Giải thích các bước giải:
Gọi $S_{MBC}=a,S_{MAC}=b,S_{MAB}=c$
$\dfrac{MA}{MD}=\dfrac{S_{MAB}}{S_{MDB}}=\dfrac{S_{MAC}}{S_{MDC}}=\dfrac{S_{MAB}+S_{MAC}}{S_{MDB}+S_{MDC}}=\dfrac{S_{MAB}+S_{MAC}}{S_{MBC}}=\dfrac{c+b}{a}$
Tương tự ta có:
$\dfrac{MB}{ME}=\dfrac{a+c}{b}$
$\dfrac{MC}{MF}=\dfrac{a+b}{c}$
$\rightarrow P= \dfrac{MA}{MD}.\dfrac{MB}{ME}.\dfrac{MC}{MF}=\dfrac{b+c}{a}.\dfrac{a+c}{b}.\dfrac{a+b}{c}$
$\rightarrow P=\dfrac{(a+b)(b+c)(c+a)}{abc}\ge\dfrac{2\sqrt{ab}.2\sqrt{bc}.2\sqrt{ca}}{abc}=8$
Dấu = xảy ra khi $S_{MAB}=S_{MAC}=S_{MBC}$
$\rightarrow $M là trọng tâm tam giác ABC
$\rightarrow \dfrac{MA}{MD}.\dfrac{MB}{ME}.\dfrac{MC}{MF}\ge 8\rightarrow dpcm $
