Prove: $\frac{r_a}{bc} + \frac{r_b}{ca} + \frac{r_c}{ab} = \frac{1}{r} - \frac{1}{2R}$, for circumradius R, inradius $r$, and exradii $r_x$ 
In $\triangle ABC$, prove: 
  $$\frac{r_a}{bc} + \frac{r_b}{ca} + \frac{r_c}{ab} = \frac{1}{r} - \frac{1}{2R}$$ 
  for circumradius $R$, inradius $r$, and exradii $r_a$, $r_b$, $r_c$ in the standard arrangement.

It is known that $r_a = \sqrt{\dfrac{s\left(s-b\right)\left(s-c\right)}{s-a}}$, where $s = \dfrac12\left(a+b+c\right)$ is the semiperimeter of $\triangle ABC$. Similar formulas exist for $r_b$, $r_c$ and $r$. But how does $R$ connect with all of this?

We will use the $\sum_{cyc}$ notation for cyclic sums:
$$
\sum_{cyc}f(a,b,c) = f(a,b,c)+f(b,c,a)+f(c,a,b)
$$
for any function $f$. Also, let $\Delta$ be the area of triangle $ABC$.
We have:
$$ \frac{r_a}{bc}=\frac{\Delta}{(s-a)bc}=\frac{\frac{1}{2}\sin(A)}{s-a}=\frac{a}{4R(s-a)}\tag{1}$$
hence we have to prove that:
$$ 2+\sum_{cyc} \frac{a}{s-a} = \frac{4R}{r} \tag{2}$$
that is equivalent to:
$$ 2r+\sum_{cyc}\frac{a}{\cot\left(\frac{A}{2}\right)} = 2R\tag{3}$$
or to:
$$ 2r+\sum_{cyc}2R(1-\cos(A)) = 2R\tag{4}$$
that follows from Carnot's theorem.
