hơi tắt xíu nha mà dễ hiểu lắm :
ta có :

`(HD)/(AD)+(HE)/(BE)+(HF)/(CF)`

`=(HD.BC)/(AD.BC)+(AC.HE)/(AC.BE)+(AB.HF)/(AB.CF)`

`=(2SBHC)/(2SABC)+(2SAHC)/(2SABC)+(2SAHF)/(2SABC)`

`=(2SBHC+2SAHC+2SAHF)/(2SABC)`

`=(2SABC)/(2SABC)`

`=1`

mặt khác

`((HD)/(AD)+(HE)/(BE)+(HF)/(CF)).((AD)/(HD)+(BE)/(HE)+(CF)/(HF))≥9`

`⇔(AD)/(HD)+(BE)/(HE)+(CF)/(HF)≥9/((HD)/(AD)+(HE)/(BE)+(HF)/(CF))`

`⇔(AD)/(HD)+(BE)/(HE)+(CF)/(HF)≥9/1`

`⇔(AD)/(HD)+(BE)/(HE)+(CF)/(HF)≥9 (ĐPCM)`

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