Baltic Way 1991 · Problem 7

Geometry

Let $A, B, C$ be the angles of an acute-angled triangle. Prove the inequality

\sin A+\sin B>\cos A+\cos B+\cos C

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Topics

Coordinates and vectors · Geometric inequalities · Triangles and centers

Solutions

Solution

Solution: In an acute-angled triangle we have $A + B > \frac{\pi}{2}$ . Hence we have $\sin A > \sin \left(\frac{\pi}{2} - B\right) = \cos B$ and $\sin B > \cos A$ . Using these inequalities we get $(1 - \sin A)(1 - \sin B) < (1 - \cos A)(1 - \cos B)$ and

\begin{aligned} \sin A + \sin B &> \cos A + \cos B - \cos A \cos B + \sin A \sin B \\ &= \cos A + \cos B - \cos (A + B) = \cos A + \cos B + \cos C \end{aligned}

Elsewhere

AoPS · Baltic Way 1991