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How can I prove this?

Given a simple graph $G=\langle V, E\rangle$ where $V$ is a set of vertices, $|V|=n$, and $E$ is a 2-element subset of $V$ with the edges of the graph. Also $\forall x,y\in V(G)$ if $\langle x,y\rangle\notin E(G)\Rightarrow\deg(x)+\deg(y)\ge n-1$. Prove there exists a simple path $p$ with length $n$.

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    $E$ is the set of edges of $G$ so $\langle x,y\rangle\in E\Rightarrow x,y\in V$ and $x\ne y$. Also $\langle x,y\rangle =\langle y,x\rangle$ because is a simple graph. – 2017-02-12
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    Yes, you are right, I made a mistake. – 2017-02-12
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    By "path of length $n$" do you mean the path has $n$ edges, or do you mean it has $n$ vertices? – 2017-02-12
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    Related: Ore's Theorem, https://en.wikipedia.org/wiki/Ore's_theorem – 2017-02-12
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    @bof I mean the path has all $n$ vertices of $G$. – 2017-02-12

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