Example 8 (1990 Japanese Spring Selection Problem) Let $x, y, z > 0$, and $x+y+z=1$, find the minimum value of $\frac{1}{x}+\frac{4}{y}+\frac{9}{z}$.
Since $x, y, z > 0$ and $x + y + z = 1$, we have
$$\frac{1}{x} + \frac{4}{y} + \frac{9}{z} \geqslant \frac{(1+2+3)^{2}}{x+y+z} = 36$$

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