16. (1990 IMO Japan First Round Selection Test) Let $x, y, z > 0$, and $x + y + z = 1$, find the minimum value of $\frac{1}{x} + \frac{4}{y} + \frac{9}{z}$.
16. Since $x, y, z > 0$, by the Cauchy-Schwarz inequality, we have:
$$
(x+y+z)\left(\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\right) \geqslant\left[\sqrt{x} \cdot \sqrt{\frac{1}{x}}+\sqrt{y} \cdot \sqrt{\frac{4}{y}}+\sqrt{z} \cdot \sqrt{\frac{9}{z}}\right]^{2}=36 \text {, }
$$

i.e., $\frac{1}{x}+\frac{4}{y}+\frac{9}{z} \geqslant 36$.
Equality holds if and only if $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$, i.e., $x=\frac{1}{6}, y=\frac{1}{3}, z=\frac{1}{2}$.
Therefore, the minimum value of $\frac{1}{x}+\frac{4}{y}+\frac{9}{z}$ is 36.