Ta co
$\sin A + \sin B + \sin C = 4 \cos \dfrac{A}{2} \cos \dfrac{B}{2} \cos \dfrac{C}{2} $(tự chứng minh)
$\sin 2A + \sin 2B + \sin 2C = 4\sin A \sin B \sin C $
Vậy $\cos \dfrac{A}{2} \cos \dfrac{B}{2} \cos \dfrac{C}{2} = \sin A \sin B \sin C = 8\cos \dfrac{A}{2} \cos \dfrac{B}{2} \cos \dfrac{C}{2} . \sin \dfrac{A}{2} \sin \dfrac{B}{2} \sin \dfrac{C}{2} $
$\Rightarrow 8 \sin \dfrac{A}{2} \sin \dfrac{B}{2} \sin \dfrac{C}{2} = 1 $
Ta có $ 8 \sin \dfrac{A}{2} \sin \dfrac{B}{2} \sin \dfrac{C}{2} = 4 \sin \dfrac{A}{2} [ \cos \dfrac{B-C}{2} - \cos \dfrac{B+C}{2}] = 4 \sin \dfrac{A}{2} [\cos \dfrac{B-C}{2}-\sin \dfrac{A}{2}] \le 4\sin \dfrac{A}{2} [1 -\sin \dfrac{A}{2}]$
(Vi $\cos \dfrac{B-C}{2} \le 1;\ \sin \dfrac{A}{2} \ge 0) $)
$= 1-[1 -2\sin \dfrac{A}{2}]^2 \le1 $
Dấu $=$ xảy ra khi chỉ khi $\sin \dfrac{A}{2} =\dfrac{1}{2}; \ \cos \dfrac{B-C}{2} =1$
$\Leftrightarrow A=B =C=\frac{\pi}{3} $