Minimizing $f(x,y,z)=\dfrac{|x|+|y|+|z|}{xyz}$ on a sphere I need to find the minimum of the function: $$f(x,y,z)=\dfrac{|x|+|y|+|z|}{xyz}$$ with the condition: $$x^2+y^2+z^2=r^2$$
Using numerical methods it's quite easy to solve the problem. How can I proceed to solve it analytically? Thanks. 

The first thing to note is that for every $(x,y,z) \in \mathbb{R}^3$, we have $$f(-|x|,-|y|,-|z|) \leq f(x,y,z),$$
It follows that we may look for a global minimum in the set
$$S := \{(x,y,z) \in \mathbb{R}^3: x,y,z \leq 0 \text{ and } x^2+y^2+z^2=r^2\}.$$
Now by the inequalities between means, we may observe that for every $(x,y,z) \in S,$ 
$$((-x)(-y)(-z))^{1/3} \leq \frac{-x-y-z}{3}\leq\frac{\sqrt{x^2+y^2+z^2}}{\sqrt{3}} = \frac{r}{\sqrt{3}}.$$
It follows that for every $(x,y,z) \in S$ 
$$ \begin{array}{rcl}f(x,y,z) &=& \frac{-x-y-z}{xyz} = \frac{x+y+z}{(-x)(-y)(-z)} \\ &\geq & \frac{x+y+z}{3^{-3/2}r^3} = -\frac{3^2}{r^2}\left(\frac{\sqrt{3}}{r}\cdot\frac{-x-y-z}{3}\right) \\ &\geq &-\frac{9}{r^2} = f\left(-\dfrac{r}{\sqrt{3}},-\dfrac{r}{\sqrt{3}},-\dfrac{r}{\sqrt{3}}\right). \end{array}$$
