Squaring the traffic intensity lowers it, so the second queue is shorter

Consider the two M/M/1 queuing systems $Q_1$ and $Q_2$, where $Q_1$: arrival rate $\lambda$, service rate $\mu$; $Q_2$: arrival rate $\lambda^2$, service rate $\mu^2$. It is known that $\lambda<\mu$. Let $L_s^{(i)}$ be the number of customers in the system in the equilibrium state, $i=1,2$. Then: