Product of inscribed circle and circumscribed circle radiuses Let a and b be the two shortest sides of a triangle.
r is the radius of the inscribed circle.
R is the radius of the circumscribed circle.
Prove that ab > 4Rr
Thanks for your help.

Let the third side be $c$. Let $2s=a+b+c$. Then according to the triangle inequality, we have $a+b>c\implies a+b+c>2c\implies2s>2c\implies s>c$
Also, we know that $r=\dfrac{\Delta}{s}$ and $\Delta=\dfrac{abc}{4R}$. Can you take it from here?
