Example Thirteen, given an integer $n \geq 2$ and a positive real number $a$, positive real numbers $x_{1}, x_{2}, \ldots, x_{n}$ satisfy $\prod_{i=1}^{n} x_{i}=1$. Find the smallest real number $M=M(n, a)$ such that $\sum_{i=1}^{n} \frac{1}{a+S-x_{i}} \leq M$ always holds, where $S=\sum_{j=1}^{n} x_{i}$.
Example Thirteen, 
Solution: Discuss by cases.
(1) When $a \geq 1$. Let $x_{i}=y_{i}^{n}(1 \leq i \leq n)$, then $\prod_{i=1}^{n} y_{i}=1$. According to the AM-GM inequality, we have:
$$
S-x_{i}=\sum_{j \neq i} y_{j}^{n} \geq(n-1) \cdot\left(\frac{\sum_{j \neq 1} y_{j}}{n-1}\right)^{n} \geq(n-1)\left(\frac{\sum_{i \neq 1} y_{i}}{n-1}\right) \prod_{j \neq i} y_{j}=\frac{\sum_{j \neq 1} y_{j}}{y_{i}}
$$

Thus, we know:
$$
\sum_{i=1}^{n} \frac{1}{a+S-x_{i}} \leq \sum_{i=1}^{n} \frac{1}{a+\frac{\sum_{j \neq i} y_{i}}{y_{i}}}=\sum_{i=1}^{n} \frac{y_{i}}{a y_{i}+\sum_{j \neq 1} y_{j}}
$$
(1) If $a=1$. Then according to (1), $\sum_{i=1}^{n} \frac{1}{a+S-x_{i}} \leq \sum_{i=1}^{n} \frac{y_{i}}{\sum_{j=1}^{n} y_{i}}=1$, equality holds when $x_{1}=x_{2}=\ldots=x_{n}=1$. Hence, $M=1$.
(2) If $a>1$, according to the homogeneity of (1), we can assume $\sum_{\mathrm{j}=1}^{\mathrm{n}} \mathrm{y}_{\mathrm{j}}=1$. Then, according to the Cauchy-Schwarz inequality, we have:
$$
\begin{array}{l}
\sum_{i=1}^{n} \frac{1}{a+S-x_{i}} \leq \sum_{i=1}^{n} \frac{y_{i}}{(a-1) y_{i}+1}=\frac{1}{a-1} \cdot\left(\sum_{i=1}^{n}\left(1-\frac{1}{(a-1) y_{i}+1}\right)\right)=\frac{1}{a-1} \cdot\left(n-\sum_{i=1}^{n} \frac{1}{(a-1) y_{i}+1}\right) \\
\leq \frac{1}{a-1} \cdot\left(n-\frac{n^{2}}{(a-1) \sum_{i=1}^{n} y_{i}+n}\right)=\frac{1}{a-1} \cdot\left(n-\frac{n^{2}}{(a-1)+n}\right)=\frac{n}{a-1+n}
\end{array}
$$

Equality holds when $x_{1}=x_{2}=\ldots=x_{n}=1$. Hence, $M=1$.
(2) When $a<1$. According to the sugar-water inequality, we have $\frac{1}{a+S-x_{i}}=\frac{1}{a} \cdot \frac{a}{a+S-x_{i}} \leq \frac{1}{a} \cdot \frac{1}{1+S-x_{i}}$. Thus, by the conclusion when $a=1$, we have:
$$
\sum_{i=1}^{n} \frac{1}{a+S-x_{i}}=\frac{1}{a} \cdot \sum_{i=1}^{n} \frac{a}{a+S-x_{i}} \leq \frac{1}{a} \cdot \sum_{i=1}^{n} \frac{1}{1+S-x_{i}} \leq \frac{1}{a}
$$

When $x_{1}=x_{2}=\ldots=x_{n-1}=\varepsilon, x_{n}=\varepsilon^{1-n}$, and let $\varepsilon \rightarrow 0^{+}$, then $\sum_{i=1}^{n} \frac{1}{a+S-x_{i}} \rightarrow \frac{1}{a}$. Hence, $M=\frac{1}{a}$. In summary, the required $M=\left\{\begin{array}{l}\frac{n}{a-1+n}, \text { if } a \geq 1 \\ \frac{1}{a}, \text { if } 0<a<1\end{array}\right.$.