Baltic Way 2021 · Shortlist problem

Geometry

Given is a triangle with real side lengths $a$ , $b$ , $c$ and real area $A$ . Prove that $3a^2 + 3b^2 - c^2 \ge 4\sqrt{3}A.$

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Review

Coordinates and vectors · Geometric inequalities · Triangles and centers

Solution

Let $\gamma$ be the angle between the sides $a$ and $b$ . We have

\begin{aligned} a^2 + b^2 &\ge 2ab \\ &\ge 2ab \sin(\gamma - 30^\circ) \\ &= 2ab [\sin(\gamma) \cos(30^\circ) - \cos(\gamma) \sin(30^\circ)] \\ &= ab \sin(\gamma)\sqrt{3} - ab \cos(\gamma) \\ &= 2\sqrt{3}A - \frac{1}{2}(a^2 + b^2 - c^2), \end{aligned}

which implies the desired inequality.