True or False: The circumradius of a triangle is twice its inradius if and only if the triangle is equilateral. Let $R$ be the circumradius and $r$ be the inradius. The if part is clear to me. For an equilateral triangle, the circumcentre, the incentre and the centroid are the same point. So, by property of cebntroid $AG:GD=2:1\Rightarrow AG=2GD$. Thus $R=2r$.

But is the converse true? Whether $R=2r$ implies that the triangle must be equilateral ? We know some relations involving circumradius and inradius, like $R=\dfrac{abc}{4\Delta}, r=\dfrac{\Delta}{s}$, where $\Delta$ is the area of the triangle and $s$ is its semi-perimeter i.e. $s=\dfrac{a+b+c}{2}$. But then how to show that the triangle is equilateral if $R=2r$.
I would be thankful if anyone can help me.

Let $a,b,c$ are the sides of a triangle, $A=$ area of the triangle, $s=$ semi-perimeter.
$R=\dfrac{abc}{4A}, r=\dfrac{A}{s}$
We have to show $R\geq 2r$.
The relation $\dfrac{abc}{4A}\geq \dfrac{2A}{s}$ holds
if $abc\geq \dfrac{8A^2}{s}$
if $abc\geq 8(s-a)(s-b)(s-c)$
if $abc\geq (b+c-a)(c+a-b)(a+b-c)$
This is true for all triangles.
When $a=b=c$, the equality holds.This is the case of an equilateral triangle.
So, $R=2r$.
