Suppose G is an $n$-vertex graph containing no triangle as a subgraph. What is the best upperbound you can find for $d(u) + d(v)$, where u and v are two different vertices of $G$. (recall: $d(u)$ means the degree of vertex $u$) .
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If $u,v$ are neighbours, then each of the other $n-2$ vertices can be connected to at most one of $u,v$. This makes $(d(u)-1)+(d(v)-1)\le n2$, i.e., $d(u)+d(v)\le n$.
If $u,v,$ are not neighbours, then nothing really prevents $u$ and $v$ from both being connected to all other $n-2$ vertices. In fact, this happens in the complete bipartite graph $K_{2,n-2}$. Here, $d(u)+d(v)=2n-4$.
We conclude that the best bound when we do not know/specify if $u,v$ are neighbours or not, is $$ d(u)+d(v)\le\max\{n,2n-4\}.$$
Say vertices $u,v$ are connected. Let $N(v)$ be set of neighbour for $v$ and $N(u)$ for $u$.
If $|N(u)\cap N(v)|>0$ then there exist $w$ which is connected to both. A contradiction.
So $|N(u)\cap N(v)|=0$ and thus $$d(u)+d(v) = |N(u)|+|N(v)| = |N(u)\cup N(v)|\leq |V(G)| = n$$