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.
 A: 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 $(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)$
A: 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 $\sqrt3=1.73$
A: 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
$$\Delta s = 0$$
