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?
On
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?
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.