(4) Let $x, y, z \in \mathbf{R}^{+}, x+y+z=1$, then the maximum value of the function $f(x, y, z)=x y^{2} z^{3}$ is $\qquad$
(4) $\frac{1}{432}$ Hint: From
$$
\begin{aligned}
1 & =x+y+z \\
& =x+\frac{y}{2}+\frac{y}{2}+\frac{z}{3}+\frac{z}{3}+\frac{z}{3} \\
& \geqslant 6 \sqrt[6]{x \cdot \frac{y}{2} \cdot \frac{y}{2} \cdot \frac{z}{3} \cdot \frac{z}{3} \cdot \frac{z}{3},}
\end{aligned}
$$

Therefore,
$$
\frac{1}{4 \cdot 27} x y^{2} z^{3} \leqslant\left(\frac{1}{6}\right)^{6},
$$

which means
$$
x y^{2} z^{3} \leqslant \frac{1}{2^{4} \cdot 3^{3}}=\frac{1}{432},
$$

Equality holds when $x=\frac{y}{2}=\frac{z}{3}=\frac{1}{6}$, i.e., $x=\frac{1}{6}, y=\frac{1}{3}, z=\frac{1}{2}$.