Why is the inradius of any triangle at most half its circumradius?

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

Compute the area of a triangle (first method):

Consider the following diagram:

$\hspace{4.5cm}$

The area of the green triangle is $$ A=\tfrac12ab\sin(\theta)\tag{1} $$ By the Inscribed Angle Theorem, the angle that $c$ subtends at the origin, $o$, is $2\theta$. Therefore, we get that $$ c=2R\sin(\theta)\tag{2} $$ Combining $(1)$ and $(2)$ yields $$ 4AR=abc\tag{3} $$ Compute the area of a triangle (second method):

Consider the following diagram:

$\hspace{4.5cm}$

Note that the areas of the the red ($\color{#C00000}{\triangle iyz}$), green ($\color{#00A000}{\triangle izx}$), and blue ($\color{#0000FF}{\triangle ixy}$) triangles are $\frac12r$ times $a$, $b$, and $c$, respectively. Therefore, we get $$ 2A=r(a+b+c)\tag{4} $$

Compute $d$:

Translate the circumcenter, $o$, of $\triangle xyz$ to the origin. Then $$ |x|=|y|=|z|=R\tag{5} $$ Furthermore, using $a=|y-z|$, $b=|z-x|$, and $c=|x-y|$, we get $$ \begin{align} 2y\cdot z&=2R^2-a^2\\ 2z\cdot x&=2R^2-b^2\\ 2x\cdot y&=2R^2-c^2 \end{align}\tag{6} $$ As mentioned above, the areas of the the red ($\color{#C00000}{\triangle iyz}$), green ($\color{#00A000}{\triangle izx}$), and blue ($\color{#0000FF}{\triangle ixy}$) triangles are proportional to $a$, $b$, and $c$, respectively. Thus, the barycentric coordinates of the incenter, $i$, are the mean of the vertices weighted by the lengths of the opposite sides: $$ i=\frac{ax+by+cz}{a+b+c}\tag{7} $$ and therefore, using $(3)$-$(7)$ yields that $d$, the distance between the incenter and circumcenter, satisfies $$ \begin{align} d^2 &=\frac{a^2R^2+b^2R^2+c^2R^2+2abx\cdot y+2bcy\cdot z+2caz\cdot x}{(a+b+c)^2}\\ &=\frac{a^2R^2+b^2R^2+c^2R^2+ab(2R^2-c^2)+bc(2R^2-a^2)+ca(2R^2-b^2)}{(a+b+c)^2}\\ &=\frac{(a+b+c)^2R^2-(a+b+c)abc}{(a+b+c)^2}\\ &=R^2-\frac{abc}{a+b+c}\\[4pt] &=R^2-2Rr\\[6pt] &=R(R-2r)\tag{8} \end{align} $$ Since $d^2\ge0$, and $R>0$, we immediately get that $$ r\le\tfrac12R $$

Expanding on Rofler's answer, just so there'll be a complete argument here:

Consider the circle which meets the midpoints of the three sides (this is secretly the 9-point circle, but we don't need to know that to finish this argument). It circumscribes a triangle which is similar to the reference triangle and scaled by $\frac{1}{2}$. Thus its radius is half the circumradius.

To show that the radius of the nine-point circle is larger than the inradius of the reference triangle, first translate it in your favorite direction until the first time it becomes tangent to a triangle side. Then slide it along that side of the triangle until the first time it becomes tangent to another triangle side. The result will be a circle which is tangent to two sides of the reference triangle and intersects the third, and any such circle must clearly be at least as large as the incircle.