Inequality in triangle $ABC$ $a=BC$ $b=AC$ $c=AB$

Question

Prove that :

$r_{a}^{4}r_{b}^{4}r_{c}^{4}≥9r^{3}p^{9}$

Where : $p=$ semiperimeter

Actually I don't know if above inequality true or no but my attempt as following :

We known : $r_{a}=\frac{S}{p-a}$, $r_{b}=\frac{S}{p-b}$, $r_{c}=\frac{S}{p-c}$ where $S=$ area

Now by multiple we obtaine :

$r_{a}^{4}r_{b}^{4}r_{c}^{4}=\frac{S^{12}}{((p-a)(p-b)(p-c))^{4}}$

But I don't know to I complete this work ?

Answer 1

We need to prove that $$\frac{S^{12}}{(p-a)^4(p-b)^4(p-c)^4}\geq\frac{9S^3p^9}{p^3}$$ or $$S^9\geq9p^6(p-a)^4(p-b)^4(p-c)^4$$ or $$S\geq9 p^2,$$ which is obviously wrong.