Minimizing the Expression Three

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Given that A , B , C A, B, C are the three angles of an acute triangle, find the greatest lower bound of sin ⁡ A + sin ⁡ B + sin ⁡ C \sin A + \sin B + \sin C


The answer is 2.

Alan Yan by Alan Yan , 20, USA

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1 solution

Alan Yan
Sep 9, 2015

A , B , C ≤ π 2 A, B, C \leq \frac{\pi}{2} f ( x ) = sin ⁡ x f(x) = \sin x f f is concave for [ 0 , π 2 ] [0, \frac{\pi}{2}] ( π 2 , π 2 , 0 ) ≻ ( A , B , C ) ⟹ sin ⁡ A + sin ⁡ B + sin ⁡ C ≥ sin ⁡ π 2 + sin ⁡ π 2 + 0 (Karamata) (\frac{\pi}{2} , \frac{\pi}{2} , 0 ) \succ (A, B, C) \implies \sin A + \sin B + \sin C \geq \sin \frac{\pi}{2} + \sin \frac{\pi}{2} + 0 \text{ (Karamata)} ⟹ sin ⁡ A + sin ⁡ B + sin ⁡ C ≥ 2 \implies \sin A + \sin B + \sin C \geq \boxed{2}

Extra Exercise: Can you find the equality case?

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You are showing that 2 is a lower bound; now why is 2 the greatest lower bound?

Otto Bretscher - 5 years, 9 months ago

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