How prove $a + b +c + \frac{1}{abc} \geq 4\sqrt{3}$?

Question

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

Possibly related... http://math.stackexchange.com/questions/180937/let-a-b-c0-and-abc-1-how-to-prove-the-inequality-frac-sqrta1-a?rq=1 — A. Thomas Yerger, Apr 25 '15 at 19:18

score 1 · Accepted Answer · answered Apr 25 '15 at 19:40

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.

How prove $a + b +c + \frac{1}{abc} \geq 4\sqrt{3}$?

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