The AM-GM inequality via optimisation

Lagrange. Maximize $\ln P = \sum \ln x_i$ under the constraint $\sum x_i = S$. The system is
$$\frac{1}{x_i} = \lambda \quad \text{for all } i.$$
So all $x_i$ are equal to $1/\lambda$, and the constraint gives $x_i = S/n$ for all $i$.

Maximum. At this point, $P = (S/n)^n$. Since the domain $x_i > 0$, $\sum x_i = S$ is bounded and the product tends to $0$ on the boundary (if some $x_i \to 0$), the maximum is indeed attained at this point.

Conclusion. For all $x_i > 0$ with sum $S$,
$$x_1 \cdots x_n \le \left(\frac{S}{n}\right)^n = \left(\frac{x_1+\cdots+x_n}{n}\right)^n.$$
Taking the $n$-th root gives the AM-GM inequality.