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.

Let $c$ be the largest of $a$, $b$, and $c$. Then \begin{align} \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}&\le\frac{a}{a+b}+\frac{b}{a+b}+\frac{c}{a+b}\\ &=1+\frac{c}{a+b}\\ &< 2 \end{align} since $c<a+b$.

For a degenerate triangle with $a=0$, we have $c=a+b$ and the sum equals $2$.