Kẻ $MH \perp BC, AK \perp BC$
Dễ thấy $\dfrac{MA_1}{AA_1} = \dfrac{MH}{AK} = \dfrac{\dfrac{1}2.BC.MA_1}{\dfrac{1}2.BC.AA_1} = \dfrac{S_{MBC}}{S_{ABC}}$
$\implies \dfrac{AA_1}{MA_1} = \dfrac{S_{ABC}}{S_{MBC}}$
Dấu '=' xảy ra khi $AA_1 \perp BC$
Tương tự : $\dfrac{BB_1}{MB_1} = \dfrac{S_{ABC}}{S_{MAC}}$
và $\dfrac{CC_1}{MC_1} = \dfrac{S_{ABC}}{S_{MAB}}$
$\implies \dfrac{AA_1}{MA_1}+\dfrac{BB_1}{MB_1}+\dfrac{CC_1}{MC_1} = S_{ABC}(\dfrac{1}{S_{MAB}}+\dfrac{1}{S_{MBC}} + \dfrac{1}{S_{MAC}})$
$= (S_{MAB} + S_{MBC}+S_{MAC})((\dfrac{1}{S_{MAB}}+\dfrac{1}{S_{MBC}} + \dfrac{1}{S_{MAC}})$
$\geqslant 9$
$\implies B = \dfrac{AA_1}{MA_1}-1+\dfrac{BB_1}{MB_1}-1+\dfrac{CC_1}{MC_1}-1$
$\geqslant 9 - 3 = 6$
Dấu '=' xảy ra $\iff M$ là trực tâm $\triangle{ABC}$