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

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 $(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)$
