Có : $\cos A \cos B \cos C \\ = \dfrac{\cos(A+B)+\cos(A-B)}2\cos C \\ =-\dfrac{\cos^2C}2-\dfrac{\cos(A-B)\cos(A+B)}2 \\ = -\dfrac{\cos^2C}2-\dfrac{\cos2A+\cos2B}4 \\ =-\dfrac{\cos^2C}2-\dfrac{\cos^2A}2-\dfrac{\cos^2B}2+\dfrac12$
\Leftrightarrow $1+\cos A \cos B \cos C \\= -\dfrac{\cos^2C}2-\dfrac{\cos^2A}2-\dfrac{\cos^2B}2+\dfra
\\ = \dfrac12(\sin^2A+\sin^2B+\sin^2C)$
Lại có: $\sin^2 A + \sin^2 B + \sin^2 C \\= 1-\cos^2A +\dfrac{1-\cos2B}2+\dfrac{1-\cos2C}2 \\ = 1-\cos^2A+1-\dfrac12(\cos2B+\cos2C) \\ = \dfrac94 - \dfrac14-\cos^2A+\cos A\cos(B-C) \\ =\dfrac14 -\dfrac14\sin^2(B-c) - \cos^2A+\cos A\cos(B-C) -\dfrac14\cos^2(B-C) \\ =\dfrac94 -\dfrac14\sin^2(B-C) - (\cos A-\dfrac{\cos(B-C)}2)^2 \\ \le \dfrac94$
Mặt khác:$\dfrac94 \ge \sin^2A+\sin^2B+\sin^2C \ge 3\sqrt[3]{\sin^2A\sin^2B\sin^2C}$
\Leftrightarrow $\sqrt[3]{\sin A\sin B \sin C} \le \dfrac{\sqrt3}2$
\Leftrightarrow $\sin A\sin B\sin C \le \dfrac{\sqrt3}2\sqrt[3]{\sin^2 A\sin^2 B \sin^2 C}$
\Leftrightarrow $\sqrt3\sin A\sin B\sin C \le \dfra
\sqrt[3]{\sin^2 A\sin^2 B \sin^2 C} \le \dfrac{\sin^2A+\sin^2B+\sin^2C}2 = 1+\cos A\cos B\cos C$
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
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