Variant on Turán's theorem Write $K_m^-$ for the complete graph on $m$ edges with an edge removed. Write $e(G)$ for the size of $G$ (number of edges). I am trying to prove the following twist on Turán's theorem:

If $e(G)>e\big(T_{m-1}(n)\big)$, with $|G|=n>m$, then there exists a $K_{m+1}^-\subseteq G$.

I am trying induction on $n$, fixing $m$:


*

*base case: $n=m+1$. Then $e(G)>e\big(T_{m-1}(n)\big)=\binom{m+1}{2}-1$, so we in fact have $G\cong K_{m+1}$.

*assume true for $m<n< N$. Remove a vertex $x$ of minimum degree. Then by assumption:


$$e(G\setminus\{x\})=e(G)-\delta(G)>e\big(T_{m-1}(N)\big)-\delta(G)$$
What I'd like is $\delta(G)≤\delta\big(T_{m-1}(N)\big)$ because then we would have $$\cdots ≥ e\big(T_{m-1}(N)\big)-\delta\big(T_{m-1}(N)\big)=e\big(T_{m-1}(N-1)\big)$$ so $K_m^-\subseteq G\setminus\{x\}$ by induction hypothesis and therefore $K_m^-\subseteq G.$
But I can't see why this would be true: we have $\sum d_G(x)>\sum d_{T_{m-1}(N)}(x)$ so there is nothing to indicate what I want. I'm coming to the conclusion that this approach is doomed, so would appreciate some help.
 A: edit: [Case 1: We contain a $T_{m-1}$.
When you add one edge to the Turan graph, it's in one of the $m-1$ partitions, between vertices $x$ and $y$, say. Then $x$ and $y$ together with one vertex from each other partition set build a $K_m$. Now take another vertex from any of the partition sets and add it to the subgraph: it's connected to all other vertices of our $K_m$ but the one it lies in the same partition set with, so the graph is $K_{m+1}^-$.
Case 2:] we have a $K_m$ in G.
Now if there was a $v \in G-K_m$ that had more than $m-2$ neighbours in the $K_m$ that would give us a $K⁻_{m+1}$. So we can assume that we have at most $(m-2)(n-m)$ edges between $G-K_m$ and $K_m$. So in total we have (letting $t_{m-1}(n)$ denote the number of edges of $T_{m-1}(n)$)
$$e(G-K_m) \ge t_{m-1}(n) + 1 - {{m}\choose{n}} - (n-m)(m-2) = t_{m-1}(n - m + 1) + (m-2).$$
This recursion can be counted directly in a Turan graph (see for instance in Diestel's book.) Now we are ready by induction (on $n$). (For $n-m+1 < m-1$ we could not have that many edges in a simple graph.)
