4.2.14 * Given $x, y, z \in \mathbf{R}^{+}$, and $x+y+z=1$, find the minimum value of $\frac{1}{x}+\frac{4}{y}+\frac{9}{z}$.
By the AM-GM inequality, we have
$$
\frac{1}{x}+36 x \geqslant 12, \frac{4}{y}+36 y \geqslant 24, \frac{9}{z}+36 z \geqslant 36,
$$

Therefore, $\frac{1}{x}+\frac{4}{y}+\frac{9}{z}+36(x+y+z) \geqslant 72$, which means $\frac{1}{x}+\frac{4}{y}+\frac{9}{z} \geqslant 36$. Equality holds if and only if $x=\frac{1}{6}$, $y=\frac{1}{3}$, $z=\frac{1}{2}$. Thus, the desired minimum value is 36.