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All-Soviet-Union Mathematical Olympiad

Geometry Difficulty 4.3 AIME Prove it Soviet Union

Problem:

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

Solution

Solution:

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

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