Example 1 Let $a_{i} \in \mathrm{R}^{+},(i=1,2, \cdots, n)$, and $\sum_{i=1}^{n} a_{i}=1$, find
$$
S=\frac{a_{1}}{1+a_{2}+\cdots+a_{n}}+\frac{a_{2}}{1+a_{1}+a_{3}+\cdots+a_{n}}+\cdots+\frac{a_{n}}{1+a_{1}+\cdots+a_{n-1}} \text { 's }
$$
minimum value.
Solution: $S=\frac{a_{1}}{2-a_{1}}+\frac{a_{2}}{2-a_{2}}+\cdots+\frac{a_{n}}{2-a_{n}}$ is symmetric with respect to $a_{1} \cdots a_{n}$. Without loss of generality, assume $1>a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}>0$, then $2-a_{1} \leqslant 2-a_{2} \leqslant \cdots \leqslant 2-a_{n}$, and
$$
\frac{1}{2-a_{1}} \geqslant \frac{1}{2-a_{2}} \geqslant \cdots \geqslant \frac{1}{2-a_{n}}>0 .
$$

By Chebyshev's inequality, we have
$$
\begin{aligned}
S & \geqslant \frac{1}{n}\left(a_{1}+a_{2}+\cdots+a_{n}\right)\left(\frac{1}{2-a_{1}}+\frac{1}{2-a_{2}}+\cdots+\frac{1}{2-a_{n}}\right) \\
& =\frac{1}{n}\left(\frac{1}{2-a_{1}}+\cdots+\frac{1}{2-a_{n}}\right) .
\end{aligned}
$$

Also, by the arithmetic mean inequality,
$$
\left[\left(2-a_{1}\right)+\left(2-a_{2}\right)+\cdots+\left(2-a_{n}\right)\right]\left(\frac{1}{2-a_{1}}+\frac{1}{2-a_{2}}+\cdots+\right.
$$
$\left.\frac{1}{2-a_{n}}\right) \geqslant n^{2}$, and $\left(2-a_{1}\right)+\left(2-a_{2}\right)+\cdots+\left(2-a_{n}\right)=2 n-1$,
so, $S \geqslant \frac{1}{n} \cdot \frac{n^{2}}{2 n-1}=\frac{n}{2 n-1}$, with equality holding if and only if $a_{1}=a_{2}=\cdots=a_{n}=\frac{1}{n}$. Therefore, the minimum value of $S$ is $\frac{n}{2 n-1}$.
Note: Another solution to this problem is to use the basic inequality after a constant transformation:
$$
\begin{array}{l} 
\sum_{i=1}^{n} x_{i} \cdot \sum_{i=1}^{n} \frac{1}{x_{i}} \geqslant n^{2} . \\
\text { Since } S+n=\left(\frac{a_{1}}{2-a_{1}}+1\right)+\left(\frac{a_{2}}{2-a_{2}}+1\right)+\cdots+\left(\frac{a_{n}}{2-a_{n}}+1\right) \\
=\frac{2}{2-a_{1}}+\frac{2}{2-a_{2}}+\cdots+\frac{2}{2-a_{n}} \\
\geqslant \frac{n^{2}}{\frac{1}{2}\left(2-a_{1}\right)+\cdots+\frac{1}{2}\left(2-a_{n}\right)}=\frac{n^{2}}{\frac{1}{2}(2 n-1)}
\end{array}
$$

Therefore, $S \geqslant \frac{2 n^{2}}{2 n-1}-n=\frac{n}{2 n-1}$.