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Erdős #1033 · graph theory

Let h(n)h(n) be such that every graph on nn vertices with >n2/4>n^2/4 many edges contains a triangle whose vertices have degrees summing to at least h(n)h(n). Estimate h(n)h(n). In particular, is it true that

h(n)≥(2(\sq…h(n)\geq (2(\sq…

Inferred from the community field; finer classification has not been reviewed.

Source reports OPENAs of 2026-09-08Source & evidence ↓

Problem statement

Let h(n)h(n) be such that every graph on nn vertices with >n2/4>n^2/4 many edges contains a triangle whose vertices have degrees summing to at least h(n)h(n). Estimate h(n)h(n). In particular, is it true that

h(n)≥(2(3−1)−o(1))n?h(n)\geq (2(\sqrt{3}-1)-o(1))n?

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Background

A conjecture of Bollobás and Erdős. In [Er82e] it is asked whether h(n)≥32nh(n)\geq \frac{3}{2}n. It is now known that

2(3−1)n+O(1)≥h(n)≥2116n.2(\sqrt{3}-1)n +O(1)\geq h(n) \geq \frac{21}{16}n.

The upper bound is due to Erdős and Laskar [ErLa85], and the lower bound is due to Fan [Fa88]. (Note that 2(3−1)≈1.4642(\sqrt{3}-1)\approx 1.464 and 21/16=1.312521/16=1.3125.)

The upper bound is not made explicit in [ErLa85], which is concerned with chordal subgraphs (i.e. subgraphs which do not contain any induced cycle on >3>3 vertices). The connection to this problem is that taking a triangle together with all incident edges forms a chordal subgraph of GG. The construction of the upper bound is made more explicit in [Fa88].

In brief, let m=⌊n2/4⌋+1m=\lfloor n^2/4\rfloor+1 and take a complete bipartite graph between kk and ll vertices where k+l=nk+l=n and k=cnk=cn. If cc is chosen such that m−kl≤k2/4m-kl\leq k^2/4 then we can add m−klm-kl edges between the kk vertices creating no triangles, while having maximum degree (amongst these vertices) ⌈2m−klk⌉\lceil 2\frac{m-kl}{k}\rceil. This creates a graph GG on nn vertices with mm edges. Every triangle uses exactly one vertex from the side with ll vertices and two vertices from the side with kk vertices, and hence the sum of degrees is at most

k+2⌈2m−klk⌉+2l=(c−1+3c−2)n+O(1).k+2\left\lceil 2\frac{m-kl}{k}\right\rceil+2l=(c^{-1}+3c-2)n+O(1).

This is minimised at c=1/3c=1/\sqrt{3}, where it is 2(3−1)n+O(1)2(\sqrt{3}-1)n+O(1) (and a short calculation verifies that with this choice of cc we do indeed have m−kl≤k2/4m-kl\leq k^2/4).

More generally, let Δr(n,m)\Delta_r(n,m) be such that every graph GG with nn vertices and mm edges contains a clique on rr vertices, say x1,…,xrx_1,\ldots,x_r, such that

d(x1)+⋯+d(xr)≥Δr(n,m).d(x_1)+\cdots+d(x_r)\geq \Delta_r(n,m).

Let r≥2r\geq 2 and let tr(n)t_r(n) be the Turán number (the maximal number of edges in a graph on nn vertices with no Kr+1K_{r+1}). One can ask more generally about the behaviour of Δr(n,m)\Delta_r(n,m) for m>tr−1(n)m>t_{r-1}(n) (when m≤tr−1(n)m\leq t_{r-1}(n) this is trivial since there may be no clique on rr vertices at all). The behaviour of Δr(n,m)\Delta_r(n,m) in the regime m≥tr(n)m\geq t_r(n) is the subject of [904].

Erdős (see [Fa92]) proved that, for every ϵ>0\epsilon>0, there exists δ>0\delta>0 such that if tr−1(n)<m<tr(n)−δn2t_{r-1}(n)<m<t_r(n)-\delta n^2 then

Δr(n,m)≤(1−ϵ)2rmn.\Delta_r(n,m)\leq (1-\epsilon)\frac{2rm}{n}.

Bollobás and Nikiforov [BoNi05] proved that, for every ϵ>0\epsilon>0, there exists δ>0\delta>0 such that if m>tr(n)−δn2m>t_r(n)-\delta n^2 then

Δr(n,m)≥(1−ϵ)2rmn.\Delta_r(n,m)\geq (1-\epsilon)\frac{2rm}{n}.

Progress & source assessment

The source labels this entry OPEN in the 2026-09-08 snapshot. Editorial progress notes are reproduced as source commentary, not independently verified results. Prize amounts and Lean badges at the source are not local awards or verification.

References
  1. [1]Thomas Bloom (ed.), Erdős Problems #1033 link

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