Cho tam giác ABC nhọn nội tiếp (O) có 3 đường cao AD,BE,CF cắt nhau tại H .cmr $\frac{EF}{BC}$+ $\frac{FD}{AC}$ +$\frac{ED}{AB}$ $\geq$ $\frac{3}{2}$

Dễ dàng chứng minh tứ giác $BCEF$ nội tiếp

$\Rightarrow \widehat{AEF}=\widehat{ABC}$

$\Rightarrow \triangle AEF\backsim \triangle ABC\ (g.g)$

$\Rightarrow \dfrac{EF}{BC}=\dfrac{AE}{AB}=\dfrac{AF}{AC}$

$\Rightarrow \dfrac{EF}{BC}=\sqrt{\dfrac{AE}{AB}\cdot \dfrac{AF}{AC}}$

$\Rightarrow \dfrac{EF}{BC}\leqslant \dfrac12\left(\dfrac{AE}{AC}+\dfrac{AF}{AB}\right)$

Hoàn toàn tương tự, ta được:

$\dfrac{FD}{AC}\leqslant \dfrac12\left(\dfrac{DB}{BC}+\dfrac{FB}{AB}\right)$

$\dfrac{ED}{AB}\leqslant \dfrac12\left(\dfrac{CD}{BC}+\dfrac{EC}{AC}\right)$

Cộng vế theo vế ta được:

$\dfrac{EF}{BC} +\dfrac{FD}{AC} +\dfrac{ED}{AB}\leqslant \dfrac12\left(\dfrac{AE}{AC}+\dfrac{AF}{AB}+ \dfrac{DB}{BC}+\dfrac{FB}{AB} + \dfrac{CD}{BC}+\dfrac{EC}{AC}\right)=\dfrac32$

Dấu $=$ xảy ra $\Leftrightarrow \triangle ABC$ đều