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A family of triangles in a graph is edge-disjoint if no edge belongs to two members. Any deletion of edges producing a triangle-free graph must remove at least one distinct edge for each member. The triangles in a graph may share vertices.

The lower shadow consists of the sets obtained by removing one element from a member:For a nonempty uniform set family, write its size in its unique decreasing binomial coefficient expansionThe Kruskal-Katona theorem states thatA colexicographic initial segment attains equality; thus this is an exact minimum, not just an asymptotic estimate.

Here is a direct proof of the requested Lovász shadow bound, avoiding any unproved numerical interpolation of the Kruskal-Katona theorem. The real binomial coefficient means $\left(\genfrac{}{}{0ex}{}{x}{r}\right)=x(x-1)\cdots (x-r+1)/r!$. It is strictly increasing for $x>r-1$. Since the set family is nonempty and its size is an integer, necessarily $x\ge r$.

Apply the usual left coordinate shifts of a set family until the set family is fixed by every shift. A shift preserves its size and does not increase its lower shadow; the witness argument is supplied in Solution 3. Termination follows because every nontrivial shift reduces ${\sum }_{A\in \mathcal{A}}{\sum }_{i\in A}i$. It suffices to prove the inequality for this shifted set family.

Induct on $r$, and, for each fixed $r$, on $|\mathcal{A}|$. The cases $r=1$ and $|\mathcal{A}|=1$ give, respectively, the single empty shadow member and the $r$ faces of an $r$-set. Split the set family at coordinate one, writing ${\mathcal{A}}_0$ for the members avoiding it and ${\mathcal{A}}_1$ for the sets obtained by deleting it from members containing it. Every $(r-1)$-subset of a member of ${\mathcal{A}}_0$ belongs to ${\mathcal{A}}_1$: replace the one omitted coordinate by coordinate one. ThereforeThe two terms count shadow members avoiding and containing coordinate one.

Put $b=|{\mathcal{A}}_1|$. This is positive. If $b<\left(\genfrac{}{}{0ex}{}{x-1}{r-1}\right)$, the Pascal's identity gives $|{\mathcal{A}}_0|>\left(\genfrac{}{}{0ex}{}{x-1}{r}\right)$. In this case $x>r$ and ${\mathcal{A}}_0$ is a smaller nonempty set family. The mathematical induction hypothesis on family size gives $|\partial {\mathcal{A}}_0|>\left(\genfrac{}{}{0ex}{}{x-1}{r-1}\right)$, contradicting its containment in ${\mathcal{A}}_1$. Consequently $b\ge \left(\genfrac{}{}{0ex}{}{x-1}{r-1}\right)$. Write $b=\left(\genfrac{}{}{0ex}{}{z}{r-1}\right)$, so $z\ge x-1$. The mathematical induction hypothesis on uniformity and the Pascal's identity now giveFor $r=2$ the degree-zero term is the constant one, which causes no difficulty.

For the graph application, identify a triangle-free graph with a subset of the $N$ possible edges. Removing an edge preserves being triangle-free, so $\partial {\mathcal{G}}_{m+1}\subseteq {\mathcal{G}}_m$. If ${\mathcal{G}}_{m+1}$ is nonempty, the Lovász shadow bound givesThus, for $1\le m<N$ with $p_m>0$,If the next set family is empty its probability is zero, so the same inequality holds without introducing $x_{m+1}$. If $p_m=0$, every later set family is empty; the divided expression is undefined and the correct assertion is this zero-probability conclusion.

Iterate the monotonicity of the effective ground-set parameter of a hereditary uniform layer. Whenever $p_{m+k}>0$, all the intermediate parameters are at least their layer sizes, and all factors below are nonnegative. We obtain the stronger estimateThe second inequality follows from $x_m\le N$ and the fact that $(x_m-t)/(N-t)$ decreases with $t$. If the target set family is empty, the requested bound follows directly when its right side is nonnegative. When $k>x_m+1$, the printed product can have signed factors and should instead be stopped at the first empty layer: its unrestricted signed form is not a meaningful probability bound. In the nonzero target range there is no such ambiguity. A uniformly safe extension, for $0\le k\le N-m$, isIt agrees with the printed expression whenever all its factors are nonnegative. In particular $0\le k\le m$ is always safe when $p_m>0$, and this includes the requested doubling case.

Taking $k=m$ in the nonzero range,If the $2m$-edge set family is empty the conclusion is immediate; if $2m>N$, interpreting the event as impossible gives the same conclusion. These qualifications distinguish the probability assertion from products or ratios outside their domains.

Here $|V(G)|=2n$ and $e(G)=n^2+1$. By Mantel theorem, a triangle-free graph of this order has at mostedges, so $G$ contains a triangle in a graph.

If $n=2$, then $G$ has four vertices and five of the six possible edges, so it is $K_4$ with one edge deleted. Exactly the two triangles containing neither both endpoints of the deleted edge remain. Hence $G$ contains two triangles.

The extremal number $\mathrm{ex}(n,H)$ is the maximum number of edges in an $n$-vertex graph containing no subgraph isomorphic to $H$.

Let $G$ be triangle-free with $e$ edges. For every edge $uv$, the neighbourhoods of $u$ and $v$ are disjoint apart from their endpoints, so
$d(u)+d(v)\le n$. Summing over edges and applying Cauchy-Schwarz inequality giveswhileThus $e\le n^2/4$, proving the required Mantel theorem bound.

Let $X$ count the triangles of the binomial random graph $G(n,p)$. ThenTwo distinct triangle indicators are dependent only when the triangles share an edge. Hence their ordered Janson dependency sum is

If $0<p\le n^{-1/2}$, then $\Delta =O(\mu )$. The first Janson inequality givesFor the reverse bound, the events that individual triangles are absent are decreasing, so Harris' inequality giveswhere $p<1/2$ keeps the logarithmic estimate uniform. Thus the probability is $e^{-\Theta (p^3{n}^3)}$.

If $n^{-1/2}\le p<1/2$, the extended Janson bound givesFor a lower bound, fix a balanced bipartition of the vertices and require every edge inside either part to be absent. The resulting graph is bipartite, hence triangle-free, and this event has probabilityCombining the bounds proves the second regime.

Fix the requested $\epsilon >0$ and choose much smaller constants $\eta ,\delta ,\rho >0$ in that order. Apply the Szemerédi regularity lemma with regularity parameter $\rho$ and a sufficiently large lower bound $m_0$ to a largest triangle-free subgraph $F\subseteq G$. Letbe the resulting equitable partition, and form the reduced graph of a regularity partition $R$ on $[m]$ by joining $i$ and $j$ when $(V_i,V_j)$ is $\rho$-uniform in $F$ and has $F$-density at least $\delta$.

Choose $\rho <\delta /2$. If $R$ contained a triangle, the triangle embedding lemma for regular pairs would give a triangle in $F$. Thus $R$ is triangle-free, and Turan theorem gives

Part (c), with error $\eta$, holds simultaneously for every pair of sets of size at least $n/\log n$. Since $m$ is bounded independently of $n$, all regularity classes and their halves satisfy this size condition for large $n$. It follows that every cross-pair contains at most $(p+\eta )|V_i||V_j|$ edges of $G$. It also givestake a bipartition of $V_i$ carrying at least half its internal edges and apply the cross-pair estimate.

Now count the edges of $F$. Pairs represented in $R$ contribute at mostThe exceptional set, the at most $\rho m^2$ irregular pairs, pairs of $F$-density below $\delta$, and edges inside classes together contribute at mostChoose $\eta$, then $\delta ,ho$, and finally $m_0^{-1}$ small enough in terms of $p\epsilon$. The total is at mostwith high probability, as required.

Apply the Shearer independence bound for a triangle-free graph. It states that an $n$-vertex triangle-free graph of maximum degree at most $d$ has an independent set of sizefor an absolute $c>0$. The result is immediate for bounded $d$ after reducing $c$, while the theorem gives the asserted logarithmic gain for $d$ large. Thus

For completeness, the key input in Shearer's proof is to expose a random independent set one degree scale at a time. Triangle-freeness makes every neighbourhood independent, so conditioning on earlier choices creates no edges inside the available neighbours. The expected gain at degree scale $j$ is $\Omega (n_j/d)$; summing over the $\Theta (\log d)$ nonempty scales gives $\Omega (n\log d/d)$. The entropy, or hard-core-model, form of the argument makes this scale calculation rigorous without losing vertices counted at adjacent scales.