}
{\sqrt{
\left(3 +
\frac{1}{a^{2}}\right)
\left(3 +\frac{1}{b^{2}}\right)}}
}
$$
$$\tag{By C-S}\geqslant \sqrt{9 +
\frac{1}{a^{2}} + 
\frac{1}{b^{2}} +
\frac{1}{c^{2}} + 
2\sum_{cyc}{3 + \frac{1}{ab}}}$$
$$= \sqrt{9 +
\frac{1}{a^{2}} + 
\frac{1}{b^{2}} +
\frac{1}{c^{2}} + 
54}$$
$$\geqslant \sqrt{9 +18 + 54} =\sqrt{81} = 9$$
Which uses $\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} = 18$ and $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \geqslant \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}$.
A: let $x=\frac{1}{a},y=\frac{1}{b},z=\frac{1}{c}$
$$xy+yz+zx=18 \tag{given}$$
$$\color{red}{x+y+z\ge 3\sqrt{6}} \tag1$$
notice that $${(x-\sqrt{6})}^2\ge 0\Rightarrow \sqrt{3+x^2}\ge \frac{\sqrt{6}}{3}x+1$$
Thus $$\sum \sqrt{3+x^2}\ge \sum \frac{\sqrt{6}}{3}x+1\ge\frac{\sqrt{6}}{3}\color{red}{3\sqrt{6}}+3=9$$
Note the inequality marked $(1)$ is left as an exercise
A: From Minkowski inequality: $\sqrt{x^2 + a^2} + \sqrt{y^2 + b^2} \ge \sqrt{(x + y)^2 + (a + b)^2}$
applied to the first two terms, with the notation of @Albus, we get:
$\sqrt{\sqrt{3}^2 +x^2} + \sqrt{\sqrt{3}^2 +y^2} \ge \sqrt{(2\sqrt{3})^2 + (x+ y)^2}$. Let's denote RHS of the inequality by $X$.
Then, apply Minkowski inequality again to $ X + \sqrt{3 + z^2}$, to obtain:
$\sqrt{27 + (x + y + z)^2} \ge 9$, which is equivalent to (1) from @Albus Dumbledore's answer.
A: 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 .
