Finding the range of $\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$, for $a$, $b$, $c$ the sides of a triangle 
If $a$, $b$, and $c$ are the three sides of a triangle, then 
  $$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$$ 
  lies in what interval?


The problem is Nesbitt's inequality. (Source: A.M. Nesbitt, Problem 15114, Educational Times (2) 3, (1903), 37-38.)
Lower Bound: Let's say $S= \dfrac{a}{b+c} + \dfrac{b}{c+a} + \dfrac{c}{a+b} $. $S+3= \dfrac{a}{b+c}+1 + \dfrac{b}{c+a}+1 + \dfrac{c}{a+b}+1=(a+b+c)\left( \dfrac{1}{b+c} + \dfrac{1}{c+a} + \dfrac{1}{a+b}\right)$. By aritmethic-harmonic mean inequality we can find that $$ (a+b+c)\left( \dfrac{1}{b+c} + \dfrac{1}{c+a} + \dfrac{1}{a+b}\right) \ge \dfrac92$$ Therefore $S\ge \dfrac32$.Equality holds when $a=b=c$.
For other solutions of first part: Nesbitt's Inequality
Upper Bound:
1. Method: By triangle inequality $b+c<a, c+a<b, a+b<c$ and hence $b+c>\dfrac{1}{2}(a+b+c)$, $c+a>\dfrac{1}{2}(a+b+c)$, $a+b>\dfrac{1}{2}(a+b+c)$. We have
$$ S = \dfrac{a}{b+c} + \dfrac{b}{c+a}+\dfrac{c}{a+b} < \dfrac{2a}{a+b+c} + \dfrac{2b}{a+b+c} + \dfrac{2c}{a+b+c} = \dfrac{2(a+b+c)}{a+b+c} = 2 $$
If $a=b$ and $c \to 0$ degenere triangle form, then $S=2$. Hence for all non-degenere triangles $S<2$.
2. Method: Without loss of generality, we assume that $a\leq b \leq c$. Thus,
$$ \dfrac{a}{b+c}\leq \dfrac{a}{a+c}, \quad  \dfrac{b}{a+c}\leq \dfrac{c}{a+c}, \quad c < a+b $$
and therefore $$S < \dfrac{a}{a+c} + \dfrac{c}{a+c} + \dfrac{a+b}{a+b} = \dfrac{a+c}{a+c} + 1 = 2 .$$
So, $S<2$.
In summary, we obtain that $$ \dfrac32 \le S <2 .$$
Also this interval is best.
