Example 22 (Mathematical Problem 1486 from "Mathematical Bulletin") Find the maximum value of the function $f(x)=\frac{x(1-x)}{(x+1)(x+2)(2x+1)}$, $x \in (0,1]$.
Let $f(x)=y$, and let $x=\frac{1-t}{1+t}(0 \leqslant t<1)$. Substituting $x$ into $f(x)$, we get $y=\frac{t\left(1-t^{2}\right)}{9-t^{2}}$.
Introducing a positive constant $\alpha$, we have
$$
\begin{aligned}
y & =\frac{\alpha t\left(1-t^{2}\right)}{\alpha\left(9-t^{2}\right)} \leqslant \frac{\left(\alpha^{2}+t^{2}\right)\left(1-t^{2}\right)}{2 \alpha\left(9-t^{2}\right)}=\frac{\alpha^{2}+\left(1-\alpha^{2}\right) t^{2}-t^{4}}{2 \alpha\left(9-t^{2}\right)} \\
& =-\frac{1}{2 \alpha}\left[\left(9-t^{2}\right)+\frac{8\left(9+\alpha^{2}\right)}{9-t^{2}}\right]+\frac{17+\alpha^{2}}{2 \alpha} \leqslant-\frac{1}{2 \alpha} \cdot 2 \sqrt{8\left(9+\alpha^{2}\right)}+\frac{17+\alpha^{2}}{2 \alpha} \\
& =\frac{17+\alpha^{2}-4 \sqrt{2\left(9+\alpha^{2}\right)}}{2 \alpha} .
\end{aligned}
$$

Solving for $t=\alpha=2 \sqrt{2}-\sqrt{5}$, and substituting into (3) (or (2)), we find:
When $x=\frac{1}{2}(1+\sqrt{10}-\sqrt{5}-\sqrt{2})$, $y=f(x)$ reaches its maximum value $\frac{1}{3}(8 \sqrt{2}-5 \sqrt{5})$.