Let $\delta > 0$. If $n$ is sufficiently large and $G$ is a graph on $n$ vertices with no $K_5$ and at least $\delta n^2$ edges, must $G$ contain a set of $\gg_\delta n$ vertices spanning no triangle?

Let $\delta > 0$. If $n$ is sufficiently large and $G$ is a graph on $n$ vertices with no $K_5$ and at least $\delta n^2$ edges, must $G$ contain a set of $\gg_\delta n$ vertices spanning no triangle?

Equivalently, writing $\mathrm{RT}_3(n, K_5, m)$ for the maximum number of edges of a $K_5$-free graph on $n$ vertices in which every triangle-free vertex set has fewer than $m$ vertices (the *triangle Ramsey–Turán number*), is $$\delta_3(5) = \lim_{\epsilon \to 0} \lim_{n \to \infty} \frac{\mathrm{RT}_3(n, K_5, \epsilon n)}{n^2} = 0?$$

This is a problem of Erdős, Hajnal, Simonovits, Sós, and Szemerédi [EHSSS94], who proved $\delta_3(5) \leq 1/12$ and the analogous $\delta_3(4) = 0$, and observed $\delta_3(7) \geq 1/4$ via a construction of Erdős and Rogers [ErRo62].

The answer is no: Balogh and Lenz [BaLe13] disproved it by showing $\delta_3(5) > 0$, and the exact value $\delta_3(5) = 1/12$ was determined by the matching lower-bound construction of Liu, Reiher, Sharifzadeh, and Staden [LRSS21] (see erdos_533.variants.lrss_lower).