 Solution

Define the downstream queue workload by starting both queues empty at time $-N$ and taking the increasing remote-past limit. For each finite $N$, part (c) makes their total queue workload exactly the single-server workload with service $d$, while the upstream queue workload is the single-server workload with service $c$. ThereforeThe downstream limit is increasing: the coupled upstream and downstream recursions are increasing in their initial state, because $D_t=\min (c,Q_{t-1}+a_t)$ is increasing in $Q_{t-1}$. With finite upstream queue workload, taking the limit givesThe result is nonnegative because reducing service cannot reduce queue workload. It can be $+\infty$ when the slower server is unstable; under $\lambda <d<c$, both quantities are finite. The finite-upstream condition matters: subtracting two infinite queue workloads would not define a downstream queue.