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?


Since the target function $f(a,b,c) = \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$ and the triangle inequalities are homogeneous in $a, b, c$, scaling $(a,b,c)$ by a common factor will not change the value of $f(a,b,c)$ nor breaking the triangle inequalities.
This means we only need to study the case $a + b + c = 1$.
When $a + b + c = 1$, the target function $( a,b,c )$ reduces to
$$f(a,b,c)\quad\rightarrow g(a,b,c)\quad\stackrel{def}{=} \frac{a}{1-a} + \frac{b}{1-b} + \frac{c}{1-c}$$
and the triangle inequalities becomes
$$\begin{cases}
0 \le a \le \frac12\\
0 \le b \le \frac12\\
0 \le c \le \frac12
\end{cases}
$$
The admissible values of $(a,b,c)$ form a triangle on the plane $a + b + c = 1$ with vertices at 
$$\left(0,\frac12,\frac12\right),\quad\left(\frac12,0,\frac12\right)\quad\text{ and }\quad\left(\frac12,\frac12,0\right)$$
Since
$$\frac{d^2}{dx^2} \frac{x}{(1-x)} = \frac{2}{(1-x)^3} > 0\quad\text{ for }\quad x \in \left[0,\frac12\right]$$
$g(a,b,c)$ is a strictly convex function over the triangle. It cannot attain local maximum at any interior points of the triangle nor its edges. This implies
$$g(a,b,c) \le \max\left\{ g\left(0, \frac12,\frac12 \right), g\left(\frac12, 0, \frac12\right), g\left(\frac12,\frac12, 0 \right) \right\} = 2$$
On the other direction, by Jensen's inequality, we have
$$g(a,b,c) \ge g\left(\frac13,\frac13,\frac13\right) = \frac{3}{2}$$ 
The possible range of $f(a,b,c)$ is $\left[\frac32,2\right]$ or $\left[\frac32,2\right)$ depends on whether degenerate triangle is allowed or not.
