Question

[HHV]

Cho tam giác ABC là tam giác nhọn, CMR: Sin^2A+Sin^2B+Sin^2C > 2?

Answer 1

[Phương An]

Kẻ đường cao AD, BE và CF.

\(\Delta AEF~\Delta ABC\left(c.g.c\right)\Rightarrow\dfrac{S_{AEF}}{S_{ABC}}=\left(\dfrac{AE}{AB}\right)^2=\cos^2A\)

\(\Delta BFD~\Delta BCA\left(c.g.c\right)\Rightarrow\dfrac{S_{BFD}}{S_{BCA}}=\left(\dfrac{BF}{BC}\right)^2=\cos^2B\)

\(\Delta CDE~\Delta CAB\left(c.g.c\right)\Rightarrow\dfrac{S_{CDE}}{S_{CAB}}=\left(\dfrac{CE}{CB}\right)^2=\cos^2C\)

\(\sin^2A+\sin^2B+\sin^2C=3-\left(\cos^2A+\cos^2B+\cos^2C\right)\)

\(=3-\left(\dfrac{S_{AEF}}{S_{ABC}}+\dfrac{S_{BFD}}{S_{BCA}}+\dfrac{S_{CDE}}{S_{CAB}}\right)>3-\dfrac{S_{ABC}}{S_{ABC}}=2\left(\text{đ}pcm\right)\)

Answer 2

[Nguyễn Thành Trương]

Ta có:
\(A + B + C = π \Rightarrow C = π - (A + B) \Rightarrow cosC = cos[π - (A + B)] = - cos(A + B) \)

\(P = Sin^2A+Sin^2B+Sin^2C = \dfrac{1 - cos2A}2 + \dfrac{1 - cos2B}2 + 1 - cos^2C\)

\(= 2 - \dfrac{cos2A + cosB}2 - cos^2(A+B)\)

\(= 2 - cos(A+B).cos(A-B) - cos^2(A+B)\)

\(= 2 - cos(A+B)[cos(A-B) + cos(A+B)]\)

\(= 2 - cos(A+B).2cosA.cosB\)

\(= 2 + 2.cosC.cosA.cosB \)
\(A ,B , C\) là các góc nhọn \(\Rightarrow\) \(cosC.cosA.cosB > 0\)

\(\Rightarrow\) \(P = Sin^2A+Sin^2B+Sin^2C > 2\)