Irska

Choose two points $X$ and $Y$ from $S$, and consider the half plane $h$ with boundary the line $XY$, and containing at least one other point of $S$. Now select a point $Z$ of $S$ in the interior of $h$ whose distance from the line $XY$ is a minimum. If there are several such points, choose an arbitrary one.

Now claim that the triangle $XYZ$ does not contain any other point of $S$, apart from $X$, $Y$, and $Z$. Suppose there is a point $A\in S$ which is in the interior of the triangle $XYZ$. Then $\mathrm{area}(AXY)<\mathrm{area}(ZXY)$ and so the altitude from $A$ of the triangle $AXY$ is less than the altitude from $Z$ of the triangle $ZXY$. This contradicts the choice of $Z$.

Suppose that there are $k$ pairs $(X,Y)$ of points in $S$ with the property that the whole of the set $S$ lies in one of the two half planes with boundary the line $XY$. Such pairs are adjacent vertices of the convex hull of $S$ and so $k\le n$. For each of these pairs $(X,Y)$ we have shown that there is at least one triangle $XYZ,Z\in S$, which is “empty”, that is it does not contain any of the remaining $n-3$ points of $S$.

The remaining $\left(\genfrac{}{}{0ex}{}{n}{2}\right)-k$ pairs $(X,Y)$ have the property that there are at least two “empty” triangles $XYZ,Z\in S$. Now sum all these triangles over all the pairs $(X,Y)$ and we get in all $$k+2\left(\left(\genfrac{}{}{0ex}{}{n}{2}\right)-k\right)=n(n-1)-k\ge n(n-1)-n=n(n-2)$$ such triangles. Each is counted at most three times so the number of “empty” triangles is at least $\frac{n(n-2)}{3}$.