Baltic Way 1992 · Problem 18

Solution:

Let $K$, $L$, $M$ be the midpoints of the sides $AB$, $BC$, $AC$ of a non-obtuse triangle $ABC$ (see Figure 2). Note that the centre $O$ of the circumcircle is inside the triangle $KLM$ (or at one of its vertices if $ABC$ is a right-angled triangle). Therefore $|AK| + |KL| + |LC| > |AO| + |OC|$ and hence $|AB| + |AC| + |BC| > 2(|AO| + |OC|) = 2d$, where $d$ is the diameter of the circumcircle.