Example 3 Let $x, y, z>0$, and $x+y+z=1$. Find the minimum value of $\frac{1}{x}+\frac{4}{y}+\frac{9}{z}$.
(1990 Japan IMO Team First Round Selection Problem)
Solve: In a $3 \times 2$ grid, fill the numbers as follows:
\begin{tabular}{|c|c|}
\hline$\frac{1}{x}$ & $x$ \\
\hline$\frac{4}{y}$ & $y$ \\
\hline$\frac{9}{z}$ & $z$ \\
\hline
\end{tabular}

Then
$$
\left(\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\right)^{\frac{1}{2}} \cdot(x+y+2)^{\frac{1}{2}} \geqslant 1+4^{\frac{1}{2}}+9^{\frac{1}{2}},
$$

i.e., $\frac{1}{x}+\frac{4}{y}+\frac{9}{z} \geqslant 36$,
with equality if and only if $\frac{1}{x}=\frac{2}{y}=\frac{3}{z}$, i.e., $x=\frac{1}{6}, y=\frac{1}{3}, z=\frac{1}{2}$.