47. Given an integer $n \geqslant 2$ and a positive integer $a$, positive real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy $x_{1} x_{2} \cdots x_{n}=1$, find the smallest real number $M=M(n, a)$, such that $\sum_{i=1}^{n} \frac{1}{a+S-x_{i}} \leqslant M$ always holds, where $S=x_{1}+x_{2}+\cdots+x_{n} \cdot(2010$ China National Training Team Test Problem)
47. First consider the case $a \geqslant 1$, let $x_{i}=y_{i}^{n}, y_{i}>0$, then $y_{1} y_{2} \cdots y_{n}=1$, we have
$$\begin{aligned}
S-x_{i}= & \sum_{j \neq i} y_{j}^{n} \geqslant(n-1)\left(\frac{\sum_{j \neq i} y_{j}}{n-1}\right)^{n}(\text { by the AM-GM inequality }) \geqslant \\
& (n-1)\left(\frac{\sum_{j \neq i} y_{j}}{n-1}\right) \prod_{j \neq i} y_{j \cdot}(\text { AM } \geqslant \text { GM })= \\
& \frac{\sum_{j \neq i} y_{j}}{y_{i}}
\end{aligned}$$

Thus,
$$\sum_{i=1}^{n} \frac{1}{a+S-x_{i}} \leqslant \sum_{i=1}^{n} \frac{y_{i}}{a y_{i}+\sum_{j \neq i} y_{j}}$$

When $a=1$, $\sum_{i=1}^{n} \frac{y_{i}}{a y_{i}+\sum_{j \neq i} y_{j}}=\sum_{i=1}^{n} \frac{y_{i}}{\sum_{j=1}^{n} y_{j}}=1$. And when $x_{1}=x_{2}=\cdots=x_{n}=1$, $\sum_{i=1}^{n} \frac{1}{a+S-x_{i}}=1$, at this point $M=1$.

Now assume $a>1$, let $z_{i}=\frac{y_{i}}{\sum_{j=1}^{n} y_{j}}, i=1,2, \cdots, n$, then $\sum_{i=1}^{n} z_{i}=1$,
$$\begin{aligned}
\frac{y_{i}}{a y_{i}+\sum_{j \neq i} y_{j}}= & \frac{y_{i}}{(a-1) y_{i}+\sum_{j=1}^{n} y_{j}}=\frac{z_{i}}{(a-1) z_{i}+1}= \\
& \frac{1}{a-1}\left(1-\frac{1}{(a-1) z_{i}+1}\right)
\end{aligned}$$

By the Cauchy-Schwarz inequality,
$$\left(\sum_{i=1}^{n}\left((a-1) z_{i}+1\right)\right)\left(\sum_{i=1}^{n} \frac{1}{(a-1) z_{i}+1}\right) \geqslant n^{2}$$

And $\sum_{i=1}^{n}\left((a-1) z_{i}+1\right)=a-1+n$, hence
$$\sum_{i=1}^{n} \frac{1}{(a-1) z_{i}+1} \geqslant \frac{n^{2}}{a-1+n}$$

Combining (1), (2), and (3), we have
$$\begin{aligned}
\sum_{i=1}^{n} \frac{1}{a+S-x_{i}} \leqslant & \sum_{i=1}^{n}\left(\frac{1}{a-1}\left(1-\frac{1}{(a-1) z_{i}+1}\right)\right) \leqslant \\
& \frac{n}{a-1}-\frac{1}{a-1} \cdot \frac{n^{2}}{a-1+n}= \\
& \frac{n}{a-1+n}
\end{aligned}$$

When $x_{1}=x_{2}=\cdots=x_{n}=1$, we have $\sum_{i=1}^{n} \frac{1}{a+S-x_{i}}=\frac{n}{a-1+n}$, hence $M=\frac{n}{a-1+n}$.
Next consider $0 < a < 1$, the function $f(x)=\frac{x}{x+\lambda}=1-\frac{\lambda}{x+\lambda}$ is strictly increasing on the interval $(0,+\infty)$, so $f(a)<f(1)$, i.e., $\frac{a}{a+\lambda}<\frac{1}{1+\lambda}$. Thus, by the result when $a=1$,
$$\sum_{i=1}^{n} \frac{1}{a+S-x_{i}}=\frac{1}{a} \sum_{i=1}^{n} \frac{a}{a+S-x_{i}}<\frac{1}{a} \sum_{i=1}^{n} \frac{1}{1+S-x_{i}} \leqslant \frac{1}{a}$$

When $x_{1}=x_{2}=\cdots=x_{n-1}=\varepsilon \rightarrow 0^{+}$, and $x_{n}=\varepsilon^{1-n} \rightarrow+\infty$,
$$\lim _{\varepsilon \rightarrow 0^{+}} \sum_{i=1}^{n} \frac{1}{a+S-x_{i}}=\lim _{\varepsilon \rightarrow 0^{+}}\left(\frac{n-1}{a+\varepsilon^{1-n}+(n-2) \varepsilon}+\frac{1}{a+(n-1) \varepsilon}\right)=\frac{1}{a}$$

Thus, $M=\frac{1}{a}$, in summary,
$$M=\left\{\begin{array}{l}
\frac{n}{a-1+n}, \text { if } a \geqslant 1 \\
\frac{1}{a}, \text { if } 0<a<1
\end{array}\right.$$

Solution: The answer is $\lambda \geqslant \mathrm{e}$.