A graph is called 10-tough and $(2P_2 \cup P_1)$-free if every vertex set whose deletion leaves at least two components has cardinality at least ten times the number of those components and if the graph has no induced subgraph consisting of two disjoint edges and an isolated vertex. We prove that every finite simple 10-tough $(2P_2 \cup P_1)$-free graph on at least three vertices is Hamiltonian. The proof splits according to whether some edge has joint neighbourhood of order at most $4n/11$. In the small-neighbourhood case, a matched path-cover is compressed to a prescribed matching. In the large-neighbourhood case, an asymmetric analysis of the two components left by a putative small cut yields the required connectivity bound. A Hamilton cycle through the prescribed edges is then expanded, and a cycle-extension lemma inserts the remaining vertices.

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