Not standard way...
Let ABC be some triangle,with AB=a, AC=b, BC=c,and with area S.
Let, p=(a+b+c)/3, hence, r=S/p.
We, also know that: R=abc/4S
I'ii leave like that, maybe someone would like to proceed somehow(if possible)...
THANKS!
oh yes!i wish to complete your proof:now we just have to show that abc/4s >= 2s/p ,which reduces to abc>= 8(p-a)(p-b)(p-c). now we know that a^2-(b-c)^2<a^2.so similarly we could writeLet ABC be some triangle,with AB=a, AC=b, BC=c,and with area S.
Let, p=(a+b+c)/2, hence, r=S/p.
We, also know that: R=abc/4S
abc>=sqrt( a^2-(b-c)^2)sqrt(b^2-(c-a)^2)sqrt(c^2-(a-b)^2)
=(a+b-c)(b+c-a)(c+a-b)=8(p-a)(p-b)(p-c) Q.E.D
please show the standard way,it will be very helpfulNot standard way...