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?

As @user236182 pointed out, we can write $a=p+q$, $b=q+r$, and $c=r+p$, with $p,q,r>0$. Letting $s = p+q+r$, we have
$$\frac{a}{b+c} = \frac{s-r}{s+r} = 1 - \frac{2r}{s+r} $$
and hence
$$\sum\limits_{\text{cyc}}{\frac{a}{b+c}}< 2 \iff \sum\limits_{\text{cyc}}{\left(1-\frac{2r}{s+r}\right)}< 2 \iff \sum\limits_{\text{cyc}}{\frac{2r}{s+r}} > 1. $$
By Cauchy-Schwarz we have
$$\left(\sum\limits_{\text{cyc}}{\frac{r}{s+r}}\right)\left(\sum\limits_{\text{cyc}}{r(s+r)}\right)\ge\left(\sum\limits_{\text{cyc}}{r}\right)^2 = s^2. $$
Furthermore,
$$\sum\limits_{\text{cyc}}{r(s+r)} = s\sum\limits_{\text{cyc}}{r}+\sum\limits_{\text{cyc}}{r^2} = s^2 + (p^2+q^2+r^2) < s^2+(p+q+r)^2 = 2s^2. $$
Note that the inequality is strict as $p,q,r>0$. It follows that
$$ 2s^2\sum\limits_{\text{cyc}}{\frac{r}{s+r}} > \left(\sum\limits_{\text{cyc}}{\frac{r}{s+r}}\right)\left(\sum\limits_{\text{cyc}}{r(s+r)}\right) \ge s^2 $$
and hence $\sum\limits_{\text{cyc}}{\frac{2r}{s+r}} > 1$, as desired.
