Let's use $u$, $v$, $w$ - method. You can get more information about this method in numerous articles, e.g. ($x = p - a$, $y = p - b$, $z = p - c$):
$$R = \frac{abc}{4S} = \frac{(x+y)(y+z)(z+x)}{4\sqrt{xyz}(x+y+z)} \quad \text{and} \quad r = \frac{S}{p} = \sqrt{\frac{xyz}{x+y+z}}$$

Then we have to prove equivalent inequality:
$$(x+y)(y+z)(z+x) + 2(3\sqrt{3}-4)xyz \geq 2(x+y+z)\sqrt{xyz}(x+y+z).$$
Let's denote:
$$\begin{cases} x + y + z = 3u, \\ xy + yz + zx = 3v^2, \quad \text{Then } 9uv^2 + (6\sqrt{3}-9)w^3 \ge 6\sqrt{3}\sqrt{uw^3}u \\ xyz = w^3. \end{cases}$$

$$(3uv^2 + (2\sqrt{3}-3)w^3)^2 \ge 12u^3w^3 \Leftrightarrow \\ f(w^3) = (2\sqrt{3}-3)^2w^6 + w^3(6(2\sqrt{3}-3)uv^2 - 12u^3) + 9u^2v^4 \ge 0.$$

If $u$, $v^2$ are fixed then $f$ is convex. That's why $f$ has a minimum value when 2 variables $\{x, y, z\}$ coincide. Without loss of generality $y = z$. Let's check our inequality. It's equivalent to the following:
$$(x + y)((2x + y) - xy) + 2(3\sqrt{3} - 4)xy \geq 2\sqrt{x(x + 2y)^3}.$$
Let's denote: $t = \frac{x}{y} \Rightarrow$
$$(t+2)(t+1) + 2(3\sqrt{3}-4)t \ge 2\sqrt{t(t+2)^3} \Leftrightarrow t^2 + 3(3\sqrt{3}-2)t + 1 \ge \sqrt{t(t+2)^3} \Leftrightarrow \\ t^3(6\sqrt{3}-10) + t^2(21-12\sqrt{3}) + t(6\sqrt{3}-12) + 1 \ge 0 \Leftrightarrow (t-1)^2((6\sqrt{3}-10)t+1) \ge 0.$$
What was to be demonstrated. The equation is true for $x = y = z \Leftrightarrow a = b = c$.