3. Let $x, y, z>0$ and $x+y+z=1$. Then $\frac{1}{x}$ $+4+9$ has the minimum value of $\qquad$

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3. 36
$$
\begin{aligned}
\text { Given } & \left(\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\right)(x+y+z) \\
& \geqslant(1+2+3)^{2}=36 \\
\quad & \frac{1}{x}+\frac{4}{y}+\frac{9}{z} \geqslant \frac{36}{x+y+z}=36 .
\end{aligned}
$$

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.