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{align*} \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{align*}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)} $.