Proving a local minimum is a global minimum. Let $f(x,y)=xy+ \frac{50}{x}+\frac{20}{y}$, Find the global minimum / maximum of the function for $x>0,y>0$
Clearly the function has no global maximum since $f$ is not bounded.
I have found that the point $(5,2)$ is a local minimum of $f$. It seems pretty obvious that this point is a global minimum, but I'm struggling with a formal proof.

By AM-GM $$f(x,y)\geq3\sqrt[3]{xy\cdot\frac{50}{x}\cdot\frac{20}{y}}=30.$$
The equality occurs for $$xy=\frac{50}{x}=\frac{20}{y}=10,$$
id est, for $(x,y)=(5,2)$, which says that $30$ is a minimal value.
The maximum does not exist. Try $x\rightarrow0^+$.
