If A, B, C are the angles of a triangle then show that $sin A + sin B + sin C \leq \frac{3 \sqrt{3}}{2}$ .

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Solution

To prove $sin A + sin B + sin C = \frac{3 \sqrt{3}}{2}$
We know $sin A + sin B = 2 sin \frac{(A + B)}{2} \frac{(A - B)}{2} sin C = sin (180 - (A + B)) = sin (A + B) = 2 sin (\frac{A + B}{2}) cos (\frac{A + B}{2}) sin A + sin B + sin C = 2 sin (\frac{A + B}{2}) cos (\frac{A - B}{2}) + 2 sin (\frac{A + B}{2}) cos (\frac{A + B}{2}) = 2 sin (\frac{A + B}{2}) [cos (\frac{A - B}{2}) + cos (\frac{A + B}{2})] 2 sin (\frac{A + B}{2}) (2 cos \frac{A}{2} cos \frac{B}{2}) 4 sin (\frac{A + B}{2}) sin (\frac{B + C}{2}) sin (\frac{A + C}{2}) I f x + y + z = 1 & x y z i s m a x t h e n x = y = z = \frac{1}{3} S o, A + B + C; A = B = C = \frac{(A + B + C)}{3} W h e r e, sin A + sin B + sin C i s m a x = 4 sin (\frac{180}{3}) = 4 (\frac{\sqrt{3}}{2})^{3})$ $= 4 {\frac{\sqrt{3}}{2}}^{2} = \frac{3 \sqrt{3}}{2} ∴ sin A + sin B + sin C \leq \frac{3 \sqrt{3}}{2}$

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