Junior Balkan Mathematical Olympiad

Solution:
Since the quadrilateral $A P M Q$ is cyclic, we have $\angle P M Q=180^\circ-\angle P A Q=180^\circ-\angle B A C=120^\circ$. Therefore $\angle P M B+\angle Q M C=180^\circ-\angle P M Q=60^\circ$.
Let the point $B'$ be the symmetric of the point $B$ with respect to the line $P M$ and the point $C'$ be the symmetric of the point $C$ with respect to the line $Q M$. The triangles $B' M P$ and $B M P$ are congruent and the triangles $C' M Q$ and $C M Q$ are congruent. Hence $\angle B' M C'=\angle P M Q-\angle B' M P-\angle C' M Q=120^\circ-\angle B M P-\angle C M Q=120^\circ-60^\circ=60^\circ$. As we also have $B' M=B M=C M=C' M$, we conclude that the triangle $B' M C'$ is equilateral and $B' C'=B C / 2$.

On the other hand, we have $P B'+B' C'+C' Q \geq P Q$ by the Triangle Inequality, and hence $P B+B C / 2+Q C \geq P Q$. This gives the inequality $A B+B C / 2+A C \geq A P+P Q+A Q$.
We get an equality only when the points $B'$ and $C'$ lie on the line segment $P Q$. If this is the case, then $\angle P Q C+\angle Q P B=2(\angle P Q M+\angle Q P M)=120^\circ$ and therefore $\angle A P Q+\angle A Q P=240^\circ \neq 120^\circ$, a contradiction.