jmc

By Cauchy-Schwarz, $$(a+2b)\left(\frac{1}{a}+\frac{1}{b}\right)\ge (1+\sqrt{2}{)}^2=3+2\sqrt{2}.$$For equality to occur, we must have $a^2=2b^2,$ or $a=b\sqrt{2}.$ Then $b\sqrt{2}+2b=1,$ or $$b=\frac{1}{2+\sqrt{2}}=\frac{2-\sqrt{2}}{(2+\sqrt{2})(2-\sqrt{2})}=\frac{2-\sqrt{2}}{2},$$and $a=b\sqrt{2}=\frac{2\sqrt{2}-2}{2}=\sqrt{2}-1.$

Hence, the minimum value is $\boxed{3+2\sqrt{2}}.$