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Let $v_1\cdots v_k$ be a longest path. Every neighbour of either endpoint lies on it. Among the $k-1$ indices $i=1,\dots ,k-1$, mark those with $v_1{v}_{i+1}$ an edge and those with $v_i{v}_k$ an edge. Their total number isso some index has both marks. The edges at that index close the path into the cycle graphThe minimum-degree hypothesis also forces the graph to be a connected graph: each component would otherwise have at least $n/2+1$ vertices. If $k<n$, an edge from this cycle to an outside vertex would give a longer path by opening the cycle at its incident vertex. Hence $k=n$, proving the Dirac theorem and Hamiltonicity.

The bound is sharp. For $n=2m$, take two disjoint copies of $K_m$, with minimum degree $m-1$. For $n=2m+1$, take two copies of $K_{m+1}$ sharing exactly one vertex. The minimum degree is $m$, but the shared vertex is a cut vertex, which no Hamiltonian graph can have. Both constructions have

Let $v_1\cdots v_r$ be a longest path in the finite simple graph. Every neighbour of either endpoint lies on it. Among indices $1\le i<r$, consider those with edges $v_1{v}_{i+1}$ and those with edges $v_i{v}_r$. Their cardinalities are the endpoint degrees, whose sum is at least $n\ge r$, exceeding the $r-1$ available indices. They intersect, producing the cycleEvery connected component has at least $\delta (G)+1>n/2$ vertices, so the graph is connected. If $r<n$, an edge joins a vertex outside this cycle to it, and breaking the cycle there gives a longer path, a contradiction. Hence $r=n$ and this is a Hamilton cycle, proving Dirac theorem.

The complete graph $K_4$ is planar, has chromatic number four and a Hamilton cycle. Adding one leaf attached to one of its vertices preserves planarity and chromatic number four but prevents a Hamilton cycle, since a leaf has degree one.

For the last claim, the given plane embedding has all vertices on the outer cycle. Add noncrossing diagonals until its polygonal interior is triangulated. A triangulated polygon has a vertex incident with only its two boundary neighbours: for example the dual graph of its triangles is a tree and a leaf triangle supplies an ear. Remove this vertex. Inductively three-colour the smaller triangulated polygon, starting with a triangle. The removed vertex's neighbours are adjacent and have different colours, so it can receive the third. Restriction to the original graph givesThis uses the specified embedding, not the false assertion that every planar Hamiltonian graph is three-colourable; $K_4$ is already a counterexample to that broader assertion.

Letbe a longest path. Every neighbour of either endpoint lies on $P$. Among the $m-1$ possible cuts between consecutive vertices, mark a cut $i$ when $v_m{v}_i$ is an edge and also mark it when $v_1{v}_{i+1}$ is an edge. There are at leastmarks, so some cut receives both marks. The path edges together with $v_m{v}_i$ and $v_{i+1}{v}_1$ form a cycle through all vertices of $P$.

The degree assumption makes $G$ connected: two components would each contain at least $n/2+1$ vertices. If $m<n$, connectedness gives an edge from a vertex outside the cycle to a vertex on it; breaking the cycle there produces a path with $m+1$ vertices, contrary to maximality. Hence $m=n$, so $G$ is Hamiltonian. This proves the Dirac theorem in this case.

This is Dirac theorem. Take a longest path $v_1\dots v_k$. All neighbours of its ends lie on it. If no index $i$ has both $v_1{v}_{i+1}$ and $v_k{v}_i$, the two end-neighbour sets have total size at most $k-1$, contradicting $2\delta \ge n\ge k$. Such an index closes a cycle through the path, and maximality plus connectedness makes it Hamiltonian. For even $n$, the disjoint union of two copies of $K_{n/2}$ has minimum degree $n/2-1$ and is not Hamiltonian. For odd $n$, $K_{(n-1)/2,(n+1)/2}$ has minimum degree $(n-1)/2$ and no Hamilton cycle because its bipartition sizes differ.