<s>[INST] 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?
 [/INST] Using Sine Rule
and
$r_a=4R\sin\dfrac A2\cos\dfrac B2\cos\dfrac C2,$
$r=4R\sin\dfrac A2\sin\dfrac B2\sin\dfrac C2,$ 
$11(v)$,$12(iii)$ respectively of Properties of triangle ,
$\dfrac{r_a}{bc}=\cdots=\dfrac{\sin^2\dfrac A2}r$
Now $\cos A=1-2\sin^2\dfrac A2\iff2\sin^2\dfrac A2=1-\cos A$
Finally use $\cos A+\cos B+\cos C=1+\dfrac rR,11(vi)$ of Properties of triangle
Can you take it from here?
</s>