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OPEN
Let $\epsilon>0$ and $n$ be large. Let $G$ be a graph on $n$ vertices containing no $K_5$ and such that every set of $\epsilon n$ vertices contains a triangle. Must $G$ have $o(n^2)$ many edges?
The best known result is that $G$ can have at most $(\tfrac{1}{12}+o(1))n^2$ many edges.
SOLVED
Does there exist some constant $c>0$ such that if $G$ is a graph with $n$ vertices and $\geq (1/8-c)n^2$ edges then $G$ must contain either a $K_4$ or an independent set on at least $n/\log n$ vertices?
A problem of Erdős, Hajnal, Simonovits, Sós, and Szemerédi [EHSSS93]. In other words, if $\mathrm{rt}(n;k,\ell)$ is the Ramsey-Turán number then is it true that $\mathrm{rt}(n; 4,n/\log n)< (1/8-c)n^2?$ Erdős, Hajnal, Sós, and Szemerédi [EHSS83] proved that for any fixed $\epsilon>0$ $\mathrm{rt}(n; 4,\epsilon n)< (1/8+o(1))n^2.$ Sudakov [Su03] proved that $\mathrm{rt}(n; 4,ne^{-f(n)})=o(n^2)$ whenever $f(n)/\sqrt{\log n}\to \infty$.

Resolved by Fox, Loh, and Zhao [FLZ15] who showed that the answer is no; in fact they prove that $\mathrm{rt}(n; 4, ne^{-f(n)})\geq (1/8-o(1))n^2$ whenever $f(n) =o(\sqrt{\log n/\log\log n})$.

See also [22] and the entry in the graphs problem collection.

Additional thanks to: Mehtaab Sawhney