Triangle-free graphs have at most n²/4 edges (Mantel)

Idea. Count the pairs (vertex, edge containing it) in two ways.

Local bound. Let $uv$ be an edge. If a vertex $w$ were adjacent to both $u$ and $v$, then $uvw$ would be a triangle. So the neighbourhoods of $u$ and $v$ are disjoint, and
$$d(u)+d(v)\le n.$$

Summation. Let $e$ be the number of edges. Summing over all edges $uv$,
$$\sum_{uv\in E}\bigl(d(u)+d(v)\bigr)\le ne.$$
On the left, each vertex $v$ appears $d(v)$ times, each with the term $d(v)$. So the left side equals $\sum_vd(v)^2$:
$$\sum_vd(v)^2\le ne.$$

Cauchy–Schwarz. Since $\sum_vd(v)=2e$,
$$\sum_vd(v)^2\ge\frac{\left(\sum_vd(v)\right)^2}n=\frac{4e^2}n.$$

Conclusion. $\dfrac{4e^2}n\le ne$, so $e\le\dfrac{n^2}4$.

Equality. For $n$ even, the complete bipartite graph $K_{n/2,n/2}$ is triangle-free with $\frac{n^2}4$ edges. For $n$ odd, the bound is $\left\lfloor\frac{n^2}4\right\rfloor$ and is attained by $K_{(n-1)/2,(n+1)/2}$.

Lesson. Double counting: when a sum over edges involves degrees, rewrite it as a sum over vertices; combine with Cauchy–Schwarz.