Example 5 Given $a>0, b>0, a+2b=1$, find the minimum value of $t=\frac{1}{a}+\frac{1}{b}$.
Let's multiply $\frac{1}{a}+\frac{1}{b}$ by 1, and replace 1 with $a+2b$.
$$\begin{array}{c}
\left(\frac{1}{a}+\frac{1}{b}\right) \cdot 1=\left(\frac{1}{a}+\frac{1}{b}\right) \cdot(a+2b)=1+\frac{2b}{a}+\frac{a}{b}+2= \\
3+\frac{2b}{a}+\frac{a}{b} \geqslant 3+2 \sqrt{\frac{2b}{a} \cdot \frac{a}{b}}=3+2 \sqrt{2}
\end{array}$$

The equality holds if and only if $\frac{2b}{a}=\frac{a}{b}$. The minimum value of $t=\frac{1}{a}+\frac{1}{b}$ is $3+2 \sqrt{2}$.