Question

Chứng minh rằng

cosA + cosB + cosC \(\le\dfrac{3}{2}\)

Answer 1

\(cosA+cosB+cosC\le\dfrac{3}{2}\)

\(\Leftrightarrow2cos\dfrac{A+B}{2}cos\dfrac{A-B}{2}+1-2sin^2\dfrac{C}{2}-\dfrac{3}{2}\le0\)

\(\Leftrightarrow-2sin^2\dfrac{C}{2}+2sin\dfrac{C}{2}cos\dfrac{A-B}{2}-\dfrac{1}{2}\le0\)

\(\Leftrightarrow-2x^2+2xcos\dfrac{A-B}{2}-\dfrac{1}{2}\le0\left(1\right)\left(x=sin\dfrac{C}{2}\right)\)

\(\Delta'=cos^2\dfrac{A-B}{2}-1\)

mà \(0\le cos^2\dfrac{A-B}{2}\le1\)

\(\Leftrightarrow\Delta'\le0\)

\(\left(1\right)\Leftrightarrow-2x^2+2xcos\dfrac{A-B}{2}-\dfrac{1}{2}\le0\left(đúng\right)\)

\(\Leftrightarrow cosA+cosB+cosC\le\dfrac{3}{2}\left(dpcm\right)\)