Example 4 Let $x_{i}>0(i=1,2, \cdots, n), \mu>0$, $\lambda-\mu>0$, and $\sum_{i=1}^{n} x_{i}=1$. For a fixed $n\left(n \in \mathbf{N}_{+}\right)$, find the minimum value of $f\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\sum_{i=1}^{n} \frac{x_{i}}{\lambda-\mu x_{i}}$.
$$
\begin{array}{l}
\text { Explanation: } f\left(x_{1}, x_{2}, \cdots, x_{n}\right)+\frac{n}{\mu} \\
=\sum_{i=1}^{n}\left(\frac{x_{i}}{\lambda-\mu x_{i}}+\frac{1}{\mu}\right)=\frac{\lambda}{\mu}\left(\sum_{i=1}^{n} \frac{1}{\lambda-\mu x_{i}}\right) \\
=\frac{\lambda}{\mu(n \lambda-\mu)}\left[\sum_{i=1}^{n}\left(\lambda-\mu x_{i}\right)\right]\left[\sum_{i=1}^{n} \frac{1}{\lambda-\mu x_{i}}\right] \\
\geqslant \frac{\lambda}{\mu(n \lambda-\mu)} \cdot n^{2} .
\end{array}
$$

When and only when $x_{i}=\frac{1}{n}(i=1,2, \cdots, n)$, the equality holds. Therefore, when $x_{i}=\frac{1}{n}(i=1,2, \cdots, n)$,
$$
f\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\frac{n}{n \lambda-\mu} \text {. }
$$

Thus, $f\left(x_{1}, x_{2}, \cdots, x_{n}\right)_{\text {min }}=\frac{n}{n \lambda-\mu}$.
Note: Using the arithmetic mean inequality and the Cauchy-Schwarz inequality is a common method for finding the extremum of multivariate functions. The key to solving this problem is to construct $\left[\sum_{i=1}^{n}\left(\lambda-\mu x_{i}\right)\right]\left(\sum_{i=1}^{n} \frac{1}{\lambda-\mu x_{1}}\right) \geqslant n^{2}$.
$$