jmc

By AM-HM, $$\frac{a+b}{2}\ge \frac{2}{\frac{1}{a}+\frac{1}{b}}=\frac{2ab}{a+b},$$so $$\frac{ab}{a+b}\le \frac{a+b}{4}.$$Similarly, $$\begin{array}{r}\frac{ac}{a+c}\le \frac{a+c}{4},\\ \frac{bc}{b+c}\le \frac{b+c}{4}.\end{array}$$Hence, $$\frac{ab}{a+b}+\frac{ac}{a+c}+\frac{bc}{b+c}\le \frac{a+b}{4}+\frac{a+c}{4}+\frac{b+c}{4}=\frac{a+b+c}{2}=\frac{1}{2}.$$Equality occurs when $a=b=c=\frac{1}{3},$ so the maximum value is $\boxed{\frac{1}{2}}.$