Proof: Let $x_{1}, x_{2}, \cdots, x_{n} \in(1,+\infty)$, then by the generalized Cauchy inequality we have
$$\left(1+x_{1}\right)\left(1+x_{2}\right) \cdots\left(1+x_{n}\right) \geqslant\left(1+\sqrt[n]{x_{1} x_{2} \cdots x_{n}}\right)^{n}$$

i.e., $\sqrt[n]{\left(1+x_{1}\right)\left(1+x_{2}\right) \cdots\left(1+x_{n}\right)} \geqslant 1+\sqrt[n]{x_{1} x_{2} \cdots x_{n}}$.
In (1), replace $x_{i}$ with $x_{i}-1$ to get
$$\sqrt[n]{\left(x_{1}-1\right)\left(x_{2}-1\right) \cdots\left(x_{n}-1\right)} \leqslant \sqrt[n]{x_{1} x_{2} \cdots x_{n}}-1$$
(1) $\div$ (2) gives
$$\sqrt[n]{\frac{\left(x_{1}+1\right)\left(x_{2}+1\right) \cdots\left(x_{n}+1\right)}{\left(x_{1}-1\right)\left(x_{2}-1\right) \cdots\left(x_{n}-1\right)}} \geqslant \frac{\sqrt[n]{x_{1} x_{2} \cdots x_{n}}+1}{\sqrt[n]{x_{1} x_{2} \cdots x_{n}}-1}$$

Take $x_{i}=r_{i} s_{i} t_{i} u_{i} v_{i}(i=1,2, \cdots, n)$ to get
$$\prod_{i=1}^{n} \frac{r_{i} s_{i} t_{i} u_{i} v_{i}+1}{r_{i} s_{i} t_{i} u_{i} v_{i}-1} \geqslant\left(\frac{\sqrt{\prod_{i=1}^{n} r_{i} s_{i} t_{i} u_{i} v_{i}}+1}{\sqrt{\prod_{i=1}^{n} r_{i} s_{i} t_{i} u_{i} v_{i}}-1}\right)$$

By the AM-GM inequality, we have
$$\sqrt[n]{\prod_{i=1}^{n} r_{i}} \leqslant \frac{1}{n} \sum_{i=1}^{n} r_{i}=R$$

Similarly, $\sqrt[n]{\prod_{i=1}^{n} s_{i}} \leqslant S, \sqrt[n]{\prod_{i=1}^{n} t_{i}} \leqslant T, \sqrt[n]{\prod_{i=1}^{n} u_{i}} \leqslant U, \sqrt[n]{\prod_{i=1}^{n} v_{i}} \leqslant V$,
so
$$\sqrt[n]{\prod_{i=1}^{n} r_{i} s_{i} t_{i} u_{i} v_{i}} \leqslant R S T U V$$

Since $y=\frac{x+1}{x-1}$ is a decreasing function on $(1,+\infty)$, we have
$$\frac{\sqrt{\prod_{i=1}^{n} r_{i} s_{i} t_{i} u_{i} v_{i}}+1}{\sqrt{\prod_{i=1}^{n} r_{i} s_{i} t_{i} u_{i} v_{i}}-1} \geqslant \frac{R S T U V+1}{R S T U V-1}$$

By (3) and (4), the original inequality holds.