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3\sqrt3(cot AXC/2+cot AXB/2 )+ cot AXC/2+cot AXB/2>5 , BX+CX <3 AX

Source: V All-Ukrainian Tournament of Young Mathematicians, Qualifying p8

May 20, 2021
geometrytrigonometrygeometric inequalityEquilateralUkrainian TYM

Problem Statement

Let XX be a point inside an equilateral triangle ABCABC such that BX+CX<3AXBX+CX <3 AX. Prove that 33(cot⁡∠AXC2+cot⁡∠AXB2)+cot⁡∠AXC2cot⁡∠AXB2>53\sqrt3 \left( \cot \frac{\angle AXC}{2}+ \cot \frac{\angle AXB}{2}\right) +\cot \frac{\angle AXC}{2} \cot \frac{\angle AXB}{2} >5
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