prove that for any triangle ,R>= 2r,where R and r are the circumradius and inradius of the triangle respectively.
post in different proofs
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
please show the standard way,it will be very helpfulNot standard way...