Ta chứng minh chiều nghịch:

Khi tam giác ABC đều, góc A=gócB=gócC=60*

Khi đó cosA+cosB+cosC=3/2(đpcm)

Ta chứng minh chiều thuận

Ta chứng minh cosA+cosB+cosC≤3/2

Thật vậy:

 Mà theo gt, cosA+cosB+cosC=3/2

nên ta có tam giác ABC đều(đpcm)

vẽ AD,BE, CF là các đường cao của tam giác ABC

\(\cos A=\sqrt{\cos BAE\cdot\cos CAF}=\sqrt{\frac{AE}{AB}\cdot\frac{AE}{AC}}=\sqrt{\frac{AF}{AB}\cdot\frac{AE}{AC}}\le\frac{1}{2}\left(\frac{AF}{AB}+\frac{AE}{AC}\right)\)

ta có \(\cos A\le\frac{1}{2}\left(\frac{AF}{AB}+\frac{AE}{AC}\right)\left(1\right)\)

tương tự \(\cos B\le\frac{1}{2}\left(\frac{BF}{AB}+\frac{BD}{BC}\right)\left(2\right);\cos C\le\frac{1}{2}\left(\frac{CD}{BC}+\frac{CE}{AC}\right)\left(3\right)\)

do đó \(\cos A+\cos B+\cos C\le\frac{1}{2}\left(\frac{AF}{AB}+\frac{AE}{AC}+\frac{BF}{AB}+\frac{BD}{BC}+\frac{CD}{BC}+\frac{CE}{AC}\right)\)

\(\Rightarrow\cos A+\cos B+\cos C\le\frac{1}{2}\left(\frac{AF}{AB}+\frac{BF}{AB}+\frac{AE}{AC}+\frac{CE}{AC}+\frac{BD}{BC}+\frac{CD}{BC}\right)\)

\(\Rightarrow\cos A+\cos B+\cos C\le\frac{3}{2}\)

dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\frac{AF}{AB}=\frac{AE}{AC}\\\frac{BF}{AB}=\frac{BD}{BC}\\\frac{CD}{BC}=\frac{CE}{AC}\end{cases}}\Leftrightarrow AB=AC=BC\)

do vậy cosA+cosB+cosC=3/₂ <=> AB=AC=BC <=> tam giác ABC đều

Cách khác khỏi phải dùng hình học :v

\(A=\cos A+\cos B+\cos C\)

\(=\left(\cos A+\cos B\right)\cdot1+\sin A\cdot\sin B-\cos A\cdot\cos B\)

\(\le\frac{1}{2}\left[\left(\cos A+\cos B\right)^2+1\right]+\frac{1}{2}\left(\sin^2A+\sin^2B\right)-\cos A\cdot\cos B\)

\(=\frac{1}{2}\left(\cos^2A+\sin^2A+\cos^2B+\sin^2B\right)+\frac{1}{2}\)

\(=\frac{3}{2}\)

ez Problem :v

\(cosA+cosB-cosC=2cos\frac{A+B}{2}.cos\frac{A-B}{2}+2sin^2\frac{C}{2}-1\)

\(=2sin\frac{C}{2}.cos\frac{A-B}{2}+2sin^2\frac{C}{2}-1\)

\(=2sin\frac{C}{2}\left(cos\frac{A-B}{2}+sin\frac{C}{2}\right)-1\)

\(=2sin\frac{C}{2}\left(cos\frac{A-B}{2}+cos\frac{A+B}{2}\right)-1\)

\(=4cos\frac{A}{2}cos\frac{B}{2}sin\frac{C}{2}-1\)

Xét tam giác ABC nhọn có \(BC^2=AB^2+AC^2-2AB\cdot AC\cdot\cos\widehat{A}\)
\(\Rightarrow\cos\widehat{A}=\dfrac{AB^2+AC^2-BC^2}{2AB\cdot AC}=\dfrac{AB^2+AC^2-BC^2}{4\cdot\dfrac{1}{2}AB\cdot AC}=\dfrac{AB^2+AC^2-BC^2}{4S_{ABC}}\)

Cmtt: \(\left\{{}\begin{matrix}\cos\widehat{B}=\dfrac{AB^2+BC^2-AC^2}{4S_{ABC}}\\\cos\widehat{C}=\dfrac{AC^2+BC^2-AB^2}{4S_{ABC}}\end{matrix}\right.\)
\(\Rightarrow\cos\widehat{A}+\cos\widehat{B}+\cos\widehat{C}\\ =\dfrac{AB^2+AC^2-BC^2+AB^2+BC^2-AC^2+AC^2+BC^2-AB^2}{4S_{ABC}}\\ =\dfrac{AB^2+AC^2+BC62}{4S_{ABC}}\)

Tham khảo nha !!!