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A false-twin class is an equivalence class of vertices with identical open neighbourhoods. In edge-triangle symmetrization, merging nonadjacent false-twin classes by cloning the better local score never decreases the objective. A secondary maximization of squared class sizes forces all distinct classes to be adjacent, giving a complete multipartite graph.

For a fixed uniform hypergraph $H$ with $h$ vertices, uniformity $\ell$, and at least one edge, define its Turán density bywhere $\mathrm{ex}(n,H)$ is the maximum edge count of an $H$-free hypergraph. The limit exists: averaging the edge count over all $m$-vertex subsets of an extremal $n$-vertex hypergraph, with $n\ge m\ge \ell$, shows that the displayed normalized extremal numbers are nonincreasing. Thus their limit is their infimum.

For hypergraph supersaturation by sampling, choose $m\ge h$ so that $\mathrm{ex}(m,H)/\left(\genfrac{}{}{0ex}{}{m}{\ell }\right)\le \pi (H)+ϵ/2$. The expected edge count in a uniform $m$-vertex subset of $G$ is at least $(\pi (H)+ϵ)\left(\genfrac{}{}{0ex}{}{m}{\ell }\right)$. A subset containing no copy of $H$ has at most $(\pi (H)+ϵ/2)\left(\genfrac{}{}{0ex}{}{m}{\ell }\right)$ edges, while any subset has at most $\left(\genfrac{}{}{0ex}{}{m}{\ell }\right)$. Hence at least an $ϵ/2$ fraction of the subsets contain $H$. Every particular copy of $H$ lies in exactly $\left(\genfrac{}{}{0ex}{}{n-h}{m-h}\right)$ such subsets, so their number is at leastfor $n\ge N=\max (m,2h)$. TakeFor $n<N$, the requested integer lower bound is zero; for $n\ge N$, the calculation proves it. This positive $\delta$ depends only on $H$ and $ϵ$. If the density hypothesis is impossible there is nothing to prove. For an edgeless $H$, every $h$-set is already a copy and the conclusion follows directly.

For the edge-triangle objective, let $N(v)$ be the open neighbourhood of $v$ and define its local scoreIt counts the contribution to $k_2-c{k}_3$ of edges and triangles in a graph containing $v$. If $u,v$ are nonadjacent, making $u$ a false twin of $v$ changes the objective by $s(v)-s(u)$, since no edge or triangle in a graph uses both. More generally, partition vertices into false-twin classes, meaning equal open neighbourhoods. If distinct classes $U,V$ have no edges between them, making every vertex of $U$ a twin of a representative of $V$ changes the objective by $|U|(s(v)-s(u))$. Choose the direction with nonnegative change.

Among graphs maximizing the objective, choose one maximizing the sum of squared false-twin class sizes. The operation cannot split any old false-twin class and merges $U,V$, so it would strictly increase that secondary quantity. Therefore no two distinct classes can be nonadjacent. An objective maximizer is a complete multipartite graph. This edge-triangle symmetrization works for every real $c$, with no assumption on its sign.

For positive part sizes $a_1,\dots ,a_q$, the correct multipartite formula isThe first sum in the printed intermediate formula has inconsistent indices; the edge count is the pairwise-product sum above. Fix two part sizes with sum $A$ and let $B$ be the sum of all other sizes. All terms varying with the pair have the formChoose a maximizing partition with the fewest nonempty parts. If $1-cB\le 0$, merging those two parts cannot decrease $F$ and reduces the number of parts, a contradiction. Therefore this coefficient is positive for every pair. If two integer sizes differ by at least two, moving one vertex from the larger to the smaller increases their product and hence $F$, again a contradiction. All sizes differ by at most one, so a maximizer is a Turan graph. This is balancing a multipartite edge-triangle objective.

Next maximize the same polynomial on the compact simplex of nonnegative real sizes summing to $n$, starting with at least two possible parts. Choose a maximizer with the smallest positive support. The merging argument still applies, and now varying two positive unequal sizes towards equality improves the product. Thus every positive size is $n/q$ for some integer $q$. A two-part choice has value $n^2/4>0$ when $n>0$, so $q\ge 2$. The optimum is attained at rational sizes, and thereforeOne may take the initial simplex to have $\max (n,2)$ coordinates, so it includes every multipartite graph obtained above; zero coordinates are discarded.

Set $c=9/(4n)$. Dividing the right side by $n^2$ givesbecause $q$ is an integer at least two. Rearranging provesEquality in the continuous calculation occurs at two or three equal parts, explaining the triangle support line between bipartite and tripartite Turan graphs.

Symmetrization for an arbitrary real coefficient. Among graphs maximizing $f(G)=e(G)-c{k}_3(G)$, choose one maximizingFor a vertex $x$, put $L(x)=d(x)-ce(G[N(x)])$. This is the contribution of edges and triangles in a graph containing $x$. If $u,v$ are nonadjacent, Zykov symmetrization replacing $u$ by a clone of $v$ changes $f$ by $L(v)-L(u)$; the opposite replacement changes it by $L(u)-L(v)$. Maximality implies $L(u)=L(v)$, so both changes preserve $f$.

Let $S=N(v)\setminus N(u)$ and $T=N(u)\setminus N(v)$. In the first replacement the vertex degrees in $S$ increase by one and those in $T$ decrease by one; in the opposite replacement these changes reverse. The changes to $d(u{)}^2+d(v{)}^2$ cancel when the two replacements are added. Since $(d+1{)}^2+(d-1{)}^2-2d^2=2$,If the two vertex neighbourhoods differ, one replacement increases $Q$, a contradiction. Thus every pair of nonadjacent vertices has the same vertex neighbourhood. Nonadjacency, with equality allowed, is an equivalence relation: if $u$ and $v$ are nonadjacent and $v$ and $w$ are nonadjacent, their identical vertex neighbourhoods forbid $uw$ as well. Its classes are independent sets and every cross-class edge is present. ThereforeThis edge-triangle symmetrization works for either sign of $c$.

The exact supporting line. For $n\ge 3$, writeWe prove the triangle support line between bipartite and tripartite Turan graphs by maximizing $e-c_{\ast }{k}_3$. First $c_{\ast }\ge 2/n$. For $n=3,4,5$, its values are respectively $1,1/2,1/2$. For $n\ge 6$, the exact balancing of the Turán graph gives $e_3\ge n^2/3-1/3$, while $e_2\le n^2/4$ and the arithmetic-geometric mean inequality gives $t_3\le n^3/27$. Hence

By edge-triangle symmetrization a maximizing graph can be taken complete multipartite graph; among such maximizers use one with the fewest nonempty parts. If there are at least four parts, let $a,b$ be the two smallest sizes. Then $a+b\le n/2$. Merging them loses $ab$ edges and $ab(n-a-b)$ triangles in a graph, so the objective changes byThis contradicts the choice of the number of parts. Thus at most three parts remain.

For three part sizes $a\ge b\ge d>0$, the objective isIf $1-c_{\ast }b\le 0$, merging the parts of sizes $a,d$ does not decrease it, again contradicting minimality. If $1-c_{\ast }b>0$ and $a-d\ge 2$, transferring one vertex from the largest to the smallest part raises $ad$ by $a-d-1>0$ and strictly increases the objective. Thus a three-part maximizer has all sizes differing by at most one, and is $T_3(n)$. A maximizer with at most two parts has objective at most $e_2$, since it has no triangles in a graph and the Mantel theorem bounds its edges. Finally,Therefore every graph satisfies $e(G)-c_{\ast }{k}_3(G)\le e_2$. At the prescribed edge count this givesFor $n\le 2$, the conclusion is simply the nonnegative lower bound zero. All rounding in the Turán graphs has been retained.

One edge above the bipartite threshold. Write $n=2m$. We prove the Triangle lower bound one edge above the Mantel threshold by induction on $m\ge 2$, allowing at least $m^2+1$ edges. For $m=2$, five edges on four vertices give two triangles in a graph, and adding edges preserves this bound.

We will also use the following consequence of an inductive bound: on $2a$ vertices, at least $a^2+q$ edges, for an integer $q\ge 1$, force at least $a+q-1$ triangles in a graph. Delete surplus edges first to leave exactly $a^2+q$. Repeatedly delete an edge in a triangle in a graph until $a^2+1$ edges remain. The Mantel theorem guarantees such a triangle in a graph at every step; each deletion destroys at least one distinct triangle in a graph. Apply the inductive bound to the remaining graph.

Now take exactly $m^2+1$ edges. If every edge lies in a triangle in a graph, the edge-triangle incidence bound gives $3k_3(G)\ge m^2+1$. This is at least $3m$ for $m\ge 3$; the base case was handled separately.

Otherwise choose an edge $uv$ in no triangle in a graph. Its endpoints have disjoint vertex neighbourhoods, so $d(u)+d(v)\le 2m$. Deleting $u,v$ removes exactly $d(u)+d(v)-1$ edges and leaves at least $(m-1{)}^2+1$ edges. If $d(u)+d(v)\le 2m-1$, it leaves at least $(m-1{)}^2+2$ edges, and the strengthened inductive bound gives at least $m$ triangles in a graph already.

If $d(u)+d(v)=2m$, induction gives at least $m-1$ triangles in a graph after deletion. There must be an additional triangle in a graph containing $u$ or $v$. Otherwise both $N(u)$ and $N(v)$ are independent sets; since they are disjoint and cover all vertices, $G$ would be bipartite graph, contradicting $e(G)>m^2$ by the Mantel theorem. HenceThe bound is sharp: add one internal edge to one class of the complete bipartite graph $K_{m,m}$. Its triangles in a graph are precisely that edge together with each of the $m$ vertices in the other class.

For $n\ge 3$, put $e_i=e(T_i(n))$, $t_3=k_3(T_3(n))$, and $c_n=(e_3-e_2)/t_3$. Every graph on $n$ vertices satisfiesThus an edge count $e_2+\theta (e_3-e_2)$ forces at least $\theta t_3$ triangles in a graph. This is an exact finite bound, with the rounding in the Turán graph retained. To prove it, use edge-triangle symmetrization. The inequality $c_n\ge 2/n$ allows merging the two smallest classes when there are at least four classes. With three classes, fixing the middle class makes the objective affine in the product of the other two sizes; either merge them or balance them. Only $T_2(n)$ and $T_3(n)$ need remain.