Example 6 (1990 Japan CMO Team First Round Selection Test) Let $x+y+z \in \mathbf{R}^{+}$, and $x+y+z=1$, find the minimum value of $\frac{1}{x}+\frac{4}{y}+\frac{9}{z}$.
Solution 1: By the power mean inequality, we have $\frac{1}{x}+\frac{4}{y}+\frac{9}{z}=\frac{1^{2}}{x}+\frac{2^{2}}{y}+\frac{3^{2}}{z} \geqslant \frac{(1+2+3)^{2}}{x+y+z}=36$, equality holds if and only if $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$. Combining with $x+y+z=1$, we get that equality holds if and only if $x=\frac{1}{6}, y=\frac{1}{3}, z=\frac{1}{2}$. Therefore, the minimum value of $\frac{1}{x}+\frac{4}{y}+\frac{9}{z}$ is 36.

Solution 2: By the Cauchy-Schwarz inequality, we have
$$
\begin{aligned}
36 & =\left(\sqrt{\frac{1}{x}} \cdot \sqrt{x}+\sqrt{\frac{4}{y}} \cdot \sqrt{y}+\sqrt{\frac{9}{z}} \cdot \sqrt{z}\right)^{2} \leqslant\left(\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\right)(x+y+z) \\
& =\frac{1}{x}+\frac{4}{y}+\frac{9}{z},
\end{aligned}
$$
equality holds if and only if $\frac{1}{x}=\frac{2}{y}=\frac{3}{z}$, the rest is the same as Solution 1.