Product of inscribed circle and circumscribed circle radiuses

Question

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.

Answer 1

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?

Answer 2

As $\triangle=rs=\dfrac{abc}{4R}$

We need $$\dfrac{4R\triangle}c>4R\cdot\dfrac{\triangle}s\iff s>c\iff a+b+c>2c\iff a+b>c$$ which is true for all triangle