# There are at least three convex vertices in a polygon

Call a vertex $A$ convex if the interior angle at the vertex is less than $180^\circ .$ Now note that a polygon with $n$-vertices and $n \geq 4$ has the sum of interior angles equal to $(n-2)180^\circ$. Therefore a polygon has at least $3$ convex vertices.

I do not understand why it has at least $3$ such vertices. I believe that the pigeonhole principle can be applied however I do not see how. Can someone explain this?

If there were less than two convex vertices, there would be at least $n-1$ non-convex vertices, which would have interior degree sum at least $(n-1)(180)$, which exceeds $(n-2)(180)$, contradiction.
If there were exactly two convex vertices, there would be exactly $n-2$ non-convex vertices, which would have interior degree sum at least $(n-2)(180)$. But then the remaining two vertices would boost the interior degree sum to greater than $(n-2)(180)$, contradiction.