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> Show that if a graph $G$ of order $n$ is triangle-free, then $$n \ge \delta(G) + \Delta(G).$$
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What I have tried so far is as follows. I wish to show $n > \Delta(G) + \delta(G)$ implies there is a triangle. Choose a vertex $u$ with degree $\Delta(G)$. Consider the neighborhood $N(u)$ of $u$. There are $n - \Delta(G) - 1$ vertices is $V(G)\setminus N(u)$. Choose a vertex $v$ in $V(G)\setminus N(u)$. It has at least $\delta(G)$ neighbors. $\delta(G) > n - \Delta(G) $ . Thus $u$ and $v$ share two vertices in their neighborhood forming a cycle of length $4$.