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?


As expression 
$$\tag{1}f(a,b,c):=\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$$
is invariant in any change of scale, we introduce the following (value) constraint:
$$\tag{2}g(a,b,c):=a+b+c=1$$
This means that the subsequent work is in this simplex, or more exactly in the subset of this simplex with supplementary (domain) constraints: 
$$\tag{3} a \leq b+c, \ \ b \leq c+a, \ \ c \leq a+b$$
(in case of equality, we have flat triangles).
Under all these constraints, $(1)$ becomes:
$$\tag{4}f(a,b,c)=\frac{a}{1-a} + \frac{b}{1-b} + \frac{c}{1-c}$$
maximizing/minimizing $(4)$ under constraint $(2)$ is done by writing Lagrange equations:
$$grad (f) \ = \ \lambda \ grad(g) \ \ \ \iff \ \ \ \cases{\dfrac{1}{(1-a)^2}=\lambda 1\\ \dfrac{1}{(1-b)^2}=\lambda 1\\ \dfrac{1}{(1-c)^2}=\lambda 1},$$
from which $a=b=c=\frac13$ (equilateral triangle), giving $f(\frac13,\frac13,\frac13)=\frac32.$ 
This extremum is a minimum because the Hessian matrix, 
diag $\left(\dfrac{2}{(1-a)^3},\dfrac{2}{(1-b)^3},\dfrac{2}{(1-c)^3}\right)$ is positive definite.
Now, the maxima(s) should be found on the borders. As all the borders play a symmetrical role, consider only boundary $c=a+b$, which implies (using $(2)$) 
$$1=a+b+c=2(a+b)\ \ \ \Longrightarrow \ \ \ a+b=\frac12$$
$$\tag{5} \Longrightarrow \ \ \ 0 \leq a\leq 1/2, b=\dfrac12-a, c=\frac12.$$
Thus:
$$f(a,b,c)=f(a,\frac12-a,\frac12)=\dfrac{a}{1-a}+2-2a.$$
A rapid study of this function for $a \in [0,\frac12]$ shows that the value of its maximum is $2$, occurring for $a=0$ or $a=\frac12$. Both associated with a "flat" triangle with sides $(\frac12,\frac12,0) $. 
Conclusion: the values taken by $f$ are situated in the interval $[\frac32,2]$. As $f$ is continuous on a compact domain, it is the whole interval $[\frac32,2].$
