Example 4 Given positive numbers $a_{1}, a_{2}, \cdots, a_{n}(n \geqslant 3)$ satisfying
$$
\begin{array}{l}
A>a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}, \\
a_{1}+a_{2}+\cdots+a_{n}=2 A .
\end{array}
$$

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

On the other hand, when
$$
A-a_{1}=a_{1}-a_{2}=\cdots=a_{n-1}-a_{n}=\frac{2(n-2) A}{n(n+1)}
$$

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

Therefore, when $a_{k}=\frac{n(n+1)-2(n-2) k}{n(n+1)}(k=$ $1,2, \cdots, n)$, $\alpha$ attains its maximum value $\frac{2(n-2) A}{n(n+1)}$.