If A, B, C be the angles of triangle ABC then $(sin A + sin B) (sin B + sin C) (sin C + sin A) > sin A sin B sin C$ .

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Solution

$sin A + sin B - sin C = 4 sin \frac{A}{2} sin \frac{B}{2} cos \frac{C}{2} > 0$
$∴ sin A + sin B > sin C$
Write similar expressions and multiply.

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