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 $2$.

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 $p,q,r>0$.

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.

\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*}