We get the least number of triangles when $n-1$ of them are collinear and one is not. Then, we have $\binom{n-1}{2} = \frac{(n-1)(n-2)}{2}$ triangles

L567 wrote: We get the least number of triangles when $n-1$ of them are collinear and one is not. Then, we have $\binom{n-1}{2} = \frac{(n-1)(n-2)}{2}$ triangles Can induction work? Oh yes induction can work:every time we add a new point($n$ point before) we can prove that there are at least $n-1$ new triangles.

Like as above do induction add a point.If it is not on any line (which contains at least 3 ponts). Then there will be $\binom{n-1}{2}$ Otherwise it is on this line.Let there are x Points on it which is not our added point and n-x-1 point which is not on this line then there are x(n-x-1) we know that $n-1>x>1$. Then the minimum is when the x and n-x-1 are farther or in other word n-x-1=1.Then we added at least n-2 point. and by induction we have $\binom{n-2}{2}$ add it then we are done.Construction:N-1 points on line 1 point not on this line.Then we are completely done

Mine is very similar to above: We claim that the answer is $\binom{n-1}{2}$. This can easily be achieved when $n-1$ of the points are collinear. Let there be $n$ points. Note that every time we add a new point, if the new point is not on any line, then the total number of triangles will increase by $\binom{n}{2}$. If the new point is on some line which contains in total $k$ points, then the total number of triangles will increase by $\binom{n+1-k}{2} + (k-1)(n+1-k) = \frac{(n+1-k)\cdot (n+k-2)}{2} = \frac{1}{2}\cdot (n^2 - n - 2 + 3k - k^2)$ but $3k-k^2$ attains it's minimum value when $k$ is as large as possible, e.g $k=n$. When $k=n$, the total number will increase by $n-1$ which is strictly less than $\binom{n}{2}$. This completes the proof.