Example 3 If $x, y, z$ are all positive real numbers, find the maximum value of $\frac{x y z}{(1+5 x)(4 x+3 y)(5 y+6 z)(z+18)}$. (2003 Singapore Mathematical Olympiad)
Solve: By the generalization of Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
(1+5 x)(4 x+3 y)(5 y+6 z)(z+18) \geqslant \\
(\sqrt[4]{1 \cdot 4 x \cdot 5 y \cdot z}+\sqrt[4]{5 x \cdot 3 y \cdot 6 z \cdot 18})^{4}=5120 x y z
\end{array}$$

Therefore, the maximum value of $\frac{x y z}{(1+5 x)(4 x+3 y)(5 y+6 z)(z+18)}$ is $\frac{1}{5120}$.
Equality holds if and only if $\frac{1}{5 x}=\frac{4 x}{3 y}=\frac{5 y}{6 z}=\frac{z}{18}=k$, thus $k^{4}=\frac{1}{5 x} \cdot \frac{4 x}{3 y} \cdot \frac{5 y}{6 z} \cdot \frac{z}{18}=\left(\frac{1}{3}\right)^{4}$, which means $k=\frac{1}{3}$.

Hence, when $x=\frac{3}{5}, y=\frac{12}{5}, z=6$, $\frac{x y z}{(1+5 x)(4 x+3 y)(5 y+6 z)(z+18)}$ attains its maximum value $\frac{1}{5120}$.