<s>[INST] 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?
 [/INST] 
Let $a,b,c$  be the lengths of the sides of a triangle. Prove that
\begin{align}\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}&<2\tag{1}\label{1}.\end{align}

Let $\rho,\ r,\ R$
be the semiperimeter,
inradius and circumradius
of the triangle $ABC$, then
we can rewrite \eqref{1} as
\begin{align} 
\frac a{2\rho-a}
+\frac b{2\rho-b}
+\frac c{2\rho-c}
&=
\frac{
4\,(a+b+c)\rho^2
-4\,(ab+bc+ca)\,\rho
+3\,abc
}{
8\rho^3-4\,(a+b+c)\,\rho^2+2\,(ab+bc+ca)\,\rho
-abc
}
<2
\tag{2}\label{2}
,
\end{align}
then, rearranging,
\begin{align} 
16\rho^3
-12(a+b+c)\rho^2
+8(ab+bc+ca)\rho-5abc
&>0
\tag{3}\label{3}
,\\
16\rho^3
-12(2\rho)\rho^2
+8(\rho^2+r^2+4rR)\rho-5\cdot4\rho r R
&>0
\tag{4}\label{4}
,\\
4r\rho(2r+3R)
&>0
\tag{5}\label{5}
,
\end{align}
which always holds.
</s>