Proof of Mantel's theorem by double counting

Proof

We double count the sum of the squares of the degrees and apply the Cauchy–Schwarz inequality.

1. Let $G$ be a finite simple graph with vertex set $V$, $\abs{V} = n$, and $m$ edges. Suppose that $G$ has no cycle of length $3$. For $v \in V$, write $d(v)$ for its degree and $N(v)$ for its neighbourhood.
2. If $m = 0$, then $m = 0 \le \floor{n^2/4}$. From now on let $m \ge 1$, so $n \ge 2$.
3. Let $xy$ be an edge. If a vertex $z$ were adjacent to both $x$ and $y$, then $z \ne x, y$ and $x, y, z, x$ would be a cycle of length $3$. So $N(x) \cap N(y) = \varnothing$.
4. Hence $d(x) + d(y) = \abs{N(x)} + \abs{N(y)} = \abs{N(x) \cup N(y)} \le n$ for every edge $xy$.
5. We double count the sum $\sum_{xy} \big(d(x) + d(y)\big)$ over all edges $xy$. A vertex $v$ is an end of exactly $d(v)$ edges, and each of them contributes $d(v)$ to the sum.
6. So $\sum_{v \in V} d(v)^2 = \sum_{xy} \big(d(x) + d(y)\big) \le m n$, by Step 4.
7. By the handshaking lemma, $\sum_{v \in V} d(v) = 2m$.
8. Apply the Cauchy–Schwarz inequality in $\R^n$, with the standard inner product, to the vectors $(d(v))_{v \in V}$ and $(1)_{v \in V}$. This gives $\left(\sum_v d(v)\right)^2 \le n \sum_v d(v)^2$.
9. By Steps 6, 7 and 8, $4m^2 \le n \cdot m n = m n^2$.
10. Divide by $4m > 0$ to get $m \le n^2/4$.
11. Since $m$ is an integer, $m \le \floor{n^2/4}$.