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 >=).

Another variant on above. It takes 3 applications of the triangle inequality but I like how symmetrical it looks.


*

*Add a segment DE between AB and BC that intersects P.

*By the triangle inequality (1) $|BD| + |BE| > |DE|$ or $|BD| + |BE| > |DP| + |EP|$

*Then again on the two middle triangles: $|AE| + |EP| > |AP|$ and $|CD| + |DP| > |CP|$

*So add $ |AE| + |CD|$ to both sides of (1) to get: $|AB| + |BC| > (|CD| + |DP|) + (|AE| + |EP|) > |CP| + |AP|$

