Given $a, b, x_{i}(i=1,2, \cdots, n) \in \mathbf{R}_{+}$, $b>1$. Try to determine the smallest real number $c$, such that
$$
c \sum_{i=1}^{n} x_{i}^{a b+n} \geqslant\left(\prod_{i=1}^{n} x_{i}\right)\left(\sum_{i=1}^{n} x_{i}^{a}\right)^{b} .
$$
According to the problem, by homogeneity, we may assume $\prod_{i=1}^{n} x_{i}=1$.
Then the original inequality
$$
\Leftrightarrow c \geqslant \frac{\left(\prod_{i=1}^{n} x_{i}\right)\left(\sum_{i=1}^{n} x_{i}^{a}\right)^{b}}{\sum_{i=1}^{n} x_{i}^{a b+n}}=-c \geqslant \frac{\left(\sum_{i=1}^{n} x_{i}^{a}\right)^{b}}{\sum_{i=1}^{n} x_{i}^{a b+n}}.
$$

Thus, we only need to find the maximum value of $S=\frac{\left(\sum_{i=1}^{n} x_{i}^{a}\right)^{b}}{\sum_{i=1}^{n} x_{i}^{a b+n}}$.
By the power mean inequality, we have
$$
\begin{array}{l}
\left(\frac{\sum_{i=1}^{n} x_{i}^{a b}}{n}\right)^{\frac{1}{b}} \geqslant \frac{\sum_{i=1}^{n} x_{i}^{a}}{n} \\
\Rightarrow n^{b-1} \sum_{i=1}^{n} x_{i}^{a b} \geqslant\left(\sum_{i=1}^{n} x_{i}^{a}\right)^{b}. \\
\text{ Hence } S \leqslant \frac{n^{b-1} \sum_{i=1}^{n} x_{i}^{a b}}{\sum_{i=1}^{n} x_{i}^{a b+n}}.
\end{array}
$$

By the Chebyshev's inequality, we get
$$
\begin{array}{l}
n \sum_{i=1}^{n} x_{i}^{a b+n} \geqslant\left(\sum_{i=1}^{n} x_{i}^{n}\right)\left(\sum_{i=1}^{n} x_{i}^{a b}\right) \\
\geqslant n\left(\prod_{i=1}^{n} x_{i}\right)\left(\sum_{i=1}^{n} x_{i}^{a b}\right) \\
\Rightarrow \sum_{i=1}^{n} x_{i}^{a b+n} \geqslant \sum_{i=1}^{n} x_{i}^{a b}.
\end{array}
$$

Then $S \leqslant n^{b-1}$, with equality when $x_{1}=x_{2}=\cdots=x_{n}=1$.
Thus, $S_{\max }=n^{b-1} \Rightarrow c_{\min }=n^{b-1}$.
(Zhang Xiangjun, Su Lin, High School Affiliated to Hunan Normal University, 410012)