The problem in my book is as follow.
In a $\Delta ABC$ , if $r=r_2+r_3-r_1$ and $\angle A >\dfrac{\pi}{3}$ , then the range of $\dfrac{s}{a}$ is equal to: (Here $r_i $ are exradii)
I used the fact that $r_1+r_2+r_3=4R+r$ , to arrive at $r_1=2R$
After I used the fact that area by exradius and inradius are as $\Delta=r_1(s-a)=rs$
After simplifying one can arrive at $\dfrac{s}{a}=\dfrac{2R}{2R-r}$
Now Maximum value is easy to find as , I assume which seems true for time being.
Distance between incentre and circumcentre is given by $d=[R(R-2r)]^{1/2}$
I can find from this that $r<\dfrac{R}{2}$ surely.
Which gives $\dfrac{s}{a}<\dfrac{4}{3}$
But , I am not having a good idea for lower bound of expression $\dfrac{s}{a}$ , for which i need to bound inradius.
As $r_1+r_2+r_3=4R+r$. $$\implies 4R+r-2r_1=r\implies r_1=2R$$
Also $r=\dfrac{\Delta}s;r_1=\dfrac{\Delta}{s-a}$. So $$\Delta =rs=r_1(s-a)\implies(r_1-r)s=r_1a\implies \frac sa=\frac{r_1}{r_1-r}=\frac{2R}{2R-r}$$
Now: $$r\le R/2\implies \frac sa\le\frac43$$ Now: $$r=4R\sin\frac A2\sin\frac B2\sin\frac C2>2R\sin\frac B2\sin\frac C2\tag{$\because \angle A >\dfrac{\pi}{3}$}$$
Now what if $A\to(\pi/3)^+$ and $B\to(2\pi/3)^-$ and $C\to0^+$, Now $r\to0$. You won't find a smaller value, because if it would exist, it needs to be negative and negative sines are not possible with half angles of a standard triangle. So: $$\frac sa>1$$