# Hamiltonian cycles and paths in a graph

I want to present you a lemma, that I've almost proved, but i'm stuck at the very end of it.

Lemma If vertex $$v$$ of a graph $$G$$ is not isolated and degree of every vertex except $$v$$ is $$\geq k$$ (for $$k \geq 2$$), and if $$|V(G)| \leq 2k-1$$, then $$v$$ is connected by hamiltonian paths with every vertex from $$G$$.

Proof: Let $$G'$$ be graph $$G-v$$. $$V(G') \leq 2k-2$$ , and for any two vertices in $$G'$$: $$deg(x)+deg(y) \geq 2k-2$$. It can be proved from the theorem that $$G'$$ is hamiltonian (it's not the point of issue). Now let us connect to graph $$G'$$ vertex $$v$$, it'll be connected with vertices on the hamiltonian cycle, and the degree of every vertex on cycle is $$\geq k$$.

And now it should be easy to show the main thesis. ($$v$$ is connected by hamiltonian paths with every vertex from $$G$$)

Maybe someone of you knows from which theorem or lemma does it result? For now, I can't see it clearly, and it seems very simple.

Thank you in advance for every suggestion.