One, (20 points) If positive numbers $a, b$ satisfy $ab=1$, find
$$
M=\frac{1}{1+a}+\frac{1}{1+2b}
$$

the minimum value.
Given $a b=1 \Rightarrow b=\frac{1}{a}$.
Thus, $M=\frac{1}{1+a}+\frac{1}{1+\frac{2}{a}}=\frac{1}{1+a}+\frac{a}{2+a}$ $=1+\frac{1}{1+a}-\frac{2}{2+a}=1-\frac{a}{a^{2}+3 a+2}$.
Let $N=\frac{a^{2}+3 a+2}{a}$. Then
$N=a+\frac{2}{a}+3 \geqslant 3+2 \sqrt{2}$.
The equality holds when $a=\sqrt{2}$.
Thus, $M=1-\frac{1}{N} \geqslant 1-(3-2 \sqrt{2})=2 \sqrt{2}-2$.
Therefore, when $a=\sqrt{2}, b=\frac{\sqrt{2}}{2}$, $M$ attains its minimum value $2 \sqrt{2}-2$.