How to prove that triangle inscribed in another triangle (were both have one shared side) have lower perimeter? This question looks really simple, but to my (and my co-workers) frustration we were not able to prove this in any way. I tried all triangle formulas known to me but I feel I'm missing the point, and proof will be or much simpler or much more complicated than what I tried.
So, the question:
Given a triangle ABC and point P inside that triangle, prove that for triangle APB the following inequality holds:
|AB| + |BC| > |AP| + |PC|
(Actually it doesn't matter for me if it's > or >=).

There is really an elementary proof of a more general result: if one polygon fits within the other (boundaries may touch) then its perimeter is smaller than that of the bigger polygon
http://www.cut-the-knot.org/m/Geometry/PerimetersOfTwoConvexPolygons.shtml
The idea is that the sides of the small polygon may be extended, trimming the bigger polygon until the sides of one are on the sides of the other so that side-by-side inequalities could be summed up to get an inequality for the perimeters
