jmc

By AM-HM, $$\frac{a+b}{2}\ge \frac{2}{\frac{1}{a}+\frac{1}{b}},$$so $$\frac{1}{a}+\frac{1}{b}\ge \frac{4}{a+b}.$$Similarly, $$\begin{array}{rl}\frac{1}{a}+\frac{1}{c}& \ge \frac{4}{a+c},\\ \frac{1}{b}+\frac{1}{c}& \ge \frac{4}{a+b}.\end{array}$$Adding these inequalities, we get $$\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\ge \frac{4}{a+b}+\frac{4}{a+c}+\frac{4}{b+c},$$so $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge 2\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right).$$Hence, $$\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}}\ge 2.$$Equality occurs when $a=b=c,$ so the minimum value is $\boxed{2}.$