Question

[Minh Nguyễn Cao]

Giúp tớ đầy đủ câu này nha.

Tam giác ABC có trực tâm H

CMR: \(\frac{HA}{BC}+\frac{HB}{AC}+\frac{HC}{AB}\ge\sqrt{3}\)

Answer 1

[Nguyễn Linh Chi]

Gọi D, E, F lần lượt là chân đường cao hạ từ A, B, C của tam giác ABC.

+) \(\Delta AHE~\Delta ACD\)( vì ^HAE =^CAD, ^HEA=^CDA )

=> \(\frac{HA}{CA}=\frac{EA}{AD}\)=> \(\frac{HA}{CA}.\frac{HB}{BC}=\frac{EA}{CA}.\frac{HB}{BC}=\frac{2.EA.HB}{2.CA.BC}=\frac{S_{\Delta AHB}}{S_{ABC}}\)(1)

+) \(\Delta CHD~\Delta CBF\)( vì ^DCH=^FCB, ^CDH=^CFB )

=> \(\frac{CH}{CB}=\frac{CD}{CF}\)=> \(\frac{CH}{CB}.\frac{AH}{AB}=\frac{CD.AH}{CF.AB}=\frac{S_{AHC}}{S_{ABC}}\)(2)

+) \(\Delta ABE~\Delta HBF\)

=> \(\frac{HB}{AB}=\frac{BF}{BE}\Rightarrow\frac{HB}{AB}.\frac{HC}{AC}=\frac{BF.HC}{BE.AC}=\frac{S_{BHC}}{S_{ABC}}\)(3)

Từ (1) ; (2) ; (3) => \(\frac{HA}{CA}.\frac{HB}{BC}+\frac{CH}{CB}.\frac{AH}{AB}+\frac{HB}{AB}.\frac{HC}{AC}=\frac{S_{ABE}}{S_{ABC}}+\frac{S_{ABE}}{S_{ABC}}+\frac{S_{ABE}}{S_{ABC}}=1\)

=> \(\frac{HA}{BC}.\frac{HB}{AC}+\frac{HB}{AC}.\frac{HC}{AB}+\frac{HC}{AB}.\frac{HA}{BC}=1\)

Đặt: \(\frac{HA}{BC}=x;\frac{HB}{AC}=y;\frac{HC}{AB}=z\); x, y, z>0

Ta có: \(xy+yz+zx=1\)

=> \(\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)=3\)

=> \(x+y+z\ge\sqrt{3}\)

"=" xảy ra khi và chỉ khi x=y=z

Vậy : \(\frac{HA}{BC}+\frac{HB}{AC}+\frac{HC}{AB}\ge\sqrt{3}\)

"=" xảy ra <=> \(\frac{HA}{BC}=\frac{HB}{AC}=\frac{HC}{AB}\)