Example 2. Non-negative real numbers $a_{1}, a_{2}, a_{3}, \cdots, a_{n}$ satisfy $a_{1}+$ $a_{2}+\cdots+a_{n}=1$. Find the minimum value of $\frac{a_{1}}{1+a_{2}+\cdots+a_{n}}+$ $\frac{a_{2}}{1+a_{3}+\cdots+a_{n}}+\cdots+\frac{a_{n}}{1+a_{2}+\cdots+a_{n-1}}$.
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, $S_{\text {min }}=\frac{n}{2 n-1}$.