prove that in a complete oriented graph there is triangle prove that in a full complete oriented graph there is a triangle .
I tried to use something like red and blue edges. But I can't figure out how to choose directions of edges and which colour is every edge.
 A: If by a "triangle" you mean a (directed) $3-$cycle, then this is false. Line the $n$ vertices up from left to right and make every edge a forward (left to right) edge. This tournament clearly does not have a $3-$cycle.
EDIT : As @Brandon pointed out in the comments, the statement is true if the graph has at least one cycle.
To see this, suppose we're given the cycle $u_k\rightarrow u_{k+1}\rightarrow u_{k+2}\cdots \rightarrow u_{k+m}\rightarrow u_{k}$, arrange the vertices of the cycle from left to right (the last edge in the sequence is a back edge).
If we have a forward edge $u_{k+1}\rightarrow u_{k+m}$ then we are done, since we then have the 3-cycle $u_k\rightarrow u_{k+1}\rightarrow u_{k+m}\rightarrow u_k$.
Otherwise, we have a smaller cycle $u_{k+1}\rightarrow u_{k+2}\cdots \rightarrow u_{k+m}\rightarrow u_{k+1}$
Continuing this reasoning, one can also see that unless all the edges from $u_{k+m}$ to the vertices $u_{k+2}, u_{k+3}\cdots$ in the cycle are back edges, you'll eventually find a 3-cycle. But if all of them are back edges, then we have the following $3-$cycle --
$u_{k+m-2}\rightarrow u_{k+m-1}\rightarrow u_{k+m}\rightarrow u_{k+m-2}$.
So we're done.
A: I will proove it using induction.
Base: For graph of size $3$ it is true.
Hypothesis: Assume for size $k$ it is true.
Induction: In graph of size $k+1$ consider any three consicutive vertices of cycle. Let $a, b, c$ such vertices with orientation $(a-b-c)$.
Case 1: If edge $ac$ has orientation $(a-c)$ it creates cycle with $k-1$ vertices and according to hypothesis this cycle contains a triangle.
Note: This cycle does not contain $b$. i.e. $cycle-b$
Case 2: If edge ac has orientation $(c-a)$ then $abc$ creates triangle.
A: An oriented complete graph (a.k.a. tournament) does not necessarily contain a directed $3$-cycle, but the counterexamples are very special.
Suppose your oriented complete graph contains no directed $3$-cycle. For vertices $a$ and $b$, define $a\lt b$ to mean that there is an arc from $a$ to $b$. Since the graph is complete and has no directed $3$-cycles, whenever we have $a\lt b$ and $b\lt c$ we must also have $a\lt c$. That is, $\lt$ is a transitive relation, and therefore it is a linear order relation, the other required properties being obvious.
So the only oriented complete graphs without directed $3$-cycles are linear orders (a.k.a. transitive tournaments), which have no directed cycles at all.
