minimum value of $\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}$ if $x^2+y^2+z^2=1$

Question

If $x^2+y^2+z^2=1$ what is the minimum value of $\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}$ for $x,y,z \gt 0$ ?

I would like to know if the minimum could be found using simpler ways.(like $AM \ge GM \ge HM$).

knowing $xy+yz+xz \geq \frac{-1}{2}$ for real $x,y,z$ might help.

Answer 1

Comment

May be this idea can help you:

Let's start with a simpler form where in two dimension we have: $x^2+y^2=1$ corresponds to a circle $r=1$ and we want to find minimum of $\frac xy+\frac yx$. So we have to find minimum of:

$f(\alpha)=tan(\alpha)+ cotan(\alpha)$

taking derivative and equating to zero we get:

$\frac1{cos^2(\alpha)}-\frac1{sin^2(\alpha)}=0$

Which gives $\alpha=\frac{\pi}4$

That is $\frac xy+\frac yx$ is minimum when $x=y=\frac1{\sqrt2}$

Now we extend this idea to three dimension where $x^2+y^2+z^2=1$ corresponds to a sphere with radius 1 and we have to find minimum of $z\cdot tan(\alpha)+x\cdot tan(\beta)+ y\cdot tan(\gamma)$.Where $\alpha$. If we apply the idea we got from two dimension we have to have:

$x=y=z=\frac1{\sqrt3}$

$\alpha=\beta=\gamma=\frac{\pi}4$

Which gives:

$\frac{xy}z+\frac{yz}x+\frac{xz}y= 3\times\frac{(\frac1{\sqrt3})^2}{\frac 1{\sqrt3}}=\sqrt3$

I ckecked this with when $x=\frac1{\sqrt2}$, $y=\frac1{\sqrt3}$ and $z=\frac1{\sqrt6}$ which gives: $1.98$ where $\sqrt3=1.73$

Answer 2

First, notice that $\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y} = \frac{xyz}{z^2}+\frac{xyz}{x^2}+\frac{xyz}{y^2}=xyz({\frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}})$

Using the inequality of arithmetic and geometric means, we have: $$ \frac{1}{x^2}+\frac{1}{y^2}\ge2\sqrt{\frac{1}{x^2}\frac{1}{y^2}}\iff \frac{1}{x^2}+\frac{1}{y^2}\ge \frac{2}{xy}$$ Do that for every pair of variables and add everything together: $$\frac{1}{x^2}+\frac{1}{y^2}\ge \frac{2}{xy}$$ $$\frac{1}{x^2}+\frac{1}{z^2}\ge \frac{2}{xz}$$ $$\frac{1}{y^2}+\frac{1}{z^2}\ge \frac{2}{yz}$$ We have: $$2({\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}})\ge2(\frac{1}{xy}+\frac{1}{xz}+\frac{1}{yz}) \iff {\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}}\ge\frac{x+y+z}{xyz}$$ Which simplifies to: $$xyz({\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}})\ge x+y+z$$ Now, we just need to find the minimum value of $x+y+z$ when $x^2+y^2+z^2=1$, which I am not sure how to do. But since knowing $xy+xz+yz>-0.5$ might help, I suggest you to try using that $(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)$

Answer 3

As mentioned in comments, the standard way to approach such constrained minimization problems is to use the method of Lagrange multipliers. There might be a simpler or more elegant solution, I don't know.

Let $$s = \frac{xy}z + \frac{yz}x + \frac{xz}y$$ That is, $$s = \frac{(xy)^2 + (yz)^2 + (xz)^2}{xyz}$$ Alternatively, $$s = xyz\left(\frac1{x^2} + \frac1{y^2} + \frac1{z^2}\right)$$ Also note that $$s^2=2 + \left(\frac{xy}{z}\right)^2 + \left(\frac{yz}{x}\right)^2 + \left(\frac{xz}{y}\right)^2$$ (That form might lead to a simpler solution).

From symmetry considerations, it seems reasonable that $x=y=z=\sqrt3/3, s=\sqrt3$ is the solution. We can verify that it is a (local) stationary point of $s$ using first order differences.

The constraint equation is $$x^2 + y^2 + z^2 = 1$$ so $$2x\Delta x + 2y\Delta y + 2z\Delta z = 0$$ If we let $x=y=z=q=\sqrt3/3$, that reduces to $$\Delta x + \Delta y + \Delta z = 0$$ or $$\Delta z=-\Delta x - \Delta y$$

In what follows, we plug $$\begin{align}\\ x=q+\Delta x\\ y=q+\Delta y\\ z=q+\Delta z\\ s=3q+\Delta s\\ \end{align}$$ into $$s = \frac{(xy)^2 + (yz)^2 + (xz)^2}{xyz}$$ The aim is to show that $\Delta s=0$

First, note that $$(q+\Delta x)^2=q^2+2q\Delta x$$ etc. And $$(q^2+2q\Delta x)(q^2+2q\Delta y)\\ = q^4 + 2q^3(\Delta x + \Delta y)$$

So the numerator of $s$ is $$\begin{align}\\ q^4 & + 2q^3(\Delta x + \Delta y)\\ + q^4 & + 2q^3(\Delta y + \Delta z)\\ + q^4 & + 2q^3(\Delta x + \Delta z)\\ = 3q^4 & + 2q^3(2\Delta x + 2\Delta y + 2\Delta z)\\ = 3q^4 &\\ \end{align}$$

And the denominator of $s$ is $$\begin{align}\\ & (q + \Delta x)(q + \Delta y)(q + \Delta z)\\ = & (q^2 + q(\Delta x + \Delta y))(q + \Delta z)\\ = & q^3 + q^2(\Delta x + \Delta y + \Delta z)\\ = & q^3\\ \end{align}$$

Thus $$3q + \Delta s = \frac{3q^4}{q^3} = 3q$$ Hence $$\Delta s = 0$$