Let $G$ be a graph on $n$ vertices with $>n^2/4$ many edges. Must there be a triangle $T$ in $G$ and vertices $y_1,\ldots,y_t$, where $t>(\frac{1}{2}-o(1))n$, such that every $y_i$ is joined to at least two vertices of $T$?

Erdős suggested that the answer is different if $G$ has no $K_4$. In the comments Ma and Tang sketch a proof that the conjecture remains false even if we assume that $G$ contains no $K_4$, constructing a graph with $n$ vertices, $>n^2/4$ edges, and no $K_4$, in which every triangle has at most $(2\sqrt{3}-3+o(1))n$ vertices adjacent to at least two of its vertices (note that $2\sqrt{3}-3\approx 0.464$).