Can somebody think of a proof (if it is geometric, better) that the incircle of a triangle is the smallest circle which passes through the three sides of a triangle?
By proving this, one can demonstrate that the circumradius of a triangle is at least twice the inradius without calculating the distance between the circumcenter and incenter. (This is because the circumradius of the median triangle is half the circumradius of the original triangle and its circumcircle is a circle which passes through the three sides of the triangle.)
Best Answer
Consider areas.
Let $A,B,C$ be the vertices of the triangle with opposite sides measuring $a,b,c$. Pick any point $P$ within the triangle and render its distance to the $a$ side as $\alpha$ distance to the $b$ side as $\beta$ and distance to the $c$ side as $\gamma$. Draw the line segments connecting $P$ to the vertices thus dividing the area of the triangle into three smaller areas.
A circle centered on $P$ and touching all sides of the triangle must have a radius $r\ge\max(\alpha,\beta,\gamma)$. Meanwhile the area of $\triangle ABC$, being the sum of the smaller areas, is given by
$\color{blue}{S=(1/2)(a\alpha+b\beta+c\gamma)\le(1/2)(a+b+c)\max(\alpha,\beta,\gamma)\le(1/2)(a+b+c)r}$
Compare this with the specific result if $P$ is chosen as the incenter $I$, whereupon the distance to all three sides is the incircle radius $\rho$ and then the sum of the small triangle areas is
$\color{brown}{S=(1/2)(a\rho+b\rho+c\rho)=(1/2)(a+b+c)\rho}$
Comparing the brown equality with the blue inequality gives $r\ge\rho$, qed.