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AB x BC + BC x CA + CA x AB>=OA^2 + OB^2 + OC^2 for tetrahedron, angles 60^o

Source: 2006 VMEO III Shortlist SL G3 Vietnamese Mathematics e - Olympiad https://artofproblemsolving.com/community/c2461015_vmeo__viet

October 29, 2021
geometry3D geometrytetrahedrongeometric inequality

Problem Statement

The tetrahedron OABCOABC has all angles at vertex OO equal to 60o60^o. Prove that AB⋅BC+BC⋅CA+CA⋅AB≥OA2+OB2+OC2AB \cdot BC + BC \cdot CA + CA \cdot AB \ge OA^2 + OB^2 + OC^2
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