Example 4 Given a positive integer $n(n \geqslant 2)$. Find the largest real number $\lambda$, such that the inequality
$$
a_{n}^{2} \geqslant \lambda\left(a_{1}+a_{2}+\cdots+a_{n-1}\right)+2 a_{n},
$$

holds for any positive integers $a_{1}$, $a_{2}, \cdots, a_{n}$ satisfying $a_{1}<a_{2}<\cdots<a_{n}$.
(2003, Girls' Mathematical Olympiad)
Explanation: First, derive the range of $\lambda$ for a special sequence $\left\{a_{n}\right\}$, then provide a proof for any sequence $\left\{a_{n}\right\}$.
Let $a_{i}=i(i=1,2, \cdots, n)$.
From equation (1), it is easy to find that $\lambda \leqslant \frac{2(n-2)}{n-1}$.
Next, we prove: For any positive integers $a_{1}<a_{2}<\cdots<a_{n}$, we have
$$
a_{n}^{2} \geqslant \frac{2(n-2)}{n-1}\left(a_{1}+a_{2}+\cdots+a_{n-1}\right)+2 a_{n} .
$$

By $a_{k}+(n-k) \leqslant a_{n}$, i.e.,
$$
a_{k} \leqslant a_{n}-(n-k)(k=1,2, \cdots, n-1),
$$

and $a_{n} \geqslant n$, we have
$$
\begin{array}{l}
\frac{2(n-2)}{n-1}\left(a_{1}+a_{2}+\cdots+a_{n-1}\right) \\
\leqslant \frac{2(n-2)}{n-1}\left\{(n-1) a_{n}-\right. \\
\quad[(n-1)+(n-2)+\cdots+2+1]\} \\
=\frac{2(n-2)}{n-1}\left[(n-1) a_{n}-\frac{1}{2} n(n-1)\right] \\
=2(n-2) a_{n}-n(n-2) \\
=(n-2)\left(2 a_{n}-n\right) .
\end{array}
$$
$$
\begin{array}{l}
\text { and }\left(a_{n}^{2}-2 a_{n}\right)-(n-2)\left(2 a_{n}-n\right) \\
=a_{n}\left[\left(a_{n}-n\right)+(2-n)\right]+n(n-2) \\
=(n-2)\left(n-a_{n}\right)+a_{n}\left(a_{n}-n\right) \\
=\left(a_{n}-n\right)\left[a_{n}-(n-2)\right] \geqslant 0,
\end{array}
$$

Thus, inequality (2) holds.
Therefore, the maximum value of $\lambda$ is $\frac{2(n-2)}{n-1}(n \geqslant 2)$.
Note: First, by the simplest special case $a_{i}=i(i=1,2, \cdots, n)$, we conjecture the maximum possible value of $\lambda$ to be $\frac{2(n-2)}{n-1}$, then prove it for the general case.