Minimizing $f(x,y,z)=\dfrac{|x|+|y|+|z|}{xyz}$ on a sphere

Question

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.

If you want to try for yourself, you might want to look up the term Lagrange multiplier (there should be loads of questions with that around here). Although in this case I would suspect you can get the answer by using symmetry and playing around with the inequalites between the different means (arithmetric, geometric, etc.) — mlk, Jun 19 '14 at 09:45
Something looks awry with your statement; it's fairly clear $f$ is not bounded below on a sphere centered at the origin. Is Jlamprong's guess correct...? — Andrew D. Hwang, Jun 22 '14 at 23:51

Surb · Accepted Answer · 2014-06-23T08:41:07.470

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

score 0 · Answer 2 · answered Jun 19 '14 at 10:11

0

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.

answered Jun 19 '14 at 10:11

ajotatxe

65,084

With $x=y=-z=r/\sqrt 3$, you'll get $f=-9/r^2$, which is smaller. – lhf Jun 19 '14 at 10:18
With $z=-\epsilon$, you can get $f$ as low as you want. – Empy2 Jun 19 '14 at 10:55
@Michael This is the reason for that I have assumed that $x$, $y$ and $z$ are positive. – ajotatxe Jun 19 '14 at 11:55
Whoops, I missed that – Empy2 Jun 19 '14 at 12:29
Because the sphere can be centered in origin of the reference frame $(x,y,z)$, the inequality: $x\ge 0,y\ge 0, z\ge 0$ doesn't hold. – Riccardo.Alestra Jun 19 '14 at 12:31

Minimizing $f(x,y,z)=\dfrac{|x|+|y|+|z|}{xyz}$ on a sphere

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