jmc

Let $r$ be the radius of the inscribed circle. Let $s$ be the semiperimeter of the triangle, that is, $s=\frac{AB+AC+BC}{2}=9$. Let $K$ denote the area of $△ABC$.

Heron's formula tells us that $$\begin{array}{rl}K& =\sqrt{s(s-AB)(s-AC)(s-BC)}\\ & =\sqrt{9\cdot 4\cdot 3\cdot 2}\\ & =\sqrt{3^3\cdot 2^3}\\ & =6\sqrt{6}.\end{array}$$The area of a triangle is equal to its semiperimeter multiplied by the radius of its inscribed circle ($K=rs$), so we have $$6\sqrt{6}=r\cdot 9,$$which yields the radius $r=\boxed{\frac{2\sqrt{6}}{3}}$.