$\square$ Example 11 Positive integers $n \geqslant 3, x_{1}, x_{2}, \cdots, x_{n}$ are positive real numbers, $x_{n+j}=x_{j}(1 \leqslant$ $j \leqslant n-1)$, find the minimum value of $\sum_{i=1}^{n} \frac{x_{j}}{x_{j+1}+2 x_{j+2}+\cdots+(n-1) x_{j+n-1}}$. (1995 Chinese National Mathematical Olympiad Training Team Question)
Let $a_{j}^{2}=\frac{x_{j}}{x_{i+1}+2 x_{i+2}+\cdots+(n-1) x_{i+1}}$,
$$b_{j}^{2}=x_{j}\left[x_{j+1}+2 x_{j+2}+\cdots+(n-1) x_{j+n-1}\right]$$

Then $a_{j}, b_{j}$ are positive numbers, and $a_{j} b_{j}=x_{j}$. The problem is to find the minimum value of $S=\sum_{j=1}^{n} a_{j}^{2}$. By the Cauchy-Schwarz inequality, $\sum_{j=1}^{n} a_{j}^{2} \cdot \sum_{j=1}^{n} b_{j}^{2} \geqslant\left(\sum_{j=1}^{n} a_{j} b_{j}\right)^{2}$. We now calculate $\sum_{j=1}^{n} b_{j}^{2}$.
$$\begin{aligned}
\sum_{j=1}^{n} b_{j}^{2}= & \sum_{j=1}^{n} x_{j}\left[x_{j+1}+2 x_{j+2}+\cdots+(n-1) x_{j+n-1}\right] \\
= & x_{1}\left[x_{2}+2 x_{3}+3 x_{4}+\cdots+(n-1) x_{n}\right]+x_{2}\left[x_{3}+2 x_{4}\right. \\
& \left.+3 x_{5}+\cdots+(n-1) x_{1}\right]+\cdots+x_{n-1}\left[x_{n}+2 x_{1}+3 x_{2}\right. \\
& \left.+\cdots+(n-1) x_{n-2}\right]+x_{n}\left[x_{1}+2 x_{2}+3 x_{3}+\cdots+(n\right. \\
& \left.-1) x_{n-1}\right] \\
= & {[1+(n-1)] x_{1} x_{2}+[2+(n-2)] x_{1} x_{3}+\cdots+[(n} \\
& -1)+1] x_{1} x_{n}+[1+(n-1)] x_{2} x_{3}+[2+(n- \\
& 2)] x_{2} x_{4}+\cdots+[(n-2)+2] x_{2} x_{n}+\cdots+[1+(n- \\
& 1)] x_{n-1} x_{n} \\
= & n \cdot \sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j},
\end{aligned}$$

Thus,
$$n S \sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j} \geqslant\left(\sum_{i=1}^{n} x_{j}\right)^{2}$$

Next, we compare $\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}$ with $\left(\sum_{i=1}^{n} x_{j}\right)^{2}$. Since
$$0 \leqslant \sum_{1 \leqslant i<j \leqslant n}\left(x_{i}-x_{j}\right)^{2}=(n-1) \sum_{i=1}^{n} x_{i}^{2}-2 \sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}$$

From the above, we have
$$\frac{2}{n-1} \sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j} \leqslant \sum_{i=1}^{n} x_{i}^{2}=\left(\sum_{i=1}^{n} x_{i}\right)^{2}-2 \sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j},$$

Thus,
$$\frac{2 n}{n-1} \sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j} \leqslant\left(\sum_{i=1}^{n} x_{i}\right)^{2}$$

From (1) and (2), we get $S \geqslant \frac{2}{n-1}$. When $x_{1}=x_{2}=\cdots=x_{n}$, the value of $S$ is $\frac{2}{n-1}$. Therefore, the minimum value of $S$ is $\frac{2}{n-1}$.