Example 6 Given $x, y, z \in \mathrm{R}^{+}, x+y+z=1$, find the minimum value of $\frac{1}{x}+\frac{4}{y}+\frac{9}{z}$.

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$$\begin{array}{l}
\text { Analysis: } \because x, y, z \in \mathbb{R}^{+}, x+y+z=1, \\
\therefore \quad \frac{1}{x}+\frac{4}{y}+\frac{9}{z} \\
=(x+y+z)\left(\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\right) \\
=\frac{y+z}{x}+\frac{4 z+4 x}{y}+\frac{9 x+9 y}{z}+14 \\
=\left(\frac{y}{x}+\frac{4 x}{y}\right)+\left(\frac{4 z}{y}+\frac{9 y}{z}\right)+\left(\frac{9 x}{z}+\frac{z}{x}\right)+14 \\
\geqslant 2 \sqrt{\frac{y}{x} \cdot \frac{4 x}{y}}+2 \sqrt{\frac{4 z}{y} \cdot \frac{9 y}{z}}+2 \sqrt{\frac{9 x}{z} \cdot \frac{z}{x}}+14 \\
=4+12+6+14=36
\end{array}$$

Therefore, when $x=\frac{1}{6}, y=\frac{1}{3}, z=\frac{1}{2}$, the minimum value of $\frac{1}{x}+\frac{4}{y}+\frac{9}{z}$ is 36.

The clever introduction of conditions makes it possible to effectively use the two-variable mean inequality.