In a group of $n$ people, each pair are friends or strangers. No set of three people are mutually friends. For any partition of the $n$ people into two groups, there exists two people in a group that are friends. Prove that there exists a person who is friends with at most $\frac{2n}{5}$ people in the group.
Assume for contradiction this is false.
Attempt 1: Let $P$ be a person, and let $A$ be the set of his friends and $B$ be the set of his strangers. Since this graph is triangle-free, $A$ is an empty graph. Every vertex in $A$ must then connect to more than $\frac{2n}{5}-1$ people in the second graph. This means every pair of vertices in $A$ is connected to more than $\frac{n}{5}-2$ people in the second graph. Then I'm stuck.
Attempt 2: The graph must have an odd cycle. Start with the minimal odd cycle $C_x$, where $x$ is odd. Then add vertices one by one, and use induction to show that at every step there must be a vertex with degree $\le\frac{2n}{5}$
Help please. I'm not sure how to proceed from either path. Any hint would be appreciated!