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3(AM^2 + B'P^2 + CN^2)>=2D'B^2, cube if (2019 Romania District VIII p2)

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May 23, 2020
parallelepipedgeometry3D geometrycubegeometric inequality

Problem Statement

Let ABCDA′B′C′D′ABCDA'B'C'D' be a rectangular parallelepiped and M,N,PM,N, P projections of points A,CA, C and B′B' respectively on the diagonal BD′BD'.
a) Prove that BM+BN+BP=BD′BM + BN + BP = BD'.
b) Prove that 3(AM2+B′P2+CN2)≥2D′B23 (AM^2 + B'P^2 + CN^2)\ge 2D'B^2 if and only if ABCDA′B′C′D′ABCDA'B'C'D' is a cube.
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