For a #DeltaABC#, exradii #r_1=Delta/(s-a)#, #r_2=Delta/(s-b)#, #r_3=Delta/(s-c)# and inradius #r=Delta/s#, where #s# is semiperimeter of the triangle.
#r_+r_2+r_3-r#
= #Delta/(s-a)+Delta/(s-b)+Delta/(s-c)-Delta/s#
= #Delta(s(s-b)(s-c)+s(s-a)(s-c)+s(s-a)(s-b)-(s-a)(s-b)(s-c))/Delta^2#
= #(s(s-c)(s-b+s-a)+(s-a)(s-b)(s-s+c))/Delta#
= #(cs(s-c)+c(s-a)(s-b))/Delta#
= #(c(s^2-sc+s^2-as-bs+ab))/Delta#
= #(c(2s^2-s*2s+ab))/Delta=(abc)/Delta=4R#
Hence #(r_1+r_2+r_3)/r=(4R)/r+1# (A)
There is another interesting result, known as Euler's Theorem, we can use here i.e.
#R^2-d^2=2Rr#, where #d# is the distance between circumceter and incenter. This gives us
#R^2/(2Rr)=1+d^2/(2Rr)# or #R/(2r)=1+d^2/(2Rr)#
i.e. #R/r=2+d^2/(Rr)#
Hence least value of #R/r# is #2#, when #d=0# and using it in (A)
Least value of #(r_1+r_2+r_3)/r# is #9#
Additional information - This arises when the triangle is equilateral and then #R=2r# and #r_1=r_2=r_3=3r#.
