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\( (0 < A, B, C < A+B+C=\pi) \Rightarrow \sin A + \sin B + \sin C \leq \frac{3\sqrt{3}}{2} \)
03-06-2010, 11:29 PM
Post: #1
\( (0 < A, B, C < A+B+C=\pi) \Rightarrow \sin A + \sin B + \sin C \leq \frac{3\sqrt{3}}{2} \)
Proof. If \(0 < A< B\), \(A+B<\pi\), then \(\displaystyle{\sin A + \sin B = 2\sin \frac{A+B}{2}\cos\frac{A-B}{2} < 2\sin\frac{A+B}{2}} \)
Thus \(\displaystyle{\sin A + \sin B + \sin C \leq 3\sin \frac{A+B+C}{3}=3\sin \frac{\pi}{3}=\frac{3\sqrt{3}}{2}}\)
The equality holds iff \(A=B=C=\pi/3\)
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