Does 3-partite graph with at least n+1 edges per vertex have a triangle? I need help for one problem.

In every of 3 schools there are n students (in total = 3n vertex, n per school).
  Every student knows at least n+1 students from the other two schools.
  Prove that there are 3 students, one from each school, who know each
  other.

Do you have some idea for this proving?

Let $G$ be a counterexample, i.e. a 3-partite graph with $n$ vertices in each partite set
and minimum degree $n+1$ that has no triangle.
Let $v$ be a vertex that maximizes the number of neighbors it has in one partite set, say $k$.
We may label the partite sets $A,B,C$, such that $v\in A$ and $v$ has $k$ neighbors in $B$.
Since $k\leq n$, $v$ has at least one neighbor in $C$, say $w$.
$w$ has at most $n-k$ neighbors in $B$ (otherwise we find a triangle involving $v,w$ and a vertex of $B$),
so $w$ has at least $(n+1)-(n-k)=k+1$ neighbors in $A$. This contradicts the choice of $v$.
Done.
(This problem awoke my interest in the sharpness question: is it true that for a given $n$ and even $k<3n(n+1)$
there always is a 3-partite graph with $n$ vertices in each partite set and total degree at most $k$
(and minimum degree at least $n$?) that has no triangle?)
