The Relation Between Average Speed and Inverse Average Speed How is the average speed $\langle v \rangle$ of gas molecules related to their average inverse speed $\langle \frac{1}{v} \rangle$?
I know that the average speed is defined as:
$$ \langle v \rangle = \sqrt{ \frac{8 k_B T}{\pi m}}$$
 A: Averages of this kind can always be written as integrals over the Maxwell-Boltzmann distribution, so this is more of a mathematical question than a physics one. But still, it may give some insight to see how these two averages can be related.
For a 3D system we integrate over three components of the velocity, but if we are only interested in functions of the speed $v$, this can be converted into a single integral:
$$
\langle f(v) \rangle = C \int_0^\infty 4\pi v^2 \, f(v) \, \exp(-\alpha v^2) \, \text{d}v ,
$$
where $C$ is a normalization constant, $\alpha = \frac{1}{2}m/k_BT$, and $f(v)$ is any function of $v$. Let's relate the two quantities that you are interested in:
$$
\frac{\langle v\rangle}{\langle v^{-1}\rangle} = 
\frac{\int_0^\infty v^3 \, \exp(-\alpha v^2) \text{d}v}{\int_0^\infty v \, \exp(-\alpha v^2) \text{d}v}
= \frac{1}{\alpha}
$$ 
You should be able to derive this last result, for example by doing an integration by parts. Alternatively, you can start with the known integral
$$
\int_0^\infty v \, \exp(-\alpha v^2) \text{d}v = \frac{1}{2\alpha}
$$
and differentiate both sides with respect to $\alpha$.
