GPT 6 Astra Ultra attempt on Math-6133
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Contents
Erdős Problem 1079
Statement and result.
For fixed $r\geq4$, suppose an $n$-vertex graph $G$ has at least $\operatorname{ex}(n;K_r)$ edges. We prove that every maximum-degree vertex $v$, of degree $d$, satisfies \[ e(G[N(v)])\geq\operatorname{ex}(d;K_{r-1}), \qquad d\geq\left(1-\frac1{r-1}\right)n-\frac{r-1}{4n}. \tag{1} \] Thus $d\gg_r n$ as required. When $e(G)>\operatorname{ex}(n;K_r)$, the first inequality is strict. The non-strict statement includes the Turán graph; it does not need to exclude that graph.Balanced multipartite edge counts.
Write $t_s(n)$ for the number of edges of the complete $s$-partite graph whose part sizes differ by at most one; empty parts are permitted. For part sizes $b_1,\ldots,b_s$ summing to $n$, the edge count is \[ \frac12\left(n^2-\sum_i b_i^2\right). \] If $b_i\geq b_j+2$, transferring one vertex from the former part to the latter reduces the square sum. Repeating proves that $t_s(n)$ is the maximum over complete $s$-partite graphs of order $n$.In particular, for any $0\leq d\leq n$, adjoining an independent part of size $n-d$ to a balanced $(s-1)$-partite graph on $d$ vertices gives \[ t_{s-1}(d)+d(n-d)\leq t_s(n). \tag{2} \]
A maximum-degree inequality.
Take any graph $G$, a maximum-degree vertex $v$, and let \[ A=N(v),\quad |A|=d,\qquad B=V(G)\setminus A,\quad |B|=n-d. \] The vertex $v$ itself lies in $B$. Every vertex of $B$ has degree at most $d$, so \[ e(A,B)+2e(B)=\sum_{b\in B}\deg(b)\leq d(n-d). \] Consequently \[ e(G)=e(A)+e(A,B)+e(B) \leq e(A)+d(n-d)-e(B) \leq e(A)+d(n-d). \tag{3} \] No independence assumption on $B$ has been made.An accompanying proof of Turán's formula.
We prove inductively on $s$ that every $K_{s+1}$-free graph has at most $t_s(n)$ edges. For $s=1$, a $K_2$-free graph has no edges. For the induction step, the graph on $A=N(v)$ is $K_s$-free. The induction hypothesis, $3$, and $2$ give \[ e(G)\leq t_{s-1}(d)+d(n-d)\leq t_s(n). \] The balanced complete $s$-partite graph has no $K_{s+1}$ and attains that count. Hence $\operatorname{ex}(n;K_{s+1})=t_s(n)$.The required neighbourhood and its size.
Now put $s=r-1$. From $3$, the assumed edge count, and $2$, \[ e(A)\geq e(G)-d(n-d) \geq t_s(n)-d(n-d) \geq t_{s-1}(d). \] If the hypothesis has $e(G)>t_s(n)$, the same chain has a strict inequality, proving the strengthened assertion.For completeness, write $n=sq+a$, where $0\leq a<s$. The balanced part sizes give \[ t_s(n)=\frac12\left(1-\frac1s\right)n^2 -\frac{a(s-a)}{2s}. \] Since a maximum degree is at least the average degree, \[ d\geq\frac{2e(G)}n\geq \left(1-\frac1s\right)n-\frac{a(s-a)}{sn} \geq\left(1-\frac1s\right)n-\frac{s}{4n}. \] This is $1$, and proves the needed positive linear lower bound for sufficiently large $n$.
Equality and attribution.
For $G=T_s(n)$, a maximum-degree vertex lies in a smallest part. Its neighbourhood is a balanced complete $(s-1)$-partite graph, so equality in the first part of $1$ holds. Thus the page's exception involving the Turán graph pertains to strict variants, not the literal non-strict inequality.The dense-neighbourhood result is due to Bollobás--Thomason, and the maximum-degree strengthening to J. A. Bondy, Large dense neighbourhoods and Turán's theorem, https://doi.org/10.1016/0095-8956(83)90012-6. The publisher's abstract states the strict maximum-degree version. Source statement: https://www.erdosproblems.com/1079, checked 24 September 2026. All edge-count inequalities and the needed Turán formula are proved above; no external extremal theorem, novelty, or local formal verification is claimed.