Inequality involving sides of a triangle How can one show that for triangles of sides $a,b,c$ that 
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} < 2$$
My proof is long winded, which is why I am posting the problem here. 
Step 1: let $a=x+y$, $b=y+z$, $c=x+z$, and let $x+y+z=1$ to get 
$\frac{1-x}{1+x}+\frac{1-y}{1+y}+\frac{1-z}{1+z}<2$
Step 2: consider the function $f(x)=\frac{1-x}{1+x}$, and note that it is convex on the interval (0,1), so the minimum of 
$\frac{1-x}{1+x}+\frac{1-y}{1+y}+\frac{1-z}{1+z}$
is reached when the function takes the extreme points. i.e. $x=y=0, z=1$.  
But going back to the fact that this is a triangle, we note that $x=y=0 \implies a=0$ which is not possible, so the inequality is strict.  

$\displaystyle\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} <
\frac{a+a}{b+c+a} + \frac{b+b}{c+a+b} + \frac{c+c}{a+b+c} = 2$
since $a<b+c$ etc.
