Question 3 Given $n, k \in \mathbf{Z}_{+}, n>k$. Given real numbers $a_{1}, a_{2}, \cdots, a_{n} \in(k-1, k)$. Let positive real numbers $x_{1}, x_{2}, \cdots$, $x_{n}$ satisfy that for any set $I \subseteq\{1,2, \cdots, n\}$, $|I|=k$, there is $\sum_{i \in I} x_{i} \leqslant \sum_{i \in I} a_{i}$. Find the maximum value of $x_{1} x_{2} \cdots x_{n}$. ${ }^{[1]}$
(33rd China Mathematical Olympiad)
Solve the maximum value as $a_{1} a_{2} \cdots a_{n}$.
Let $U=\{1,2, \cdots, n\}, d_{i}=a_{i}-x_{i}$.
From the problem's condition, we know that for any $I \subseteq U,|I|=k$ we have
$$
\sum_{i \in I} d_{i} \geqslant 0 .
$$

Notice that,
$$
\begin{array}{l}
\prod_{i=1}^{n} x_{i}=\left(\prod_{i=1}^{n} a_{i}\right)\left(\prod_{i=1}^{n} \frac{x_{i}}{a_{i}}\right)^{n} \\
\leqslant\left(\prod_{i=1}^{n} a_{i}\right)\left(\frac{1}{n} \sum_{i=1}^{n} \frac{a_{i}-d_{i}}{a_{i}}\right)^{n} \\
=\left(\prod_{i=1}^{n} a_{i}\right)\left(1-\frac{1}{n} \sum_{i=1}^{n} \frac{d_{i}}{a_{i}}\right)^{n} .
\end{array}
$$

For the set $I \subseteq U,|I|=k$, let
$$
\begin{array}{l}
B_{I}=\left(\sum_{i \in I} d_{i}\right)\left(\sum_{i \in I} \frac{1}{a_{i}}\right), \\
C_{I}=\left(\sum_{i \in I} d_{i}\right)\left(\sum_{i \in I} \frac{1}{a_{i}}\right) \text {, } \\
T=\sum_{\substack{I \subseteq U \\
|I|=k}}\left(\sum_{i \in I} d_{i}\right)\left(\sum_{i \in I} \frac{1}{a_{i}}-\frac{k-1}{n-k} \sum_{i \in I} \frac{1}{a_{i}}\right) \\
=\sum_{I \subseteq U}\left(B_{I}-\frac{k-1}{n-k} C_{I}\right) \text {. } \\
|I|=k \\
\text { By } \sum_{i \in I} \frac{1}{a_{i}}-\frac{k-1}{n-k} \sum_{i \in \bar{I}} \frac{1}{a_{i}} \\
>k \frac{1}{k}-\frac{k-1}{n-k}(n-k) \frac{1}{k-1}=0 \text {, } \\
\end{array}
$$

we get $T \geqslant 0$.
For each $\frac{d_{i}}{a_{i}}$ coefficient: $\sum_{\substack{I \subseteq U \\ |I|=k}} B_{I}$ contributes $\mathrm{C}_{n-1}^{k-1}$, $\sum_{\substack{I \subseteq U \\ |I|=k}} C_{I}$ contributes 0;
For each $\frac{d_{i}}{a_{j}}(i \neq j)$ coefficient:
$\sum_{\substack{I \subseteq U \\ |I|=k}} B_{I}$ contributes $\frac{(k-1) \mathrm{C}_{n-1}^{k-1}}{n-1}$,
$\sum_{\substack{I \subseteq U \\ |I|=k}} C_{I}$ contributes $\frac{(n-k) \mathrm{C}_{n-1}^{k-1}}{n-1}$.
And $\frac{(k-1) \mathrm{C}_{n-1}^{k-1}}{n-1}-\frac{k-1}{n-k} \cdot \frac{(n-k) \mathrm{C}_{n-1}^{k-1}}{n-1}=0$, hence
$$
\begin{array}{l}
T=\mathrm{C}_{n-1}^{k-1} \sum_{i=1}^{n} \frac{d_{i}}{a_{i}} \geqslant 0 \Rightarrow \sum_{i=1}^{n} \frac{d_{i}}{a_{i}} \geqslant 0 \\
\Rightarrow \prod_{i=1}^{n} x_{i} \leqslant\left(\prod_{i=1}^{n} a_{i}\right)\left(1-\frac{1}{n} \sum_{i=1}^{n} \frac{d_{i}}{a_{i}}\right)^{n} \leqslant \prod_{i=1}^{n} a_{i} .
\end{array}
$$