If $G$ is a graph with diameter $\ge 5$ then $\Delta(G)+\delta(G)\le \left|V(G)\right|-2$

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Problem: Let $G$ be a graph with $k$ vertices and diameter $\ge 5$. Why then must the sum of its smallest and greatest degree be smaller or equal to $k-2$?

I have tried arguing by induction (assuming a minimal counterexample and then removing edges or vertices to use the induction hypothesis) but this did not lead anywhere. Any hints?

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Let $a$ be a vertex of degree $\Delta(G)$. Some vertex $b$ has distance $3$ from $a$ as otherwise the diameter is $\le 4$. Then the at least $\delta(G)$ neighbours of $b$, the $\Delta(G)$ neighbourts of $a$, and $a,b$ themselves are all pairwise distinct, hence $\delta(G)+\Delta(G)+2\le |V|$.