59. Let $a, b$ be positive constants. $x_{1}, x_{2}, \cdots, x_{n}$ are positive real numbers. $n \geqslant 2$ is a positive integer. Find the maximum value of $y=$ $\frac{x_{1} x_{2} \cdots x_{n}}{\left(a+x_{1}\right)\left(x_{1}+x_{2}\right) \cdots\left(x_{n-1}+x_{n}\right)\left(x_{n}+b\right)}$. (1999 Polish Mathematical Olympiad Problem)
59. Since
$$\begin{aligned}
a y & =\frac{a}{a+x_{1}} \cdot \frac{x_{1}}{x_{1}+x_{2}} \cdot \cdots \cdot \frac{x_{n-1}}{x_{n-1}+x_{n}} \cdot \frac{x_{n}}{x_{n}+b} \\
b y & =\frac{x_{1}}{a+x_{1}} \cdot \frac{x_{2}}{x_{1}+x_{2}} \cdot \cdots \cdot \frac{x_{n}}{x_{n-1}+x_{n}} \cdot \frac{b}{x_{n}+b}
\end{aligned}$$

By the AM-GM inequality, we have
$$\begin{array}{l} 
\sqrt[n+1]{a y} \leqslant \frac{1}{n+1}\left(\frac{a}{a+x_{1}}+\frac{x_{1}}{x_{1}+x_{2}}+\cdots+\frac{x_{n-1}}{x_{n-1}+x_{n}}+\frac{x_{n}}{x_{n}+b}\right) \\
\sqrt[n+1]{b y} \leqslant \frac{1}{n+1}\left(\frac{x_{1}}{a+x_{1}}+\frac{x_{2}}{x_{1}+x_{2}}+\cdots+\frac{x_{n}}{x_{n-1}+x_{n}}+\frac{b}{x_{n}+b}\right)
\end{array}$$
Adding (1) and (2) gives
$$(\sqrt[n+1]{a}+\sqrt[n+1]{b}) \sqrt[n+1]{y} \leqslant 1$$

Thus,
$$\begin{array}{l}
y \leqslant \frac{1}{(\sqrt[n+1]{a}+\sqrt[n+1]{b})^{n+1}} \\
\text { with equality if and only if }\left\{\begin{array}{l}
\frac{a}{a+x_{1}}=\frac{x_{1}}{x_{1}+x_{2}}=\cdots=\frac{x_{n-1}}{x_{n-1}+x_{n}}=\frac{x_{n}}{x_{n}+b} \\
\frac{x_{1}}{a+x_{1}}=\frac{x_{2}}{x_{1}}=\cdots=\frac{x_{n}}{x_{n}}=\frac{b}{x_{n}+b} \text { holds. In this case, } \frac{a}{x_{1}}=
\end{array}\right. \\
\frac{x_{1}}{x_{2}}=\cdots=\frac{x_{n-1}}{x_{n}}=\frac{x_{n}}{b}=k \text {. }
\end{array}$$

Clearly, \( k^{n+1}=\frac{a}{b}, k=\sqrt[n+1]{\frac{a}{b}} \), so when \( x_{k}=a\left(\frac{b}{a}\right)^{\frac{k}{n+1}}(k=1,2, \cdots, n) \), \( y \) attains its maximum
value \(\frac{1}{(\sqrt[n+1]{a}+\sqrt[n+1]{b})^{n+1}}\).