AoPS - Problem

A polygonal line is simply a set of line segments connected end to end, in this case the length of the polygonal line $DSE$ is just $DS+SE$. Let $O_1,O_2$ be the circumcenters of $\triangle ACD,\triangle BCE$, then $O_1$ is just the reflection of $S$ about $AC$. Let $L,M,N$ be the midpoints of $BC,CA,AB$. Then $DS=DO_1+O_1S=R+2SM$, where $R$ is the circumradius of $\triangle ABC$. Similarly $ES=R+2SL$. So now we want to have $2R+2SM+2SL=AB+BC+CA$. If $\angle BCA=90$, then it is easy to see that. If $\angle BCA>90$, then $2R>AB,2SM>BC,2SL>AC$. If $\angle BCA<90$, we want to show $R+SM+SL<AN+NM+NL$. Note that $\triangle LMN$ is acute and $S$ is its orthocenter. Let $NXSY$ be a parallelogram with $X$ on $MN$ and $Y$ on $NL$. Then $R+SM+SL<AN+NS+MX+LY<AN+NX+XS+MX+LY=AN+NM+NL$, so we are done.