Proof First, it is clear that for $N=1,2, \cdots$, and $x, y \geqslant 0$, we have the relation

Thus we get
$$\begin{array}{c} \mathrm{e}^{x y}>\frac{x^{N} y^{N}}{N!} \\ \mathrm{e}^{y \sum_{i} x_{i}}>\frac{\left(x_{1} x_{2} \cdots x_{n}\right)^{N} y^{n N}}{(N!)^{n}} \end{array}$$

Therefore, by comparing the coefficients of $y^{n N}$, we obtain
$$\frac{\left(\sum_{i} x_{i}\right)^{n N}}{(n N)!} \geqslant \frac{\left(x_{1} x_{2} \cdots x_{n}\right)^{N}}{(N!)^{n}}$$

Or
$$\frac{\left(\sum_{i} x_{i}\right)^{n}}{x_{1} x_{2} \cdots x_{n}} \geqslant\left(\frac{(n N)!}{(N!)^{n}}\right)^{\frac{1}{N}}$$

which holds for all positive integers $N$.
Letting $N \rightarrow+\infty$, by Stirling's formula, we get
$$\lim _{N \rightarrow+\infty}\left(\frac{(n N)!}{(N!)^{n}}\right)^{\frac{1}{N}}=n^{n}$$

Thus, equation (1) is established. This proof does not provide the condition for equality to hold.