Thấy các tam giác $HDB;ADB$ chung đáy $DB$
nên $\dfrac{S_{ADB}}{S_{HDB}}=\dfrac{AD}{HD}$
tương tự $\dfrac{S_{ADC}}{S_{HDC}}=\dfrac{AD}{HD}$
Nên $\dfrac{S_{ADB}}{S_{HDB}}=\dfrac{S_{ADC}}{S_{HDC}}=\dfrac{S_{ADB}+S_{ADC}}{S_{HDB}+S_{HDC}}$
Hay $\dfrac{AD}{HD}=\dfrac{S_{ABC}}{S_{BHC}}$
Hoàn toàn tương tự ta có:
$\dfrac{BE}{HE}=\dfrac{S_{ABC}}{S_{CHA}}$
$\dfrac{CF}{HF}=\dfrac{S_{ABC}}{S_{AHB}}$
Vì vậy
$\dfrac{AD}{HD}+\dfrac{BE}{HE}+\dfrac{CF}{HF}=\dfrac{S_{ABC}}{S_{BHC}}+\dfrac{S_{ABC}}{S_{CHA}}+\dfrac{S_{ABC}}{S_{AHB}}=S_{ABC}.(\dfrac{1}{S_{BHC}}+\dfrac{1}{S_{CHA}}+\dfrac{1}{S_{AHB}})$
Đặt $S_{BHC}=a;S_{CHA}=b;S_{AHB}=c⇒S_{ABC}=a+b+c$
Ta có $\dfrac{AD}{HD}+\dfrac{BE}{HE}+\dfrac{CF}{HF}=(a+b+c)(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})$
Thấy $(a+b+c)(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})=3+\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{b}{c}+\dfrac{c}{b}+\dfrac{a}{c}+\dfrac{c}{a}$
Áp dụng bđt Cô si có : $(a+b+c)(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})=3+\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{b}{c}+\dfrac{c}{b}+\dfrac{a}{c}+\dfrac{c}{a}≥3+2.\sqrt[]{\dfrac{a}{b}.\dfrac{b}{a}}+2.\sqrt[]{\dfrac{b}{c}.\dfrac{c}{b}}+2.\sqrt[]{\dfrac{a}{c}.\dfrac{c}{a}}=3+2+2+2=9$
Nên ta có đpcm
