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?

Since $a,b,c$ are the sides of a triangle, there exist $p,q,r\in\mathbb R^+$ such that $a=p+q$, $b=q+r$, $c=r+p$ (Ravi substitution).
$$\sum_{\text{cyc}}\frac{a}{b+c}<2\iff \sum_{\text{cyc}}\frac{p+q}{p+q+2r}<2$$
Now multiply both sides by $(2p+q+r)(p+2q+r)(p+q+2r)$, expand, rearrange.
See WolframAlpha (link) if you want.
$$\iff 3 p^2 q+3 p^2 r+3 p q^2+14 p q r+3 p r^2+3 q^2 r+3 q r^2
>0,$$
which is trivial, because $p,q,r>0$.
