Prove: Let $\sum_{i=1}^{m} a_{i j}^{n}=A_{j}^{n}$.
By the arithmetic-geometric mean inequality, we have
$$\begin{array}{l} \sum_{i=1}^{m} \prod_{j=1}^{n} \frac{a_{i j}}{A_{j}} \leqslant \sum_{i=1}^{m}\left(\frac{1}{n} \sum_{j=1}^{n} \frac{a_{i j}^{n}}{A_{j}^{n}}\right) \\ =\frac{1}{n} \sum_{j=1}^{n} \sum_{i=1}^{m} \frac{a_{j}^{n}}{A_{j}^{n}}=\frac{1}{n} \sum_{j=1}^{n} 1=1, \end{array}$$

Thus, $\left(\sum_{i=1}^{m} \prod_{j=1}^{n} a_{i j}\right)^{n} \leqslant A_{1}^{n} A_{2}^{n} \cdots A_{n}^{n}=\prod_{j=1}^{n} \sum_{i=1}^{m} a_{i j}^{n}$.
Note: Taking $n=2$ yields the Cauchy inequality, and taking $m=2$ yields
$$\left(\sum_{i=1}^{2} \prod_{j=1}^{n} a_{i j}\right)^{n} \leqslant \prod_{j=1}^{n} \sum_{i=1}^{2} a_{i j}^{n},$$

which is a transformation of a problem from the 64th Putnam Mathematical Competition.

Corollary: Let $a_{1}, a_{2}, \cdots, a_{n}$ and $b_{1}, b_{2}, \cdots, b_{n}$ be non-negative real number sequences. Then
$$\begin{array}{l} {\left[\left(a_{1}+b_{1}\right)\left(a_{2}+b_{2}\right) \cdots\left(a_{n}+b_{n}\right)\right]^{\frac{1}{n}}} \\ \geqslant\left(a_{1} a_{2} \cdots a_{n}\right)^{\frac{1}{n}}+\left(b_{1} b_{2} \cdots b_{n}\right)^{\frac{1}{n}} \end{array}$$

Prove: Let $\sum_{i=1}^{m} a_{i j}^{n}=A_{j}^{n}$.
By the Arithmetic-Geometric Mean Inequality, we have
$$\begin{array}{l} \sum_{i=1}^{m} \prod_{j=1}^{n} \frac{a_{i j}}{A_{j}} \leqslant \sum_{i=1}^{m}\left(\frac{1}{n} \sum_{j=1}^{n} \frac{a_{i j}^{n}}{A_{j}^{n}}\right) \\ =\frac{1}{n} \sum_{j=1}^{n} \sum_{i=1}^{m} \frac{a_{j j}^{n}}{A_{j}^{n}}=\frac{1}{n} \sum_{j=1}^{n} 1=1 \end{array}$$

Thus, $\left(\sum_{i=1}^{m} \prod_{j=1}^{n} a_{i j}\right)^{n} \leqslant A_{1}^{n} A_{2}^{n} \cdots A_{n}^{n}=\prod_{j=1}^{n} \sum_{i=1}^{m} a_{i j}^{n}$.