Question

Cho tam giác ABC có ba góc nhọn

Chứng minh rằng : cos A + cos B + cos C \(\le\dfrac{3}{2}\)

Answer 1

\(cosA+cosB+cosC\le\dfrac{3}{2}\\ \Leftrightarrow2cos\dfrac{A+B}{2}.cos\dfrac{A-B}{2}+1-2sin^2\dfrac{C}{2}\le\dfrac{3}{2}\\ \Leftrightarrow Sin^2\dfrac{C}{2}-sin\dfrac{C}{2}.cos\dfrac{A-B}{2}+\dfrac{1}{4}\ge0\\ \Leftrightarrow\left(sin\dfrac{C}{2}-\dfrac{1}{2}cos\dfrac{A-B}{2}\right)^2-\dfrac{1}{4}cos^2\dfrac{A-B}{2}+\dfrac{1}{4}\ge0\\ \Leftrightarrow\left(sin\dfrac{C}{2}-\dfrac{1}{2}cos\dfrac{A-B}{2}\right)^2+\dfrac{1}{4}\left(1-cos^2\dfrac{A-B}{2}\right)\Leftrightarrow\left(sin\dfrac{C}{2}-\dfrac{1}{2}cos\dfrac{A-B}{2}\right)^2+\dfrac{1}{4}sin^2\dfrac{A-B}{2}\ge0\)=> Luôn đúng

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