Is there any geometrically simple reason why the inradius of a triangle should be at most half its circumradius? I end up wanting the fact for this answer.
I know of two proofs of this fact.
Proof 1:
The radius of the nine-point circle is half the circumradius. Feuerbach's theorem states that the incircle is internally tangent to the nine-point circle, and hence has a smaller radius.
Proof 2:
The Steiner inellipse is the inconic with the largest area. The Steiner circumellipse is the circumconic with the smallest area, and has 4 times the area of the Steiner inellipse. Hence the circumcircle has at least 4 times the area of the incircle.
These both feel kind of sledgehammerish to me; I'd be happier if there were some nice Euclidean-geometry proof (or a way to convince myself that no such thing is likely to exist, so the sledgehammer is necessary).
EDIT for ease of future searching: The internet tells me this is often known as "Euler's triangle inequality."
Best Answer
So Proof #1 can be modified to be completely elementary.
First, it is easy to show that the incircle is the smallest circle touching all 3 sides. The circle passing through the midpoints (the nine-point circle) obviously has circumradius half that of the larger circle. No need to invoke Feuerbach's theorem for this.
Cheers,
Rofler