How prove $a + b +c + \frac{1}{abc} \geq 4\sqrt{3}$? Let $a, b, c >0$ and $a^2+b^2+c^2 =1$ How prove $a + b +c + \frac{1}{abc} \geq 4\sqrt{3}$?

Let's consider the function
$$f(a,b,c) = a+b+c+\frac{1}{abc}.$$
We want to minimize it.
Then 
$$f'_a = 1 - \frac{1}{(abc)a}, f'_a = 1 - \frac{1}{(abc)b}, f'_c = 1 - \frac{1}{(abc)c}$$
Imposing all of them to be $0$, we have:
$$\left\{\begin{array}{l}(abc)a = 1\\(abc)b = 1\\(abc)c = 1\end{array}\right.$$
This means that $a=b=c$. Now, consider the function:
$$f(x) = 3x+\frac{1}{x^3},$$
where $a=b=c=x$.
Since $a^2+b^2+c^2 = 3x^2 = 1$, then $x = \frac{\sqrt{3}}{3} > 0$ and get that the minimum is
$$f\left(\frac{\sqrt{3}}{3}\right) = \sqrt{3}+3\sqrt{3} = 4 \sqrt{3}$$
Addition
It is easy to show that the function $f(a,b,c)$ is convex and thus it has a relative minimum.
