Smallest relative velocity driving a two-stream instability The physical picture is a two-stream system of cold electrons and ions (i.e. $T_i=T_e=0$) with realtive velocities $V_i$ and $V_e$. The dispersion relation obtained is
$$D(\omega)=1-\frac{\omega_{pi}^2}{(\omega-kV_i)^2}-\frac{\omega_{pe}^2}{(\omega-kV_e)^2}=0$$
where $\omega_{pi}$, $\omega_{pe}$ respectevely correspond to electron and ion plasma frequency.
I am asked to give the minimum relative velocity $V_e-V_i$ that may drive an instability. What I know is that there can be an instability if $D(\omega)<0$, because such situation give two real and two imaginary roots, the second case give exponential solutions if $\text{Im}(\omega)>0$.
 A: Background
The first thing to do is use a different reference frame to simplify things by going into the ion rest frame.  Thus, the dispersion relation goes to:
$$
D\left( \omega \right) = 1 - \left( \frac{ \omega_{pi} }{ \omega } \right)^{2} - \left( \frac{ \omega_{pe} }{ \left( \omega - \mathbf{k} \cdot \mathbf{V}_{o} \right) } \right)^{2} = 0 \tag{1}
$$
where $\mathbf{V}_{o} = \mathbf{V}_{e} - \mathbf{V}_{i}$.  We can further simplify things by rewriting $D\left( \omega \right) = 1 - F\left( \omega \right)$, where this new term is given by:
$$
F\left( \omega \right) = \left( \frac{ \omega_{pi} }{ \omega } \right)^{2} + \left( \frac{ \omega_{pe} }{ \left( \omega - \mathbf{k} \cdot \mathbf{V}_{o} \right) } \right)^{2} = 1 \tag{2}
$$
We can see that $F\left( \omega \right)$ has two poles at $\omega = 0$ and $\omega = \mathbf{k} \cdot \mathbf{V}_{o}$ and a minimum at $\partial F/\partial \omega = 0$, given by:
$$
\begin{align}
  \frac{ \partial F }{ \partial \omega } & = - \frac{ 2 }{ \omega } \left( \frac{ \omega_{pi} }{ \omega } \right)^{2} - - \frac{ 2 }{ \left( \omega - \mathbf{k} \cdot \mathbf{V}_{o} \right) } \left( \frac{ \omega_{pe} }{ \left( \omega - \mathbf{k} \cdot \mathbf{V}_{o} \right) } \right)^{2} = 0 \tag{3a} \\
  & = \frac{ \omega_{pi}^{2} }{ \omega^{3} } + \frac{ \omega_{pe}^{2} }{ \left( \omega - \mathbf{k} \cdot \mathbf{V}_{o} \right)^{3} } = 0 \tag{3b} \\
  \omega_{pe}^{2} \ \omega^{3} & = - \omega_{pi}^{2} \ \left( \omega - \mathbf{k} \cdot \mathbf{V}_{o} \right)^{3} \tag{3c}
\end{align}
$$
After we make a few substitutions (i.e., $\zeta = \tfrac{\omega}{k \ V_{o}}$ and $\alpha = \tfrac{\omega_{pi}^{2}}{\omega_{pe}^{2}}$) and assume everything is one-dimensional (i.e., $\mathbf{k}$ is parallel to $\mathbf{V}_{o}$), Equation 3c can be reduced to:
$$
\zeta^{3} + \alpha \ \left( \zeta - 1 \right)^{3} = 0 \tag{4}
$$
The three roots of Equation 4 are messy but when we note that if this is a proton-electron plasma, then $\alpha = \tfrac{m_{e}}{m_{p}}$, where $m_{s}$ is the mass of species $s$.  Thus, we can expand in a Taylor series for small $\alpha$ to find more simplified results.  We can also take advantage of the fact that we are looking for the real part of the frequency, so we can rearrange Equation 4 to get:
$$
\zeta^{3} = \alpha \left( 1 - \zeta \right)^{3} \tag{5}
$$
If we take the cubic root of Equation 5, we can solve for $\zeta$ to find:
$$
\begin{align}
  \zeta & = \frac{ \alpha^{1/3} }{ 1 + \alpha^{1/3} } \tag{6a} \\
  & = \frac{ 1 }{ 1 + \alpha^{-1/3} } \tag{6b} \\
  \omega_{sol} & = \frac{ k \ V_{o} }{ 1 + \left( \frac{ m_{p} }{ m_{e} } \right)^{1/3} } \tag{6c}
\end{align}
$$
where we replaced our normalized parameters with the original inputs.
Threshold Drift Velocity
To find the threshold for instability, we use the results in Equation 6c and impose an additional constraint that $F\left( \omega_{sol} \right) > 1$, which gives us:
$$
\begin{align}
  F\left( \omega_{sol} \right) & = \left( \frac{ \omega_{pe} }{ k \ V_{o} } \right)^{2} \left[ 1 + \left( \frac{ m_{e} }{ m_{p} } \right)^{1/3} \right]^{3} > 1 \tag{7a} \\
  \left( k \ V_{o} \right)^{2} & < \omega_{pe}^{2} \ \left[ 1 + \left( \frac{ m_{e} }{ m_{p} } \right)^{1/3} \right]^{3} \tag{7b}
\end{align}
$$
