Circumference of Convex Shapes Here is a puzzle I found in Mitteilungen der DMV (roughly, "Letters of the German Society of Mathematicians"), issue 19/2011.  It was posed by Alfred Schreiber in "Wie man Hasen fangt" (How to catch rabbits), and he claims that less than 5% of his subjects could solve it in under 1 hour.  He tested it on students of mathematics and professors of mathematics, computer science, and engineering.
See if you have more luck. The problem is deceptively simple:
Suppose that you have a triangle $ABC$ and a point $D$ inside the triangle.  Prove that the perimeter of $ABC$ is larger than the perimeter of $ABD$.
I am currently working on a generalization: Given two convex shapes $S$ and $T$, where $T$ totally encloses $S$.  Prove that the perimeter of $S$ is no bigger than the perimeter of $T$.
Alternatively, for a shapes with straight edges: Prove that the perimeter of the convex hull of a set of points increases monotonically (but not strictly monotonically) as points are added to the set.
Please try to find an elementary proof for the special case of the triangle.
Edit: Thanks for all the nice answers.  By now, I have found a really elementary proof on my own that just uses the triangle inequality twice.
 A: For the triangle problem: Euclid I, 21:  

If from the ends of one of the sides of a triangle two straight lines are constructed meeting within the triangle, then the sum of the straight lines so constructed is less than the sum of the remaining two sides of the triangle, but the constructed straight lines contain a greater angle than the angle contained by the remaining two sides.

A: This also follows immediately from Crofton's formula, which says that, up to a fixed constant factor, the length of a reasonable curve (and polygons are certainly reasonable) is the integral, over straight lines (with respect to a suitable invariant measure), of the number of intersections between the straight line and the curve.  Almost every straight line meets the perimeter of a convex polygon either twice or not at all, and shrinking the polygon will only reduce the set of lines that meet it.
A: Motivated by my thoughts about a recent question I want to comment that perhaps this fact is intrinsic to our very definition of arc length.  As correctly noted, the result for convex polygons is the triangle inequality and induction (along with some nice reasoning). In the general case we consider the limit of polygonal paths. This works because  the definition of arc length (for a convex curve), is the least upper bound of the length of inscribed polygonal paths. My comment is that (part of) the justification for that definition is that we take it as obvious that each inscribed polygonal path should be shorter (or at least no longer) than the curve. 
Archimedes in his rectification of the circumference of the circle (the first use of this limiting polygonal path method) gives a careful definition of convexity  and takes as a postulate that 

If two plane  curves  C  and  D  with the same  endpoints  are  concave  in the same  direction,  and  C  is  included  between  D  and  the  straight line joining  the  endpoints,  then  the  length  of  C  is less  than  the  length  D. 

There may be more to it than that (in the case that the curve C has endpoints interior to the segment joining the endpoints of D.)
A: That's a standard inequality, though perhaps not well enough known for the usual "take it to stackexchange" comment.  Denote by $\partial \Sigma$ the boundary of any set $\Sigma$ in the plane.  In the case that $s$ is a polygon we can use induction on the number $k$ of edges of $\partial s$ not contained in the $\partial S$.  If $k=0$ we're done.  If $k>0$, choose an edge $e$ of $s$ not in $\partial S$, and let $H$ be the closed half-plane such that $H \supset s$ and $\partial H \supset e$.  Then $S' := S \cap H$ is a convex set containing $s$ whose boundary is shorter than $\partial S$ because we've replaced part of $\partial S$ with the line segment joining the same two points.  Moreover $\partial s$ has $k-1$ edges not contained in $\partial S'$.  This completes the induction step and the proof.
That argument applies more generally when $\partial s \setminus \partial S$ is polygonal.  If it's not, we can reduce to that case via a limiting argument, replacing each component of $\partial s \setminus \partial S$ by an arbitrarily close polygonal path.
A: Your result, including your proposed generalization and the further generalization to surface area in higher dimensions, is an immediate consequence of Cauchy's surface area formula.  This states that, up to a constant depending only on the dimension, the surface area of a convex body is the average of the areas of its 1-codimensional projections.  I don't know a good reference in a web page, but see for example Klain and Rota's book.
A: For the triangle problem: Let $E$ on $CA$ and $F$ on $CB$ be such  that $CEDF$ is a parallelogram. Then the triangle law and the equality of opposite sides of a parallelogram gives $$DA<EA+ED=EA+CF\quad\text{and}\quad DB<FD+FB=CE+FB.$$By addition,$$DA+DB<(CE+EA)+(CF+FB)=CA+CB,$$as required. 
