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By AM-HM, $$\frac{(a+b)+(a+c)+(b+c)}{3}\ge \frac{3}{\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}}.$$Then $$\frac{2a+2b+2c}{a+b}+\frac{2a+2b+2c}{a+c}+\frac{2a+2b+2c}{b+c}\ge 9,$$so $$\frac{a+b+c}{a+b}+\frac{a+b+c}{a+c}+\frac{a+b+c}{b+c}\ge \frac{9}{2}.$$Hence, $$\frac{c}{a+b}+1+\frac{b}{a+c}+1+\frac{a}{b+c}+1\ge \frac{9}{2},$$so $$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge \frac{3}{2}.$$Equality occurs when $a=b=c.$ This inequality is satisfied for all positive real numbers $a,$ $b,$ and $c,$ and is known as Nesbitt's Inequality.

Now, since $a,$ $b,$ $c$ are the sides of a triangle, $$b+c>a.$$Then $2b+2c>a+b+c,$ so $b+c>\frac{a+b+c}{2}.$ Therefore, $$\frac{a}{b+c}<\frac{a}{(a+b+c)/2}=\frac{2a}{a+b+c}.$$Similarly, $$\begin{array}{rl}\frac{b}{a+c}& <\frac{b}{(a+b+c)/2}=\frac{2b}{a+b+c},\\ \frac{c}{a+b}& <\frac{c}{(a+b+c)/2}=\frac{2c}{a+b+c}.\end{array}$$Adding these inequalities, we get $$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}<\frac{2a+2b+2c}{a+b+c}=2.$$Let $$S=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b},$$so $S<2.$ Furthermore, if we let $a$ and $b$ approach 1, and let $c$ approach 0, then $S$ approaches $$\frac{1}{1+0}+\frac{1}{1+0}+\frac{0}{1+1}=2.$$Thus, $S$ can be made arbitrarily close to 2, so the possible values of $S$ are $\boxed{\left[\frac{3}{2},2\right)}.$