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By the AM-HM inequality on the numbers $a+b$, $a+c$, and $b+c$, we have $$\frac{(a+b)+(a+c)+(b+c)}{3}\ge \frac{3}{\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}}.$$Cross-multiplying and simplifying gives us $$\frac{1}{3}(2a+2b+2c)\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)\ge 3,$$and hence $$(a+b+c)\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)\ge \boxed{\frac{9}{2}}.$$This value is achieved when $a=b=c=1$.