Minimum Value in a traingle

Question

In any triangle, what is minimum possible value of $\frac{r_1 r_2 r_3}{r^3}$?

I reduced its value to $ (s^4)/(Area^2) $, But I don't know how to proceed now?

Where $ r_1 r_2 r_3 $ are exradii, r is the inradiusvand s is the semi-perimeter of the triangle.

Answer 1

$A^2=s(s-a)(s-b)(s-c)$ By AM-GM inequality, $\dfrac{(s-a)+(s-b)+(s-c)}{3}\geqslant((s-a)(s-b)(s-c))^{\dfrac{1}{3}}\\ \dfrac{s}{3}\geqslant\left ( \dfrac{A^2}{s} \right )^{\dfrac{1}{3}}\\ \dfrac{s^3}{27}\geqslant\dfrac{A^2}{s}\\ \dfrac{s^4}{A^2}\geqslant{27}$

Answer 2

HINT: use that $$r_1=\sqrt{\frac{s(s-b)(s-c)}{s-a}}$$ etc and $$A=\sqrt{s(s-a)(s-b)(s-c)}$$ with $s=\frac{a+b+c}{2}$

Answer 3

I reduced its value to $s^4/Area^2$, but I don't know how to proceed now.

Hint: For a given perimeter, what is the triangle with the greatest area ? $($Obviously, if the perimeter is fixed, you need to maximize the area, for the the ratio in question to be minimal$)$.