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sum of distances >= 4S^2/ 9R^2

Source: 2011 XIV All-Ukrainian Tournament of Young Mathematiciansm Qualifying p9

May 12, 2021
geometrygeometric inequalityUkrainian TYM

Problem Statement

Prove that for any interior point of a triangle the sum of squares of distances from it to the sides of a triangle is not less than 4S29R2\frac{4S^2}{9R^2}.
[hide=about S,R]they are not defined, but I suppose they mean it's area and circumradii respectively