Six. (20 points) Let two positive real numbers $x, y, z$. Try to find the maximum value of the algebraic expression $\frac{16 x+9 \sqrt{2 x y}+9 \sqrt[3]{3 x y z}}{x+2 y+z}$.


$$
\begin{array}{l}
\text { Six, because } \frac{16 x+9 \sqrt{2 x y}+9 \sqrt[3]{3 x y z}}{x+2 y+z} \\
=\frac{16 x+\frac{9 \sqrt{x \cdot 18 y}}{3}+\frac{3 \sqrt[3]{x \cdot 18 y \cdot 36 z}}{2}}{x+2 y+z} \\
\leqslant \frac{16 x+\frac{3(x+18 y)}{2}+\frac{x+18 y+36 z}{2}}{x+2 y+z}=18,
\end{array}
$$

Therefore, the equality holds if and only if $x: y: z=36: 2: 1$.
(Provided by Yang Yinghui)
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