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.

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
$$\Delta s = 0$$
