Giải thích các bước giải:
Gọi $S_{IBC}=S_b,S_{IAC}=S_b,S_{IAB}=S_c$
Ta có :
$\dfrac{AI}{IM}=\dfrac{S_{BAI}}{S_{BIM}}=\dfrac{S_{CAI}}{S_{IMC}}=\dfrac{S_{BAI}+S_{CAI}}{S_{BIM}+S_{IMC}}=\dfrac{S_b+S_c}{S_a}$
Tương tự :
$\dfrac{BI}{IN}=\dfrac{S_{a}+S_{c}}{S_{b}}$
$\dfrac{CI}{ID}=\dfrac{S_a+S_b}{S_c}$
$\to\dfrac{AI}{IM}+\dfrac{BI}{IN}+\dfrac{CI}{IP}=\dfrac{S_b+S_c}{S_a}+\dfrac{S_a+S_c}{S_b}+\dfrac{S_b+S_a}{S_c}$
$\to\dfrac{AI}{IM}+\dfrac{BI}{IN}+\dfrac{CI}{IP}=\dfrac{S_b}{S_a}+\dfrac{S_c}{S_a}+\dfrac{S_a}{S_b}+\dfrac{S_c}{S_b}+\dfrac{S_b}{S_c}+\dfrac{S_a}{S_c}$
$\to\dfrac{AI}{IM}+\dfrac{BI}{IN}+\dfrac{CI}{IP}=6\sqrt[6]{\dfrac{S_b}{S_a}.\dfrac{S_c}{S_a}.\dfrac{S_a}{S_b}.\dfrac{S_c}{S_b}.\dfrac{S_b}{S_c}.\dfrac{S_a}{S_c}}$
$\to\dfrac{AI}{IM}+\dfrac{BI}{IN}+\dfrac{CI}{IP}=6$
Dấu = xảy ra khi $S_a=S_b=S_c\to I$ là trọng tâm $\Delta ABC$
