A simple proof that a polygon circumscribing a circle overestimates its perimeter Looking at the picture below, it's easy to see why the perimeter of a polygon inscribed in a circle is an underestimation of the circle's perimeter. This follows from the triangle inequality: Any side (say $AB$) of the polygon is shorter than the circular arc with the same endpoints ($\stackrel{\frown}{AB}$). Summing all these inequalities shows the perimeter of the inscribed polygon is indeed smaller than that of the circle.
I'm wondering if there is proof that the perimeter of a circumscribed polygon always overestimates the perimeter of the circle, which is as simple as that of the inscribed polygon case. Thanks!

 A: You may use a general fact:

If $A,B\subset\mathbb{R}^2$ are two convex bounded shapes and $A\subset B$, the perimeter of $A$ is less than the perimeter of $B$.

Proof: if $A\neq B$, you may "cut out" a slice of $B$ without touching $A$. By convexity, the perimeter of the "reduced set" $B$ is less than the perimeter of the original set $B$. If $A$ is a polygon, by iterating this argument a finite number of times you get that $A$ is a reduced version of $B$, hence $\mu(\partial A)<\mu(\partial B)$ as wanted.
A: Pick a point $F$ on one of the arcs of the circle, let us say it's on arc that faces $C$ in your circumscribed quadrilateral.  Construct a line segment $GFH$ with $G$ on $BC$ and $H$ on $CD$, tangent to the circle at $F$.  $GFH$,being a straight segment, is shorter than $GC+CH$, so the circumscribed pentagon $ABGHD$ has less perimeter than the quadrilateral $ABCD$. Keep adding sides to the polygon by drawing additional tangents and the polygon perimeters will constitute a strictly monotonic decreasing sequence.  So the terms of that sequence must be greater than the limiting value which is the circumference of the circle.
A: This expands
Yves Daoust's comment.
Call the point that
the tangent from $D$
touches the circle
$P$,
and the point where
$DE$ intersects the circle
$Q$.
Then
$DEQ$ is a right triangle.
Let
$t = \angle DEP$.
Then
$\tan(t)
=\dfrac{DP}{PE}
$
so
$DP
=PE \tan(t)
$.
We also have
the length of 
arc $QP= t\,PE$.
Therefore
$\dfrac{DP}{arc\ QP}
=\dfrac{PE \tan(t)}{t\,PE}
=\dfrac{\tan(t)}{t}
\gt 1
$
for
$t > 0$.
Since the same holds on both sides,
the sum of the lengths
of the two tangents
is greater than
the length of arc
$DE$
by a factor
$\dfrac{\tan(t)}{t}
$.
A: This proof assumes the fact (sometimes used as a definition) that the circumference of a circle is the least upper bound of the perimeters of all polygons inscribed within it.
Def.: If $A$ is a polygon, $P(A)$ is its perimeter. If $C$ is a circle, $P(C)$ is its circumference.
Lemma: If $A$ and $B$ are convex polygons with $A \subset B$, then $P(A) \leq P(B)$. (See http://www.cut-the-knot.org/m/Geometry/PerimetersOfTwoConvexPolygons.shtml for a proof.)
Now suppose that $B$ is a polygon circumscribed about circle $C$. Let $A$ be a polygon inscribed in $C$. By the lemma, $P(A) \leq P(B)$. Taking the sup over all such polygons $A$ and using our 'fact', we have that $P(C) \leq P(B)$. Now form a second circumscribed polygon $D$ within and smaller than $B$ (one additional side is enough). Then $P(C) \leq P(D) < P(B)$, showing that $P(B)$  does indeed overestimate $P(C)$.
