4.5.12 ᄎ* Given a positive integer $n \geqslant 2$, find the largest real number $\lambda$ such that the inequality $a_{n}^{2} \geqslant \lambda\left(a_{1}+\right.$ $\left.a_{2}+\cdots+a_{n-1}\right)+2 a_{n}$. holds for any integers $a_{1}, a_{2}, \cdots, a_{n}$ satisfying $0<a_{1}<a_{2}<\cdots<a_{n}$.
When $a_{i}=i, i=1,2, \cdots, n$, $\lambda \leqslant(n-2) \div \frac{n-1}{2}=\frac{2 n-1}{n-1}$.
Below, we prove the inequality $a_{n}^{2} \geqslant \frac{2 n-4}{n-1}\left(a_{1}+a_{2}+\cdots+a_{n-1}\right)+2 a_{n}$ holds for any integers $a_{1}, a_{2}, \cdots, a_{n}$ satisfying $0<a_{1}<$ $a_{2}<\cdots<a_{n}$.
Since $a_{k} \leqslant a_{n}-(n-k), k=1,2, \cdots, n-1, a_{n} \geqslant n$, we have
$$
\begin{aligned}
\frac{2 n-4}{n-1} \sum_{k=1}^{n-1} a_{k} & \leqslant \frac{2 n-4}{n-1}\left((n-1) a_{n}-\frac{n(n-1)}{2}\right) \leqslant(2 n-4) a_{n}-n(n-2) \\
& =(n-2)\left(2 a_{n}-n\right) \leqslant\left(a_{n}-2\right) a_{n},
\end{aligned}
$$

which means
$$
a_{n}^{2} \geqslant \frac{2 n-4}{n-1}\left(a_{1}+a_{2}+\cdots+a_{n-1}\right)+2 a_{n} \text {. }
$$

Therefore, the maximum value of $\lambda$ is $\frac{2 n-4}{n-1}$.