Example 1 Let $a, b$ be positive constants, $x_{1}, x_{2}, \cdots, x_{n}$ be positive real numbers, and $n \geqslant 2$ be a positive integer. Find:
$$
y=\frac{x_{1} x_{2} \cdots x_{n}}{\left(a+x_{1}\right)\left(x_{1}+x_{2}\right) \cdots\left(x_{n-1}+x_{n}\right)\left(x_{n}+b\right)}
$$

the maximum value.
Notice
$$
\begin{array}{c}
y=\frac{\frac{1}{b}}{\left(\frac{a}{x_{1}}+1\right)\left(\frac{x_{1}}{x_{2}}+1\right) \cdots\left(\frac{x_{n-1}}{x_{n}}+1\right)\left(\frac{x_{n}}{b}+1\right)} . \\
\text { Let } a_{1}=\frac{a}{x_{1}}, a_{2}=\frac{x_{1}}{x_{2}}, \cdots \cdots a_{n}=\frac{x_{n-1}}{x_{n}},
\end{array}
$$
$a_{n+1}=\frac{x_{n}}{b}$. Then $a_{1} a_{2} \cdots a_{n+1}=\frac{a}{b}$.
By the extended version 1, we know
$$
\begin{array}{l}
\left(\frac{a}{x_{1}}+1\right)\left(\frac{x_{1}}{x_{2}}+1\right) \cdots\left(\frac{x_{n-1}}{x_{n}}+1\right)\left(\frac{x_{n}}{b}+1\right) \\
=\prod_{i=1}^{n+1}\left(a_{i}+1\right) \geqslant\left(\sqrt[n+1]{\frac{a}{b}}+1\right)^{n+1},
\end{array}
$$

Equality holds if and only if $a_{1}=a_{2}=\cdots=a_{n+1}=\sqrt[n+1]{\frac{a}{b}}$, that is,
$$
\frac{a}{x_{1}}=\frac{x_{1}}{x_{2}}=\cdots=\frac{x_{n-1}}{x_{n}}=\frac{x_{n}}{b}=\sqrt[n+1]{\frac{a}{b}},
$$

which means $x_{k}=a\left(\frac{b}{a}\right)^{\frac{k}{n+1}}$ for the equality to hold.
Thus, $y \leqslant \frac{\frac{1}{b}}{\left(\sqrt[n+1]{\frac{a}{b}}+1\right)^{n+1}}=\frac{1}{(\sqrt[n+1]{a}+\sqrt[n+1]{b})^{n+1}}$.
Equality holds if and only if $x_{k}=a\left(\frac{b}{a}\right)^{\frac{k}{n+1}}(k=1,2, \cdots, n)$

when $y$ reaches its maximum value $\frac{1}{(\sqrt[n+1]{a}+\sqrt[n+1]{b})^{n+1}}$.