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?

\begin{align*}
\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} & = \frac{2a}{2(b+c)}+\frac{2b}{2(c+a)}+\frac{2c}{2(a+b)} \\
&< \frac{2a}{a+b+c} + \frac{2b}{c+a+b} + \frac{2c}{a+b+c} \\
&= 2
\end{align*}
