Show that if a convex polygon is entirely inside another convex polygon, then the outer polygon has perimeter greater than or equal to the inner one. Can anyone help me out with proving this statment?
"Show that if a convex polygon is entirely inside another convex polygon, then the outer polygon has perimeter greater than or equal to the inner one."
 A: 
Lemma: If you slice through a convex polygon with a straight line, you get two convex polygons, both of which have perimeter less than or equal to the original polygon's perimeter.

Proof: Consider one of the smaller polygons. Its boundary consists of a piece of the original polygon's boundary and a straight line, which is not longer (by triangle inequality) than the other piece of the original polygon's boundary.

Problem: If a convex polygon is entirely inside another convex polygon, then the outer polygon has perimeter greater than or equal to the inner one.

Proof: Choose as a straight line the extension of one of the inner polygon's sides. Use it to cut away part of the outer polygon that does not contain the inner polygon. Cutting the outer polygon in this way makes its perimeter smaller (by the lemma), yet it still contains the inner polygon.
Now repeat this for all other sides of the inner polygon. Each time, the perimeter of the remaining outer polygon decreases, yet it still contains the inner polygon.
In fact, after you're done with all sides, the remainder of the outer polygon is exactly the inner polygon, and by construction it has a perimeter not greater than the original outer polygon's perimeter. $~~\square$
