I've got two similar problems:
Find
\begin{align} &W:= \frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y} \rightarrow \inf \\ &x + y + z = 1 \\ &x, y, z > 0 \end{align} and \begin{align} &W:= \frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y} \rightarrow \inf \\ &x^2 + y^2 + z^2 = 1 \\ &x, y, z > 0 \end{align} I got simple solutions for both of them, but i want to be sure i'm right: $\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{xz}{y} = \dfrac{xy}{2z}+\dfrac{yz}{2x}+\dfrac{xz}{2y} + \dfrac{xy}{2z}+\dfrac{yz}{2x}+\dfrac{xz}{2y} \overset{AM-GM}{\ge} \sqrt{y^2} + \sqrt{x^2} + \sqrt{z^2} = x + y + z.$
For the first problem $\inf$ has to be $1$. For the second problem second constrain defines the sphere, which infimum is in $(0, 0, 0)$ point. So answers are $1$ and $0$. Am i right?
UPD: I'm definetly not right with the second problem.