XVI Junior Macedonian Mathematical Olympiad

From the triangle inequality we have: $$2L=\overline{AB}+\overline{BC}+\overline{CD}+\overline{DA}+\overline{AB}+\overline{CD}+\overline{DA}>\overline{AC}+\overline{BD}+\overline{AC}+\overline{BD}$$ from which we get one of the inequalities. Let us denote the point of intersection of the diagonals by $R$, and the length of the diameter of the circle by $d$. Then we have $$\begin{array}{rl}L& =\overline{AB}+\overline{BC}+\overline{CD}+\overline{DA}<\overline{AR}+\overline{BR}+\overline{BR}+\overline{CR}+\overline{CR}+\overline{DR}+\overline{DR}+\overline{AR}=\\ & =\overline{AC}+\overline{BD}+\overline{AC}+\overline{BD}\le \overline{AC}+\overline{BD}+2d=\overline{AC}+\overline{BD}+4\end{array}$$ from which we get the other inequality.