Example 5 Given positive numbers $a_{1}, a_{2}, \cdots, a_{n}(n \geqslant 3)$ satisfying
$$
\begin{array}{l}
m>a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}, \\
a_{1}+a_{2}+\cdots+a_{n}=A<n m<2 A+m .
\end{array}
$$

Let $\alpha$ be the minimum of $m-a_{1}, a_{1}-a_{2}, \cdots, a_{n-1}-a_{n}$. Find the maximum value of $\alpha$.
Solution: From the given, we have
$$
\begin{array}{l}
\alpha=\frac{n \alpha+(n-1) \alpha+\cdots+2 \alpha+\alpha}{n+(n-1)+\cdots+2+1} \\
\leqslant \frac{n\left(m-a_{1}\right)+(n-1)\left(a_{1}-a_{2}\right)+\cdots+\left(a_{n-1}-a_{n}\right)}{n+(n-1)+\cdots+2+1} \\
=\frac{2}{n(n+1)}\left[n m-\left(a_{1}+a_{2}+\cdots+a_{n}\right)\right] \\
\quad=\frac{2(n m-A)}{n(n+1)} .
\end{array}
$$

On the other hand, when
$$
\begin{array}{l}
m-a_{1}=a_{1}-a_{2}=\cdots=a_{n-1}-a_{n} \\
=\frac{2(n m-A)}{n(n+1)}>0
\end{array}
$$

$a_{0}=m, a_{1}, \cdots, a_{n}$ form an arithmetic sequence with the first term $m$ and common difference $d=-\frac{2(n m-A)}{n(n+1)}$. Thus, we have
$$
\begin{array}{l}
a_{k}=m+k\left[-\frac{2(n m-A)}{n(n+1)}\right] \\
=\frac{n m(n-2 k+1)+2 k A}{n(n+1)} .
\end{array}
$$

This satisfies (1) $m>a_{1}>a_{2}>\cdots \geqslant a_{n}$, and
$$
\begin{array}{l}
a_{n}=\frac{n m(n-2 n+1)+2 n A}{n(n+1)} \\
=\frac{(m+2 A)-n m}{n+1}>0 .
\end{array}
$$
$$
\begin{array}{l}
\text { (2) } a_{1}+a_{2}+\cdots+a_{n}=\frac{\left(a_{1}+a_{n}\right) n}{2} \\
=\frac{n}{2}\left[\frac{n^{2} m-n m+2 A}{n(n+1)}+\frac{(m+2 A)-n m}{n+1}\right] \\
=A .
\end{array}
$$

Therefore, there exist positive numbers
$$
a_{k}=\frac{n m(n-2 k+1)+2 k A}{n(n+1)}
$$
$(k=1,2, \cdots, n)$, such that $\alpha$ attains its maximum value $\frac{2(n m-A)}{n(n+1)}$.