Let $S=\frac{a_{1}}{1+a_{2}+\cdots+a_{n}}$
$$\begin{array}{l} +\frac{a_{2}}{1+a_{3}+\cdots+a_{n}}+\cdots \\ +\frac{a_{n}}{a_{1}+a_{2}+\cdots+a_{n-1}}, \end{array}$$

then $S:=\sum_{i=1}^{n} \frac{a_{i}}{2-a_{i}}=2 \sum_{i=1}^{n} \frac{1}{2-a_{i}}-n$.
Since $2-a_{i}>0$, by (*) we get
$\frac{1}{2-a_{i}} \geqslant 2 \lambda-\lambda^{2}\left(2-a_{i}\right)$.
Therefore, $S \geqslant 2 \sum_{i=1}^{n}\left[2 \lambda-\lambda^{2}\left(2-a_{i}\right)\right]-n$
$$=2\left[2 n \lambda-\lambda^{2}(2 n-1)\right]-n .$$

Taking $\lambda=\frac{n}{2 n-1}$ and substituting it into the above equation, we have
$$S \geqslant \frac{n}{2 n-1},$$

with equality holding if and only if $\lambda=\frac{1}{2-a_{i}}=\frac{n}{2 n-1}$, i.e., $a_{i}=\frac{1}{n}$. Thus, $S_{\text {min }}=\frac{n}{2 n-1}$.