cho tam giác ABC nhọn. AA', BB', CC' là đường cao của tam giác ABC cắt nhau tại H . chứng minh rằng `(HA')/(HA) + (HB')/(HB) +(HC')/(HC) >= 3/2`

$S_{BHC}=\dfrac{1}{2}.HA' . BC$

$S_{ABC}=\dfrac{1}{2}.AA' . BC$

$=>\dfrac{HA'}{AA'}=\dfrac{S_{HBC}}{S_{ABC}}$

$=>1+\dfrac{HA'}{HA}=\dfrac{S_{HBC}}{S_{HAB}+S_{HAC}}+1$

$=>\dfrac{HA'}{HA}=\dfrac{S_{HBC}}{S_{HAB}+S_{HAC}}$

Tương tự :

$\dfrac{HB'}{HB}=\dfrac{S_{HAC}}{S_{HAB}+S_{HBC}}$

$\dfrac{HC'}{HC}=\dfrac{S_{HAB}}{S_{HAC}+S_{HBC}}$

Đặt $x=S_{HAB},y=S_{HBC}, z=S_{HAC}$

$=>x,y,z>0$

Do đó :

$\dfrac{HA'}{HA}+\dfrac{HB'}{HB}+\dfrac{HC'}{HC}=\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}$

$=\dfrac{x^2}{xy+xz}+\dfrac{y^2}{xy+yz}+\dfrac{z^2}{xz+yz}$

Áp dụng BĐT Cộng mẫu ta được:

$\dfrac{HA'}{HA}+\dfrac{HB'}{HB}+\dfrac{HC'}{HC}>=\dfrac{(x+y+z)^2}{2(xy+yz+xz)}$

Áp dụng BĐT phụ $a^2+b^2+c^2>=ab+bc+ac$ ta được:

$xy+yz+xz=< x^2+y^2+z^2$

$=> 3(xy+yz+xz)=<(x+y+z)^2$

$=>xy+yz+xz=<\dfrac{(x+y+z)^2}{3}$

$=> \dfrac{HA'}{HA}+\dfrac{HB'}{HB}+\dfrac{HC'}{HC}>=\dfrac{1}{\dfrac{2}{3}}=\dfrac{3}{2}$