[Example 2.4.4] Let 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 $S=\sum_{i=1}^{n} \frac{a_{i}}{2-a_{i}}$.
First, simplify $\frac{a_{i}}{2-a_{i}}, i=1,2, \cdots, n$,
$$
\frac{a_{i}}{2-a_{i}}=\frac{a_{i}-2+2}{2-a_{i}}=\frac{2}{2-a_{i}}-1 .
$$

For $\frac{2}{2-a_{i}}$, we still use the parameterized inequality
$$
\begin{array}{c}
\frac{x^{2}}{y} \geqslant 2 \lambda x-\lambda^{2} y . \\
\frac{1}{2-a_{i}} \geqslant 2 \lambda-\lambda^{2}\left(2-a_{i}\right) . \quad(\lambda>0)
\end{array}
$$

Equality holds if and only if $\frac{1}{2-a_{i}}=\lambda^{2}\left(2-a_{i}\right)$, i.e., $\lambda=\frac{1}{2-a_{i}}$.
$$
\begin{aligned}
S & =\sum_{i=1}^{n} \frac{a_{i}}{2-a_{i}} \\
& =\sum_{i=1}^{n}\left(\frac{2}{2-a_{i}}-1\right) \\
& =2 \sum_{i=1}^{n} \frac{1}{2-a_{i}}-n \\
& \geqslant 2\left[2 n \lambda-\lambda^{2} \sum_{i=1}^{n}\left(2-a_{i}\right)\right]-n \\
& =4 n \lambda-2 \lambda^{2} \sum_{i=1}^{n} \cdot 2+2 \lambda^{2} \sum_{i=1}^{n} a_{i}-n \\
& =4 n \lambda-4 n \lambda^{2}+2 \lambda^{2}-n .
\end{aligned}
$$

From $\lambda=\frac{1}{2-a_{1}}=\frac{1}{2-a_{2}}=\cdots=\frac{1}{2-a_{n}}$ and $a_{1}+a_{2}+\cdots+a_{n}=1$, we have
$$
\begin{array}{c}
a_{1}=a_{2}=\cdots=a_{n}=\frac{1}{n}, \\
\lambda=\frac{1}{2-\frac{1}{n}}=\frac{n}{2 n-1} .
\end{array}
$$

Substituting (2) into (1) yields
$$
\begin{aligned}
S & \geqslant 4 n \cdot \frac{n}{2 n-1}-2(2 n-1) \cdot \frac{n^{2}}{(2 n-1)^{2}}-n \\
& =\frac{2 n^{2}}{2 n-1}-n \\
& =\frac{n}{2 n-1}
\end{aligned}
$$

Thus, the minimum value of $S$ is $\frac{n}{2 n-1}$.