It has no global maximum, since, for example, you can let $x_3=x_4=\cdots=x_n$ and $x_2=1/x_1$, and then the product is $1$ and the sum is bigger than $x_1$, and you can make $x_1$ as big as you want.
If you draw the picture in the case $n=2$, you'll probably expect it does have a global minimum. When $x_1=\cdots=x_n=1$, then $x_1+\cdots+x_n=n$, and if you're not sure that's the minimum, just consider the set of all points where the sum is $\le 1$, and try to reason to the conclusion that that's a compact set. The whole graph of $x_1\cdots x_n=1$ is a closed set since it's the inverse image of a single point under a continuous function. Once a set in Euclidean space is closed and bounded, it's compact.