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.

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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 $(x,y,z) \in S$
$$ \begin{array}{rcl}f(x,y,z) &=& \frac{-x-y-z}{xyz} = \frac{x+y+z}{(-x)(-y)(-z)} \\ &\geq & \frac{x+y+z}{3^{-3/2}r^3} = -\frac{3^2}{r^2}\left(\frac{\sqrt{3}}{r}\cdot\frac{-x-y-z}{3}\right) \\ &\geq &-\frac{9}{r^2} = f\left(-\dfrac{r}{\sqrt{3}},-\dfrac{r}{\sqrt{3}},-\dfrac{r}{\sqrt{3}}\right). \end{array}$$

I’ll assume that $x$, $y$ and $z$ are positive.

$$\begin{align}
f(x,y,z)&=\frac{x+y+z}{xyz}\\
&\geq\frac{3\sqrt[3]{xyz}}{xyz}\\
&=\frac3{\sqrt[3]{x^2y^2z^2}}\\
&\geq\frac9{x^2+y^2+z^2}\\
&=\frac9{r^2}
\end{align}$$

Letting $x=y=z=r/\sqrt 3$ we have

$$f\left(\frac r{\sqrt 3},\frac r{\sqrt 3},\frac r{\sqrt 3}\right)=\frac{\sqrt3r}{r^3/(3\sqrt3)}=\frac9{r^2}$$

So $9/r^2$ is the minimum.