Croatian Mathematical Olympiad

It suffices to show that
$$\frac{|AM|}{|MD|} + \frac{|BM|}{|ME|} + \frac{|CM|}{|MF|} \ge \frac{3}{2}.$$
Quadrilaterals $MBDC$ and $MCEA$ are cyclic, so
$$\angle BCD = \angle BMD = \angle EMA = \angle ECA.$$
Also, we have
$$\angle DBC = \angle DMC = 180^\circ - \angle CMA = \angle CEA.$$
Hence we conclude that the triangles BDC and EAC are similar. Analogously, we prove that the triangle BAF is similar to them.

By Ptolemy's theorem for quadrilaterals MBDC, MCEA and MAFB, and by using the ratios from the similarity of BDC, EAC and BAF, we have:
$$\begin{align*} |MD| &= |BM| \cdot \frac{|CD|}{|BC|} + |CM| \cdot \frac{|DB|}{|BC|}, \\ |ME| &= |CM| \cdot \frac{|AE|}{|CA|} + |AM| \cdot \frac{|EC|}{|CA|} = |CM| \cdot \frac{|DB|}{|CD|} + |AM| \cdot \frac{|BC|}{|CD|}, \\ |MF| &= |AM| \cdot \frac{|BF|}{|AB|} + |BM| \cdot \frac{|FA|}{|AB|} = |AM| \cdot \frac{|BC|}{|DB|} + |BM| \cdot \frac{|CD|}{|DB|}, \end{align*}$$
from where it follows that
$$\frac{|AM|}{|MD|} + \frac{|BM|}{|ME|} + \frac{|CM|}{|MF|} = \frac{|AM| \cdot |BC|}{|BM| \cdot |CD| + |CM| \cdot |DB|} \\ + \frac{|BM| \cdot |CD|}{|CM| \cdot |DB| + |AM| \cdot |BC|} \\ + \frac{|CM| \cdot |DB|}{|AM| \cdot |BC| + |BM| \cdot |CD|}$$
Let us denote $x = |AM| \cdot |BC|$, $y = |BM| \cdot |CD|$, $z = |CM| \cdot |DB|$. We need to prove the inequality
$$\frac{x}{y+z} + \frac{y}{z+x} + \frac{z}{x+y} \geq \frac{3}{2}.$$
This is Nesbitt’s famous inequality, so the proof is finished.