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.