For four vertices $x_1,x_2,x_3,x_4$, write $d_{ij}=\operatorname{dist}(x_i,x_j)$ and define

$$S_1=d_{12}+d_{34},\quad S_2=d_{13}+d_{24},\quad S_3=d_{14}+d_{23}.$$

Lemma. In a connected graph with $N$ vertices, if one of these three sums is strictly greater than both others, it is at most $N$. The four vertices may repeat.

We prove the lemma by induction on $N$. The one-vertex case is immediate.

First suppose the graph has no cut vertex, where a cut vertex is a vertex whose removal disconnects the graph. Take two vertices at distance $\ell>0$, and group the vertices by their distance from the first.

Each layer $1,\ldots,\ell-1$ contains at least two vertices. If a layer contained just one vertex, every path between the chosen endpoints would pass through it, making it a cut vertex. Counting the endpoints and these layers gives

$$N\ge1+2(\ell-1)+1=2\ell.$$

Thus every distance is at most $N/2$, and every $S_i$ is at most $N$. This proves the lemma in this case.