Parallelo-hexagon with Obtuse Angles Inequality

Below is shown a dynamic sketch of a hexagon with opposite sides equal and parallel, i.e. a parallelo-hexagon. More-over, it has been constructed in such a way that all its angles are obtuse. Explore the sketch by dragging any of the red vertices or the points H and I to change the obtuse angles at vertices A and F.
Theorem: If ABCDEF is a convex parallelo-hexagon with all its angles obtuse, then the following inequality holds: AD² + BE² + CF² > 3(AB² + BC² + CD²).

Parallelo-hexagon with Obtuse Angles Inequality

Explaining
Can you EXPLAIN WHY (prove that) this theorem is true?

Explore More
To explore more about the relations between the sides and diagonals of parallelo-hexagons, or general hexagons, click on Relations between the sides and diagonals of parallelo-hexagons, and the general theorem of Douglas to navigate to a separate, accurately constructed dynamic sketch.

Reference
De Villiers, M. (2012). Relations between the sides and diagonals of a set of hexagons. Mathematical Gazette, July, 96(536), pp. 309-315.

Related Links
A generalization of a Parallelogram Theorem to Parallelo-hexagons, Hexagons and 2n-gons in general
Easy Hexagon Explorations
The 3D parallelo-hexagon
Triangle Centroids of a Hexagon form a Parallelo-Hexagon: A generalization of Varignon's Theorem
Some Parallelo-hexagon Area Ratios
Area Parallelogram Partition Theorem: Another Example of the Discovery Function of Proof
Some Quadrilateral Inequalities involving Sides & Diagonals

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