Chavatal's theorem

 Let $G$ be a simple graph with vertex degrees $d_1\le d_2\le\dots\le d_n$, where $n\ge 3$. If for every $i<n/2$ we have

$$i< d_i\text{ or } n-i\le d_{n-i}$$ then $G$ is Hamiltonian.

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Proof

By Ore's lemma (for nonadjacent vertices $x,y$ satisfying $\deg(x)+\deg(y)\ge n$, $G$ is Hamiltonian if and only if $G+xy$ is Hamiltonian), it suffices to prove that the Hamiltonian closure $C(G)$ of $G$ is Hamiltonian. Hence below we assume that $G=C(G)$, i.e., if $\deg(x)+\deg(y)\ge n$ then $x$ and $y$ are adjacent.

We first claim that $1<d_1$. If not, then $n-1\le d_{n-1}\le d_n$ and vertex $1$ has at least two neighbors, a contradiction. 

Then we show that there does not exist $i+1<n/2$ such that $i<d_i$ and $i+1\ge d_{i+1}$. Suppose that such $i$ exists, then $d_i=d_{i+1}=i+1$ and $d_{n-i-1}\ge n-i-1$. So vertex $i$ is adjacent to vertices $n-i-1,n-i,\dots,n$ and thus has degree at least $i+2$, a contradiction.

Now we have $i<d_i$ for every $i<n/2$, and we will show that $G=K_n$, i.e., $d_1=n-1$. We aim to find an $i<n/2+1$ with $d_i\ge n-i$. This, together with $d_{i-1}\ge i$, implies that vertex $i-1$ is adjacent to vertices $i,i+1,\dots,n$ and thus has degree at least $n-i+1$, and ultimately we will have $d_1\ge n-1$. If $n$ is even, then $n/2\le d_{n/2-1}\le d_{n/2}$, so $i=n/2$. If $n$ is odd, then $(n+1)/2\le d_{(n-1)/2}$, so $i=(n-1)/2$.