18. If $x, y, z$ are all positive real numbers, find the maximum value of $\frac{x y z}{(1+5 x)(-4 x+3 y)(5 y+6 z)(z+18)}$ and prove your conclusion. (2003 Singapore Mathematical Olympiad)
18. Given a fixed $y$
$$\begin{array}{l}
\frac{x}{(1+5 x)(4 x+3 y)}=\frac{x}{20 x^{2}+(15 y+4) x+3 y}= \\
\frac{1}{20 x+\frac{3 y}{x}+(15 y+4)} \leqslant \\
\frac{1}{2 \sqrt{20 x \cdot \frac{3 y}{x}}+(-15 y+4)}= \\
\frac{1}{(\sqrt{15 y}+2)^{2}}
\end{array}$$

Equality holds if and only if $x=\sqrt{\frac{3 y}{20}}$. Similarly, we have
$$\frac{z}{(5 y+6 z)(z+18)} \leqslant \frac{1}{2 \sqrt{6 z \cdot \frac{90 y}{z}}+(5 y+108)}=\frac{1}{(\sqrt{5 y}+6 \sqrt{3})^{2}}$$

Equality holds if and only if $z=\sqrt{15 y}$. Therefore,
$$\begin{array}{l}
\frac{x y z}{(1+5 x)(4 x+3 y)(5 y+6 z)(z+18)} \leqslant \frac{y}{(\sqrt{15 y}+2)^{2}(\sqrt{5 y}+6 \sqrt{3})^{2}}= \\
{\left[-\frac{\sqrt{y}}{(-\sqrt{15 y}+2)(-\sqrt{5 y}+6 \sqrt{3})}\right]^{2}=} \\
\left(\frac{1}{5 \sqrt{3 y}+\frac{12 \sqrt{3}}{\sqrt{y}}}+20 \sqrt{5}\right)^{2} \leqslant \\
{\left[\frac{1}{2 \sqrt{5 \sqrt{3} \times 12 \sqrt{3}}+20-\sqrt{5}}\right]^{2}=} \\
{\left[\frac{1}{32 \sqrt{5}}\right]^{2}=\frac{1}{5120}}
\end{array}$$

Equality holds if and only if $x=\frac{3}{5}, y=\frac{12}{5}, z=6$, at which point $\frac{x y z}{(1+5 x)(4 x+3 y)(5 y+6 z)(z+18)}$ attains its maximum value $\frac{1}{5120}$.