Example 8 Given positive integers $r, s, t$, satisfying $1<r<s<t$. For all positive real numbers $x_{1}, x_{2}, \cdots, x_{n}$ that satisfy the condition
$$
\frac{x_{j}}{x_{j+1}} \leqslant 1+\frac{s-t}{j+t}(j=1,2, \cdots, n)
$$

find
$$
M=\frac{\sum_{j=1}^{n} j(j+1) \cdots(j+s-1) x_{j}}{\sum_{j=1}^{n}(j+r) \cdots(j+s-1) x_{j}}
$$

the minimum value.
Analysis: Note that $\frac{x_{j}}{x_{j+1}} \leqslant \frac{j+s}{j+t}$.
First, fill in the missing terms, and transform it into
$$
\frac{x_{j}}{x_{j+1}} \leqslant \frac{(j+s)(j+s+1) \cdots(j+t-1)}{(j+s+1) \cdots(j+t)} .
$$

Guess that the minimum value is achieved when all equalities hold. Thus, take $x_{j}=(j+s)(j+s+1) \cdots(j+t-1)$, and then verify that the value of the fraction is minimized at this point.
Solution: Let $x_{j}=(j+s)(j+s+1) \cdots(j+t-1)$ $(j=1,2, \cdots, n)$. Then
$$
M=\frac{\sum_{j=1}^{n} j(j+1) \cdots(j+t-1)}{\sum_{j=1}^{n}(j+r) \cdots(j+t-1)}.
$$

We will prove that for positive real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying the given conditions, we have
$$
M \geqslant \frac{\sum_{j=1}^{n} j(j+1) \cdots(j+t-1)}{\sum_{j=1}^{n}(j+r) \cdots(j+t-1)}.
$$

In fact, let
$$
\begin{array}{l}
x_{j}=(j+s)(j+s+1) \cdots(j+t-1) a_{j}, \\
m_{j}=(j+r) \cdots(j+t-1). \\
b_{j}=j(j+1) \cdots(j+r-1) \quad (j=1,2, \cdots, n).
\end{array}
$$

Thus, it suffices to prove:
$$
\sum_{j=1}^{n} m_{j} a_{j} b_{j} \cdot \sum_{j=1}^{n} m_{j} \geqslant \sum_{j=1}^{n} m_{j} a_{j} \cdot \sum_{j=1}^{n} m_{j} b_{j}.
$$

Notice that
$$
\begin{array}{l}
00 \quad (j=1,2, \cdots, n).
\end{array}
$$

By the weighted form of Chebyshev's inequality, inequality (1) holds.
Therefore, the minimum value of $M$ is
$$
\frac{\sum_{j=1}^{n} j(j+1) \cdots(j+t-1)}{\sum_{j=1}^{n}(j+r) \cdots(j+t-1)},
$$

which is $\frac{t-r+1}{t+1} \cdot \frac{n(n+1) \cdots(n+t)}{(n+r) \cdots(n+t)-r(r+1) \cdots t}$.
Note: (1) The final step of simplification can be achieved through the method of partial fraction decomposition.
(2) Replacing $x_{j}$ with $a_{j}$ is a crucial step, and determining $m_{j}$ and $b_{j}$ also requires strong insight. The weighted form of Chebyshev's inequality is not often used, and this problem can be considered a classic example.