Given positive numbers $a, b (a < b)$ and a positive integer $n$, and a sequence of numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying $x_{i} \in [a, b], i=1,2, \cdots, n$. Find
$$
f=\frac{x_{1} x_{2} \cdots x_{n}}{\left(a+x_{1}\right)\left(x_{1}+x_{2}\right) \cdots\left(x_{n-1}+x_{n}\right)\left(x_{n}+b\right)}
$$

the maximum value.
Solution: (1) First, prove an inequality.
Let $b_{i} \in \mathbf{R}_{+}, i=1,2, \cdots, m$, then
$$
\begin{array}{l}
\left(1+b_{1}\right)\left(1+b_{2}\right) \cdots\left(1+b_{m}\right) \\
\geqslant\left(1+\sqrt[m]{b_{1} b_{2} \cdots b_{m}}\right)^{m} .
\end{array}
$$

Consider $f(x)=\lg \left(1+10^{x}\right)(x \in \mathbf{R})$, it is clear that
$$
\frac{f\left(x_{1}\right)+f\left(x_{2}\right)}{2} \geqslant f\left(\frac{x_{1}+x_{2}}{2}\right) \text {. }
$$

Thus, by Jensen's inequality for convex functions, we have
$$
\begin{array}{l}
\frac{f\left(x_{1}\right)+f\left(x_{2}\right)+\cdots+f\left(x_{m}\right)}{m} \\
\geqslant f\left(\frac{x_{1}+x_{2}+\cdots+x_{m}}{m}\right) .
\end{array}
$$

Therefore, $\lg \left(1+10^{x_{1}}\right)+\lg \left(1+10^{x}\right)+\cdots+\lg \left(1+10^{x} m\right)$
$$
\geqslant \lg \left(1+10^{\frac{x_{1}+x_{2}+\cdots+x_{m}}{m}}\right)^{m} \text {. }
$$

Let $b_{i}=10_{i} x_{i}(i=1,2, \cdots, m)$, then we have
$$
\begin{array}{l}
\left(1+b_{1}\right)\left(1+b_{2}\right) \cdots\left(1+b_{m}\right) \\
\geqslant\left(1+\sqrt[m]{b_{1} b_{2} \cdots b_{n}}\right)^{m} .
\end{array}
$$
(2) From inequality (1), we have
$$
\begin{array}{l}
\frac{1}{a f}=\frac{\left(a+x_{1}\right) \cdots\left(x_{n-1}+x_{n}\right)\left(x_{n}+b\right)}{a x_{1} x_{2} \cdots x_{n}} \\
=\left(1+\frac{x_{1}}{a}\right)\left(1+\frac{x_{2}}{x_{1}}\right) \cdots\left(1+\frac{x_{n}}{x_{n-1}}\right)\left(1+\frac{b}{x_{n}}\right) \\
\geqslant\left(1+\sqrt[n+1]{\frac{b}{a}}\right)^{n+1} .
\end{array}
$$

Therefore, $f \leqslant\left(a^{\frac{1}{n+1}}+b^{\frac{1}{n+1}}\right)^{-(n+1)}$.
From the condition for equality in inequality (1), we know that the condition for equality in inequality (2) is
$$
\frac{x_{1}}{a}=\frac{x_{2}}{x_{1}}=\cdots=\frac{b}{x_{n}},
$$

i.e., $\frac{x_{1}}{a}=\frac{x_{2}}{x_{1}}=\cdots=\frac{b}{x_{n}}=\left(\frac{b}{a}\right)^{\frac{1}{n+1}}$.
At this point, $x_{i}=a^{1-\frac{1}{1+n}} b^{\frac{1}{n+1}}, i=1,2, \cdots, n$.
(Ye Songgao, Nankai University School of Mathematical Sciences, Applied Mathematics, 2003, 300071)