Write $f=PB=QC$. By the Law of Cosines, we have $PQ^2=AP^2+AQ^2-2AP\cdot AQ\cos(\angle BAC)=(AB-f)^2+(AC-f)^2-2(AB-f)(AC-f)\cos(\angle BAC)$. Expanding, we find that $PQ^2=(AB^2+AC^2-2AB\cdot AC\cos(\angle BAC))-2f(AB+BC-(AB+BC)\cos(\angle BAC))+f^2(2-2\cos(\angle BAC))$. Note that $BC^2=AB^2+AC^2-2AB\cdot AC\cos(\angle BAC)$, so the problem is reduced to proving that $-2f(AB+BC-(AB+BC)\cos(\angle BAC))+f^2(2-2\cos(\angle BAC))>0$. But we factor this as $2f(1-\cos(\angle BAC))(AB+BC-f)>0$, so the conclusion follows.