Mantel's theorem: $\mathrm{ex}(n,K_3)=\lfloor n^2\mathrm{/}4\rfloor$, uniquely attained by $T_{n,2}$

The assertion is immediate for $n=0,1,2$. Assume it, including uniqueness, for $n-2$, and let $G$ be a triangle-free $n$-vertex graph. If $G$ has no edge its bound is immediate. Otherwise choose an edge $uv$. No vertex other than $u,v$ is adjacent to both ends, so $d(u)+d(v)\le n$.

Delete $u,v$ to obtain a triangle-free graph $G^{\mathrm{\prime }}$. The removed edges number $d(u)+d(v)-1$, so $e(G)\le \lfloor (n-2{)}^2\mathrm{/}4\rfloor +n-1=\lfloor n^2\mathrm{/}4\rfloor$. The graph $T_{n,2}$ is triangle-free and has the last edge count, proving the exact maximum.

Suppose equality holds. Then $G^{\mathrm{\prime }}\cong T_{n-2,2}$, $d(u)+d(v)=n$, and every other vertex is adjacent to exactly one of $u,v$. In each part of $G^{\mathrm{\prime }}$, triangle-freeness forces all vertices to choose the same endpoint: two vertices in opposite parts choosing the same endpoint would form a triangle with their cross edge. Hence adjoining $u$ to one part and $v$ to the other makes $G$ complete bipartite.

Its two part sizes sum to $n$ and its product is $\lfloor n^2\mathrm{/}4\rfloor$; the balancing equality in the preceding lemma forces them to differ by at most $1$. Thus $G\cong T_{n,2}$. Conversely that graph has equality, completing the induction and the uniqueness proof.

Steps 1.1-4.1 prove Mantel's theorem independently of Turán's theorem, for all $n$ and with equality fully characterized.