I want to prove the following result:
Let $G$ a simple graph $n$ vertices, $n$ even, and $d(v) > n/2~\forall v \in V(G)$. Prove that $G$ has three perfect matches two on two disjoint.
I must say I am a bit clueless in this question. I am able do most others questions with not too much trouble. I thought about doing it by induction in the number of vertices, but I do not think it will work. I know for example, a $K_4$ works and that it could be my basis for induction, but I cannot find a way build a larger case (n+2) based on the previous (n).
I do not exactly know how to think about a contradiction either.
By Dirac theorem $G$ has a hamiltonian cycles $C$. But $C = M_1 \cup M_2$ where $M_1$ and $M_2$ are disjoint matchings. Now consider the graph $H = G - M_1$. We have $d(v) \gt \frac{n}{2} - 1$ for all $v \in V(H)$ (which means $d(v) \ge \frac{n}{2}$). Therefore Dirac theorem applies again and $H$ has a hamiltonian cycle $C_2$ which again is the union of two matchings $M_3$ and $M_4$. Thus $M_1$, $M_3$ and $M_4$ are three disjoint matching of $G$