Example 38 (1982 West German Mathematical Olympiad) Non-negative real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy $\sum_{k=1}^{n} a_{k}=1$, find the minimum value of $\frac{a_{1}}{1+a_{2}+a_{3}+\cdots+a_{n}}+\frac{a_{2}}{1+a_{1}+a_{3}+\cdots+a_{n}}+\cdots+\frac{a_{n}}{1+a_{1}+a_{2}+\cdots+a_{n-1}}$.
Solution 1: From the given conditions, the original problem can be transformed into finding the minimum value of $S=\sum_{k=1}^{n} \frac{a_{k}}{2-a_{k}}$ under the condition $\sum_{k=1}^{n} a_{k}=1$.
Assume $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$, then $\frac{1}{2-a_{1}} \leqslant \frac{1}{2-a_{2}} \leqslant \cdots \leqslant \frac{1}{2-a_{n}}$. By Chebyshev's and weighted power mean inequalities, we get $S \geqslant \frac{1}{n} \sum_{k=1}^{n} a_{k} \sum_{k=1}^{n} \frac{1}{2-a_{k}}=\frac{1}{n} \sum_{k=1}^{n} \frac{1^{2}}{2-a_{k}} \geqslant \frac{1}{n} \cdot \frac{\left(\sum_{k=1}^{n} 1\right)^{2}}{\sum_{k=1}^{n}\left(2-a_{k}\right)}=$
$$
\frac{1}{n} \cdot \frac{n^{2}}{2 n-1}=\frac{n}{2 n-1} .
$$

Equality holds if and only if $a_{1}=a_{2}=\cdots=a_{n}=\frac{1}{n}$, hence the minimum value of $S$ is $\frac{n}{2 n-1}$.
Solution 2: By the weighted power mean inequality, we have
$$
\begin{array}{c}
S=\sum_{k=1}^{n} \frac{a_{k}}{2-a_{k}}=\sum_{k=1}^{n}\left(\frac{2}{2-a_{k}}-1\right)=2 \sum_{k=1}^{n} \frac{1^{2}}{2-a_{k}}-n \geqslant \\
2 \frac{\left(\sum_{k=1}^{n} 1\right)^{2}}{\sum_{k=1}^{n}\left(2-a_{k}\right)}-n=\frac{2 n^{2}}{2 n-1}-n=\frac{n}{2 n-1} .
\end{array}
$$

The rest is the same as Solution 1.
Solution 3: By Cauchy's inequality, we have
$$
n^{2}=\left(\sum_{k=1}^{n} \sqrt{\frac{1}{2-a_{k}}} \cdot \sqrt{2-a_{k}}\right)^{2} \leqslant \sum_{k=1}^{n} \frac{1}{2-a_{k}} \sum_{k=1}^{n}\left(2-a_{k}\right)=(2 n-1) \sum_{k=1}^{n} \frac{1}{2-a_{k}},
$$

i.e., $\sum_{k=1}^{n} \frac{1}{2-a_{k}} \geqslant \frac{n^{2}}{2 n-1}$, so $S=\sum_{k=1}^{n} \frac{a_{k}}{2-a_{k}}=\sum_{k=1}^{n}\left(\frac{2}{2-a_{k}}-1\right) \geqslant \frac{2 n^{2}}{2 n-1}-$ $n=\frac{n}{2 n-1}$.
The rest is the same as Solution 1.