This is pretty standard. See here for an even stronger bound: https://math.stackexchange.com/questions/3575265/lower-bound-for-number-of-triangles-in-simple-graph-very-hard-exercise?fbclid=IwAR3VbBPeeqv5J1a-Dk8QzMQiGRAoxMQLCNVSjlYabhxx205WuIlmZKcrIdc

Consider a graph $G=(V, E)$ with $V=A$ and $E$ containing the $n^2+k$ segments. Let $u, v$ be two adjacent vertices, then the number of triangles (suppose is $f(u, v)$) that contains $u, v$ is the number of common neighbors of $u, v$. $\Rightarrow f(u, v)\geq(\deg(u)-1)+(\deg(v)-1)-(2n-2)=\deg(u)+\deg(v)-2n$. Since $\sum_{v\in V}\deg(v)=2(n^2+k)$, we get \begin{align*} \sum_{(u, v)\in E}f(u, v)&\geq\sum_{(u, v)\in E}\deg(u)+\deg(v)-2n\\ &=-2(n^2+k)n+\sum_{v\in V}\deg(v)^2\text{ ($\because v$ will be counted for $\deg(v)$ times) }\\ &=-2n^3-2nk+\frac1{2n}(\sum_{v\in V}1)(\sum_{v\in V}\deg(v)^2)\\ &\geq-2n^3-2nk+\frac1{2n}(\sum_{v\in V}\deg(v))^2\text{ (by Cauchy inequality) }\\ &=-2n^3-2nk+\frac1{2n}(2n^2+2k)^2\\ &=-2n^3-2nk+2n^3+4nk+\frac{2k^2}n\\ &=2nk+\frac{2k^2}n\\ &\geq2\sqrt{2nk\cdot\frac{2k^2}n}\text{ (by AM$\geq$GM) }\\ &=4\sqrt{k^3} \end{align*}Since each triangle contains $3$ edges. $\therefore$ the number of triangles $=\frac13\sum_{(u, v)\in E}f(u, v)\geq\frac43\sqrt{k^3}$.