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 >=).
 A: One way to see that this must be true is to look at the ellipse through $B$ with foci $A$ and $C$. The interior of this ellipse is precisely all those point $P$ such that $|AP| + |PC| < |AB| + |BC|$. And if $P$ is inside the triangle, then it is inside the ellipse too.
A: A late answer but I think it's the simplest...
Extend AP to intersect BC, call the intersection point D.  
Using triangle inequality |AB| + |BD| > |AD|, similarly |PD| + |DC| > |PC|.  Adding two inequalities:
|AB| + |BD| + |PD| + |DC| > |AD| + |PC|
|AB| + |BC| + |PD| > |AD| + |PC|
|AB| + |BC| > |AD| - |PD| + |PC| = |AP| + |PC|

A: This is a special case of "Archimedes' axiom", that if one convex curve $\gamma_1$ is inside another ($\gamma_2$), then $\gamma_1$ is shorter than $\gamma_2.$ Archimedes needed this to justify his computing the perimeter of a circle by inscribed/circumscribed polygons, and could not prove it, so made it an axiom. In general, this is a nontrivial fact, in this case it is a tedious computation (if you don't want to use the general machine).
EDIT the most elegant proof is via Crofton's formula, which says that the length of a convex curve is equal (up to normalizing constant) to the measure of the lines which intersect the interior -- that measure is obviously monotonic under containment...
A: Since $P$ is inside $\triangle ABC$, the convex hull span by the three vertices $A$, $B$, $C$, there exists 3 numbers $\alpha, \beta, \gamma \ge 0$ such that
$$\alpha+\beta+\gamma = 1\quad\text{ and }\quad 
\vec{P} = \alpha \vec{A} + \beta \vec{B} + \gamma \vec{C}$$
This implies
$$
|AP| = |(1-\alpha)\vec{A} - (1-\alpha-\gamma)\vec{B} - \gamma\vec{C}|
     = |(1-\alpha)(\vec{A}-\vec{B}) + \gamma (\vec{B}-\vec{C})|\\
     \le (1-\alpha) |AB| + \gamma |BC|
$$ and $$
|CP| = |-\alpha \vec{A} - (1-\alpha-\gamma)\vec{B} + (1-\gamma)\vec{C}|
     = |\alpha (\vec{B}-\vec{A}) + (1-\gamma)(\vec{C}-\vec{B})|\\
     \le \alpha |AB| + (1-\gamma)|BC|
$$
Summing these two inequalities immediately gives us $|AP| + |CP| \le |AB| + |BC|$.
Since numbers like $\alpha$ are ratios of distance of $P$ to $BC$ versus that of $A$ to $BC$, it is easy to translate above vector based inequalities to a pure geometric proof.
Construct a line through $P$ parallel to $BC$ and let it intersect $AB$ at $D$. Construct another line through $P$ parallel to $AB$ and let it intersect $BC$ at $E$. It is easy to
see $|BE| = |PD|$ and $|BD| = |EP|$. Apply triangle inequalities to $\triangle APD$ and $\triangle CPE$, we have
$$|AP| + |CP| \le \Big(|AD| + |PD|\Big) + \Big(|CE| + |EP|\Big) 
= \Big(|AD| + |BE|\Big) + \Big(|CE| + |BD|\Big)\\
= \Big(|AD| + |DB|\Big) + \Big(|CE| + |BE|\Big) = |AB| + |BC|$$

A: 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
A: 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: $|AB| + |BC| > (|CD| + |DP|) + (|AE| + |EP|) > |CP| + |AP|$

