Prove that every tournament contains at least one Hamiltonian path. A tournament is a directed graph with exactly one edge between every pair of vertices. (So for each pair  (u,v)  of vertices, either the edge from  u to  v  or from  v  to  u  exists, but not both.) You can think of the nodes as players in a tournament, where each player has to play against every other player. The edge points from the winner to the loser of a game. 
A Hamiltonian path (not cycle) is a sequence of consecutive directed edges that visits every vertex exactly once.
How can i prove that every tournament contains at least one Hamiltonian path?
thanks your help!  

A bit complicated for this problem, but the idea is often useful for other problems.
Let $V=\{v_1,v_2,\ldots,v_n\}$ be the set of vertices and $E$ be the set of edges of the graph $G$. Consider all permutations $v_{\pi(1)},v_{\pi(2)},\ldots,v_{\pi(n)}$ (here $\pi\in S_n$) and for each one count the number of "left-right" arrows, i. e. denote
$$
f(\pi):=|\{(i,j)\mid 1\leq i<j\leq n, \overrightarrow{v_iv_j}\in E\}|.
$$
Now choose such a permutation $\pi$ for which the $f(\pi)$ is maximal. We claim that $v_{\pi(1)},v_{\pi(2)},\ldots,v_{\pi(n)}$ is a desired Hamiltonian path. Indeed, if for some index $i\in\{1,2,\ldots,n-1\}$ we have $\overrightarrow{v_{\pi(i)}v_{\pi(i+1)}}\notin E$, then $\overrightarrow{v_{\pi(i+1)}v_{\pi(i)}}\in E$. Hence, if we "swap" $\pi(i)$ and $\pi(i+1)$ the value $f(\pi)$ will increase, which is impossible. 
To be precise, consider another permutation $\pi'\in S_n$ defined as follows:
$\pi(j)=\pi'(j)$ for $j\neq i, i+1$ and $\pi'(i)=\pi(i+1)$, $\pi'(i+1)=\pi(i)$. Now, note that $f(\pi')=f(\pi)+1>f(\pi)$.
Therefore, our tournament contains Hamiltonian path $v_{\pi(1)},v_{\pi(2)},\ldots,v_{\pi(n)}$, as desired.
