hint

rkm0959 wrote: Let $T$ be a point not inside $\triangle ABC$ and on the opposite side of $A$ with respect to $BC$, such that $BT-CT=c-b$. In the original problem $T$ was on the same side with $A$ respect to $BC$, so that $BT\geq c$, $CT\geq b$.

Isogonics wrote: $BT\geq c$, $CT\geq b$. Assuming this, $(m-b)\cdot BP+(n-c)\cdot CP\geq (m-b)BC$ holds, where the equality holds on $P\in BC$ or $m-b=n-c=0$. Note that by Ptolemy's inequality, $-a \cdot AP + b \cdot BP + c \cdot CP \geq 0$ for all $P$, where the equality holds on $P\in \overarc{BC}$. So the answer is $\overarc{BC}$ when $T=A$, and $B,C$ otherwise.