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Geometry Difficulty 4.3 AIME Prove it Soviet Union

Problem:

AA, BB, CC are the angles of a triangle. Show that
2sin⁡AA+2sin⁡BB+2sin⁡CC≤(1B+1C)sin⁡A+(1C+1A)sin⁡B+(1A+1B)sin⁡C. 2\frac{\sin A}{A} + 2\frac{\sin B}{B} + 2\frac{\sin C}{C} \leq \left(\frac{1}{B} + \frac{1}{C}\right) \sin A + \left(\frac{1}{C} + \frac{1}{A}\right) \sin B + \left(\frac{1}{A} + \frac{1}{B}\right) \sin C.

Solution

Solution:

Assume A≤B≤CA \leq B \leq C. Then sin⁡A≤sin⁡B\sin A \leq \sin B. Also A≤C<180∘−AA \leq C < 180^{\circ} - A, so sin⁡A≤sin⁡C\sin A \leq \sin C. Similarly sin⁡B≤sin⁡C\sin B \leq \sin C. Hence (1/A−1/B)(sin⁡B−sin⁡A)(1/A - 1/B)(\sin B - \sin A), (1/B−1/C)(sin⁡C−sin⁡B)(1/B - 1/C)(\sin C - \sin B) and (1/C−1/A)(sin⁡A−sin⁡C)(1/C - 1/A)(\sin A - \sin C) are all non-negative. Hence their sum is also non-negative, which gives the result.

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