59th Ukrainian National Mathematical Olympiad

Let $N$ be the midpoint of $AC$. Since $K$ is the midpoint of polygonal chain $BAC$, the following holds (Fig. 15): $$\frac{1}{2}(AB+AC)=KC=KN+NC=KN+\frac{1}{2}AC\Rightarrow KN=\frac{1}{2}AB=NM.$$ By the cosine theorem for $△KNM$: $$\begin{array}{rl}& K{M}^2=K{N}^2+N{M}^2-2\cdot KN\cdot NM\cdot \cos \angle KNM\\ & \Rightarrow K{M}^2=2K{N}^2(1-\cos \angle KNM)=\frac{1}{2}A{B}^2(1+\cos \angle CNM)\end{array}$$ Since $△ABC$ is acute angled, $\angle CNM=\angle CAB<90^{\circ }\Rightarrow K{M}^2>\frac{1}{2}A{B}^2$, that completes the proof.