Proving a local minimum is a global minimum. Let f(x,y)=xy+ (50)/(x)+(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.

Proving a local minimum is a global minimum.
Let $f\left(x,y\right)=xy+\frac{50}{x}+\frac{20}{y}$, 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 $\left(5,2\right)$ 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.
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Julianne Mccoy
By AM-GM
$f\left(x,y\right)\ge 3\sqrt[3]{xy\cdot \frac{50}{x}\cdot \frac{20}{y}}=30.$
The equality occurs for
$xy=\frac{50}{x}=\frac{20}{y}=10,$
id est, for $\left(x,y\right)=\left(5,2\right)$, which says that $30$ is a minimal value.
The maximum does not exist. Try $x\to {0}^{+}$