Inequality with mean inequality

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

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

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.