Example 1. Let positive numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy $a_{1}+a_{2}+\cdots+a_{n}=1$. Find
$$\begin{array}{l}
\frac{a_{1}}{1+a_{2}+\cdots+a_{n}}+\frac{a_{2}}{1+a_{1}+a_{3}+\cdots+a_{n}} \\
+\cdots+\frac{a_{n}}{1+a_{1}+\cdots+a_{n-1}}
\end{array}$$

the minimum value. (82 West Germany Competition Problem)
Let the sum of the original expression be $S$, then
$$S=\sum_{k=1}^{n} \frac{a_{k}}{2-a_{k}}=2 \sum_{k=1}^{n} \frac{1}{2-a_{k}}-n \cdot(1)$$

Since $2-a_{k}>0(k=1,2 ; \cdots, n)$, by the Cauchy-Schwarz inequality,
$$\sum_{k=1}^{n} \frac{1}{2-a_{k}} \geqslant \frac{n^{2}}{\sum^{n}}=\frac{n^{2}}{2 n-1}$$

Combining (1) and (2), we get
$$S \geqslant \frac{2 n^{2}}{2 n-1}-n=\frac{n}{2 n-1}$$

The equality holds if and only if the equality in (2) holds, which in turn holds if and only if $a_{1}=a_{2}=\cdots=a_{n}=\frac{1}{n}$. Therefore, the minimum value of $S$ is $\frac{n}{2 n-1}$.