2. Non-negative real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy $a_{1}+a_{2}+\cdots+a_{n}=1$.
Find the minimum value of $\frac{a_{1}}{1+a_{2}+\cdots+a_{n}}+\frac{a_{2}}{1+a_{1}+a_{3}+\cdots+a_{n}}+\cdots+$ $\frac{a_{n}}{1+a_{1}+\cdots+a_{n-1}}$. (1982, China Mathematical Competition)
Let $S=\frac{a_{1}}{1+a_{2}+\cdots+a_{n}}+\frac{a_{2}}{1+a_{1}+a_{3}+\cdots+a_{n}}+\cdots$ $+\frac{a_{n}}{1+a_{1}+\cdots+a_{n-1}}$. Given $\sum_{i=1}^{n} a_{i}=1$, we have $S=\sum_{i=1}^{n} \frac{a_{i}}{2-a_{i}}$.

Let $\lambda<0$. From the condition $\sum_{i=1}^{n} a_{i}=1$, we get $\lambda\left(\sum_{i=1}^{n} a_{i}-1\right)=0$. Therefore,
$$
\begin{aligned}
S & =\sum_{i=1}^{n} \frac{a_{i}}{2-a_{i}}=\sum_{i=1}^{n} \frac{a_{i}}{2-a_{i}}+\lambda\left(\sum_{i=1}^{n} a_{i}-1\right) \\
& =\sum_{i=1}^{n}\left[\frac{2}{2-a_{i}}+(-\lambda)\left(2-a_{i}\right)\right]+(2 n-1) \lambda-n \\
& \geqslant \sum_{i=1}^{n} 2 \sqrt{\frac{2}{2-a_{i}}(-\lambda)\left(2-a_{i}\right)}+(2 n-1) \lambda-n \\
& =2 n \sqrt{-2 \lambda}+(2 n-1) \lambda-n .
\end{aligned}
$$

The equality holds if and only if $\frac{2}{2-a_{i}}=-\lambda\left(2-a_{i}\right)$,
i.e., $\lambda=-\frac{1}{2}\left(\frac{2 n}{2 n-1}\right)^{2}$. At this point,
$$
\begin{array}{l}
2 n \sqrt{-2 \lambda}+(2 n-1) \lambda-n \\
=2 n \cdot \frac{2 n}{2 n-1}+(2 n-1)\left(-\frac{1}{2}\right)\left(\frac{2 n}{2 n-1}\right)^{2} \\
=\frac{n}{2 n-1} . \\
\therefore S_{\text {min }}=\frac{n}{2 n-1} .
\end{array}
$$