Given that $a, b, c, d$ are positive numbers, and $a+2b=1, c+2d=1$, find the minimum value of $\frac{1}{a}+\frac{1}{bcd}$.
From the inequality of arithmetic and geometric means, we know from the given conditions that
$2 c d \leqslant\left(\frac{c+2 d}{2}\right)^{2}=\frac{1}{4}$, i.e., $c d \leqslant \frac{1}{8}$,
with equality holding if and only if $c=\frac{1}{2}, d=\frac{1}{4}$. 
At this point, from (1) we get
$$
\frac{1}{a}+\frac{1}{b c d} \geqslant \frac{1}{a}+\frac{8}{b} \text {. }
$$

Furthermore, from the conditions and the Cauchy-Schwarz inequality, we have
$$
\frac{1}{a}+\frac{8}{b}=\left(\frac{1}{a}+\frac{8}{b}\right)(a+2 b) \geqslant(1+4)^{2}=25 \text {. }
$$
(3) Equality holds if and only if $b=2 a$, and combining this with the conditions, we find that $a=\frac{1}{5}, b=\frac{2}{5}$.

From (2) and (3), we conclude that $\frac{1}{a}+\frac{1}{b c d} \geqslant 25$.
When $a=\frac{1}{5}, b=\frac{2}{5}, c=\frac{1}{2}, d=\frac{1}{4}$, $\frac{1}{a}+\frac{1}{b c d}$ achieves its minimum value of 25.