Why is the inradius of any triangle at most half its circumradius? 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."
 A: I came up with the following (almost) purely algebraic proof for the simple reason that I'm not much of a geometry guy and it's too much mental work for me to keep track of the Nine-Point Circle.
We use the following formulae for the area of a triangle:
$A = \sqrt{s(s-a)(s-b)(s-c)} = rs = \dfrac{abc}{4R}$,
where $A$ is the area, $r$ is the inradius, $R$ is the circumradius and $s$ is the semiperimeter.
Hence,
$R \geq 2r \iff \dfrac{abc}{4A} \geq \dfrac{2A}{s} \iff %
\dfrac{abc}{8} \geq \dfrac{A^2}{s} \iff \dfrac{abc}{8} \geq (s-a)(s-b)(s-c)$.
To simplify the last inequality, use Ravi Substitution. i.e. Let $a = y + z$; $b = z + x$; $c = x + y$. (Note that this can always be done for any arbitrary triangle $ABC$. Also, $x,y,z$ are strictly positive. More information can be found if you google Ravi Substitution). It is obvious that $s = x + y + z$. Thus, last inequality is equivalent to 
$\dfrac{(x+y)(y+z)(z+x)}{8} \geq xyz$.
This can be proved if we multiply the following three inequalities:
$\dfrac{x+y}{2} \geq \sqrt{xy}; \quad \dfrac{y+z}{2} \geq \sqrt{yz}; \quad \dfrac{z+x}{2} \geq \sqrt{zx}$.
(Note that $x,y,z$ are positive and $AM \geq GM$ can be applied safely).
A: 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.
A: 
It's enough to show that $OI^2=R(R-2r) \iff R^2-OI^2=2Rr$, where $O$,$I$,$R$, and $r$ is the circumcenter, incenter, circumradius, and inradius respectively. By power of a point, $R^2-OI^2=AI\times IL$, so it's enough to show that $AI\times IL =2Rr \iff \frac{AI}{r}=\frac{2R}{IL}$. This is trivial since $\triangle AFI \sim \triangle KBL$ and $LI=BL$ (By Incenter-Excenter lemma)
A: 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
A: 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
$$
\begin{align}
2y\cdot z&=2R^2-a^2\tag{6a}\\
2z\cdot x&=2R^2-b^2\tag{6b}\\
2x\cdot y&=2R^2-c^2\tag{6c}
\end{align}
$$
Explanation:
$\text{(6a)}$: $a^2=(y-z)\cdot(y-z)=|y|^2+|z|^2-2y\cdot z$, then apply $(5)$
$\text{(6b)}$: $b^2=(z-x)\cdot(z-x)=|z|^2+|x|^2-2z\cdot x$, then apply $(5)$
$\text{(6c)}$: $c^2=(x-y)\cdot(x-y)=|x|^2+|y|^2-2x\cdot y$, then apply $(5)$
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
$$
\begin{align}
d^2
&=\frac{a^2R^2+b^2R^2+c^2R^2+2abx\cdot y+2bcy\cdot z+2caz\cdot x}{(a+b+c)^2}\tag{8a}\\
&=\frac{a^2R^2+b^2R^2+c^2R^2+ab(2R^2-c^2)+bc(2R^2-a^2)+ca(2R^2-b^2)}{(a+b+c)^2}\tag{8b}\\
&=\frac{(a+b+c)^2R^2-(a+b+c)abc}{(a+b+c)^2}\tag{8c}\\[3pt]
&=R^2-\frac{abc}{a+b+c}\tag{8d}\\[6pt]
&=R^2-2Rr\tag{8e}\\[12pt]
&=R(R-2r)\tag{8f}
\end{align}
$$
Explanation:
$\text{(8a)}$: take the dot product of $(7)$ with itself and apply $(5)$
$\text{(8b)}$: apply $(6)$ to the dot products
$\text{(8c)}$: collect terms
$\text{(8d)}$: simplify
$\text{(8e)}$: apply $(3)$ and $(4)$
$\text{(8f)}$: factor
Since $d^2\ge0$, and $R>0$, we immediately get that
$$
r\le\tfrac12R\tag9
$$
