If $a$, $b$, and $c$ are sides of a triangle, then $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2$. 
Let $a,b,c$ be the lengths of the sides of a triangle. Prove that
$$\sum_{\text{cyc}}\frac{a}{b+c}=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2\,.$$

Attempt.  By clearing the denominators, the required inequality is equivalent to
$$a^2(b+c)+b^2(c+a)+c^2(a+b)>a^3+b^3+c^3\,.$$
Since $b+c>a$, $c+a>b$, and $a+b>c$, the inequality above is true.  Is there a better, non-bruteforce way?

Hint:  Without loss of generality, suppose $c$ is the largest side.  Hence, $c<a+b$.  Also,
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\leq\frac{a}{a+b}+\frac{b}{a+b}+\frac{c}{a+b}\,.$$
Note that the bound is sharp.  In the limit $a\to 0$ and $b\to c$, we have the sum goes to $2$.
