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.

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 $\sqrt3=1.73$
