Example 11 For a positive integer $n \geqslant 3, x_{\mathrm{F}}, x_{2}, \cdots, x_{n}$ are positive real numbers, $x_{n+j}=x_{j}(1 \leqslant j \leqslant n-\mathrm{F})$, 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 National Mathematical Olympiad Training Team Question)
Let $a_{j}^{2}=\frac{x_{j}}{x_{j+1}+2 x_{j+2}+\cdots+(n-1) x_{j+n-1}}, b_{j}^{2}=x_{j}\left[x_{j+1}+2 x_{j+2}+\cdots+\right.$ $\left.(n-1) x_{j+n-1}\right]$, where $b_{j}$ is positive, then $a_{j} b_{j}=x_{j}$, by the Cauchy-Schwarz inequality
$$\sum_{j=1}^{n} a_{j}^{2} \cdot \sum_{j=1}^{n} b_{j}^{2} \geqslant\left(\sum_{j=1}^{n} a_{j} b_{j}\right)^{2}$$

This problem is to find the minimum value of $S=\sum_{j=1}^{n} a_{j}^{2}$, we now calculate $\sum_{j=1}^{n} b_{j}^{2}$.
$$\begin{array}{l}
\sum_{j=1}^{n} b_{j}^{2}=\sum_{j=1}^{n} x_{j}\left(x_{j+1}+2 x_{j+2}+\cdots+(n-1) x_{j+n-1}\right)= \\
x_{1}\left[x_{2}+2 x_{3}+3 x_{4}+\cdots+(n-1) x_{n}\right]+ \\
x_{2}\left[x_{3}+2 x_{4}+3 x_{5}+\cdots+(n-1) x_{1}\right]+\cdots+ \\
x_{n-1}\left[x_{n}+2 x_{1}+3 x_{2}+\cdots+(n-1) x_{n-2}\right]+ \\
x_{n}\left[x_{1}+2 x_{2}+3 x_{3}+\cdots+(n-1) x_{n-1}\right]= \\
{[1+(n-1)] x_{1} x_{2}+[2+(n-2)] x_{1} x_{3}+\cdots+} \\
{[(n-1)+1] x_{1} x_{n}+[1+(n-1)] x_{2} x_{3}+} \\
{[2+(n-2)] x_{2} x_{4}+\cdots+[(n-2)+2] x_{2} x_{n}+\cdots+} \\
{[1+(n-1)] x_{n-1} x_{n}=} \\
n \sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}
\end{array}$$

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 $\sum_{1 \leq i<j \leq n} x_{i} x_{j}$ 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 (2) and (3), we get $S \geqslant \frac{2}{n-1}$. When $x_{1}=x_{2}=\cdots=x_{n}$, the value of $S$ is indeed $\frac{2}{n-1}$, so the minimum value of $S$ is $\frac{2}{n-1}$.

Note: For $n=4$, this problem is the 34th IMO shortlist problem:
For all positive real numbers $a, b, c, d$, prove that $\frac{a}{b+2 c+3 d}+\frac{b}{c+2 d+3 a}+\frac{c}{d+2 a+3 b}+$
$$\frac{d}{a+2 b+3 c} \geqslant \frac{2}{3}$$