jmc

By AM-HM, $$\frac{(a+b)+(b+c)}{2}\ge \frac{2}{\frac{1}{a+b}+\frac{1}{b+c}},$$so $$\frac{1}{a+b}+\frac{1}{b+c}\ge \frac{4}{a+2b+c}=\frac{4}{b+5}.$$Since $b\le 2,$ $\frac{4}{b+5}\ge \frac{4}{7}.$ Equality occurs when $a=c=\frac{3}{2}$ and $b=2,$ so the minimum value is $\boxed{\frac{4}{7}}.$