The relative velocity is $$v_{rel}=\sqrt2v_{mean}$$ where $v_{mean}=(\frac{8KT}{\pi m})^{\frac12}$$$v_{rel}=(\frac{16KT}{\pi \frac m2})^{\frac12}$$ because the reduced mass of two of the same particles is half the mass of one. In general: $$v_{rel}=(\frac{8KT}{\pi \mu})^{\frac12}$$
However, my question is: what is the need for reduced mass? If $v_{rel}=\sqrt2v_{mean}$ where $v_{mean}=(\frac{8KT}{\pi m})^{\frac12}$. Wouldn't that just leave you with: $v_{mean}=(\frac{16KT}{\pi m})^{\frac12}$ Where does reduced mass come into it?