Existence of vertex in a triangle-free graph whose non-neighbours have at most $m(1-4m/n^2)$ edges.

Question

Given triangle-free graph $G=(V,E)$ and $n:=|V|$, $m:=|E|$. I would like to show that there is a vertex whose non-neighbours have at most $m(1-4m/n^2)$ edges, i.e. the graph $G[V\setminus N[v]]$ has at most $m(1-4m/n^2)$ edges. I tried approaching by contradiction and assuming that each such subgraph has at least the number of edges and maybe show that the graph has more than $n^2/4$ edges which would contradict Mantel's theorem but got nowhere.

Answer 1

Briefly reason like this.

Assume the contrary.

1. Suppose that for any vertex $v$ of graph $G$ we have $$ |E(G-N(v))|>m-\frac{4m^2}{n^2}. $$
2. Since the graph $G$ has no triangles, then $$ |E(G-N(v))|=m-\sum_{x\in N(v)}\operatorname{deg}(x). $$
3. It follows from 1) and 2) that $$ \sum_{x\in N(v)}\operatorname{deg}(x)<\frac{4m^2}{n^2}. $$
4. Let us sum the last inequality over all vertices of graph $G$ $$ \sum_{v\in V(G)}\sum_{x\in N(v)}\operatorname{deg}(x)<\frac{4m^2}{n}. $$
5. Then $$ \sum_{v\in V(G)}\sum_{x\in N(v)}\operatorname{deg}(x)=\sum_{v\in V(G)}(\operatorname{deg}(v))^2<\frac{4m^2}{n}. $$
6. The Cauchy–Schwarz inequality gives $$ \frac{(\sum_{v\in V(G)}\operatorname{deg}(v))^2}{n}\leq\sum_{v\in V(G)}(\operatorname{deg}(v))^2<\frac{4m^2}{n}. $$
7. It follows that $\frac{(2m)^2}{n}< \frac{4m^2}{n}$.

Contradiction.