\documentclass{article}
\usepackage{latexsym,eepic,makeidx,fancybox}
\usepackage[dvips]{graphicx} \usepackage[latin1]{inputenc} \usepackage[english]{babel} \usepackage{amstext,amssymb,amsmath,amsthm,youngtab}
\title{}
\date{}
\author{}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}{Proposition}[section]
\newtheorem{lemma}{Lemma}[section]
\newtheorem{conjecture}{Conjecture}[section]
\newtheorem{definition}{Definition}[section]
\newtheorem{remark}{Remark}[section]
\newtheorem{claim}{Claim}[section]
\newcommand{\p}{^\prime}
\newcommand{\pp}{^{\prime\prime}}
\newcommand{\lra}{\leftrightarrow}
\newcommand{\mbf}{\mathbf}
\newcommand{\mbb}{\mathbb}
\newcommand{\mcl}{\mathcal}
\begin{document} 
\maketitle

Let $G$ be a graph such that $d(x) \geq n/2$ for each $x \in V(G)$, where $n = |V(G)|$. Then $G$ has a Hamiltonian cycle.

We prove this by finding a Hamiltonian path. This is sufficient: suppose $x_1\dots x_n$ is a Hamiltonian path in $G$ and consider the two sets $S = \{i:x_1x_i\in E(G)\}$ and $T = \{i:x_{i-1}x_n \}$. We can assume $n\notin S$ and $2\notin T$. Then there is an $i \in S\cap T$, which gives us a Hamiltonian cycle $x_1x_i\to x_nx_{i-1} \leftarrow x_1$.

Now we find a Hamiltonian path. Clearly, there is a path of length $\geq n/2$, since all degrees are $\geq n/2$. Let $P = x_1\dots x_r$ be a longest path, where $r \geq n/2$.
Let $S$ be the set of $i$ such that $x_i$ is connected to some vertex outside $P$.
We may assume that $1, r\notin S$.
It is now sufficient to consider the following picture: A vertex in $S$ adjacent to a vertex outside $S$ on $P$, together with their edges to vertices in $P$, with special attention to $x_1$ and $x_r$.


\end{document}
