, within $n$ points, $\triangle XYZ$ is a triangle formed by such $3$ points where the area is maximal. Here,surely, $[XYZ] \le 1$. Now, we construct a triangle $\triangle PQR$ whose medial triangle is $\triangle XYZ$. So, now, $[PQR]=4[XYZ] \le 4$. We claim
, all the points are contained in $\triangle PQR$ . We suppose
, A point named $A$ lies outside $\triangle PQR$. If we add $'A'$ with two vertices of $\triangle XYZ$, then that will form a triangle larger than $\triangle XYZ$, contradicts the maximality of $[XYZ]$.