Prove that mean time spent in an M/M/1 system having arrival rate $\lambda$ and service rate $2μ$ is less than the mean time spent in an M/M/2 system with arrival rate $\lambda$ and each service rate $μ$.
When I solve this question, I get $E(N_1)=\frac{1}{2-p}$ and $E(N_2)=\frac{4p}{4-p^2}$, where $p=\frac{\lambda}{μ}$ (I can just compare $E(N_i)$ since it'll be directly proportional to time spent).
I now get $E(N_2)-E(N_1)=\frac{3p-2}{4-p^2}$. This should be more than $0$ for the given statement to be true.
But what now? Like where did I miss that p should be more than $\frac{2}{3}$?
Any help would be appreciated. Thanks!
$E(N_1)={1\over \mu(2-p)},\ \ E(N_2)={4\over \mu(4-p^2)}.$
We have
$E(N_2)-E(N_1)={1\over \mu(2+p)}.$
This is more than zero for any p (but everywhere above $p<2$).