A Minimum Over Triangle Angles

Problem

Let A,B,CA, B, C be the interior angles of a triangle. Find the minimum value of

m=1sin⁡2A+1sin⁡2B+41+sin⁡C.m = \frac{1}{\sin^2 A} + \frac{1}{\sin^2 B} + \frac{4}{1 + \sin C}.

Answer

Solution

Difficulty8/10
Topicstrigonometry, Extrema, Trigonometric Identities, AM-GM, inequality

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