With N teams there are
a total of $N(N-1)/2$ total games to be played. Any particular team will play $N-1$ games. Assume a team is awarded 2 points for a win, 1 point for a draw, and 0 points for a loss. Thus each game awards a total of 2 points, and there are a total of $N(N-1)$ points to be awarded.
If I draw twice and win the rest of my games, I will get $2(N-3) + 2$ points. The remainder of the points need to be distributed amongst the remaining $N-1$ teams.
If there are fewer than 7 teams, there is no way to allocate the remaining points in such a way that all of the teams would have fewer than my number of wins and my number of draws.
With 7 teams, if I draw twice and win the other four games, I'll have a total of 10 points. There are 32 points left to distribute among the remaining 6 teams, meaning an average of 5.33 points per team. This can be done with 2 teams with 3 wins and 0 draws each, and 4 teams with 2 wins and 1 draw each.
Edited to improve solution:
It actually is possible to allocate the points when N = 6, I made an arithmetical error previously which led me to believe it wasn't. For N = 6, you can draw twice and win three times for a total of 8 points. The remaining 22 points can be distributed between the other 5 teams by having two teams with two wins and a draw each, and three teams with two wins.
For N = 5, if you win 2 and draw 2 you have 6 points. There would be 14 points remaining to distribute among the other four teams, meaning they'd have 3.5 points on average. At least one of the teams would have 4+ points, which is only possible if they have at least two wins or at least two draws.