I was reading a result where the following proposition appears as a preliminary step (and left as exercise):
Claim: Suppose $G$ is a graph on $n$ vertices ($n$ even and $n \geqslant 3$) with minimum degree at least $n/2$. Show that $G$ contains a perfect matching.
Proof: By Dirac's theorem, $G$ has a Hamiltonian cycle $C$. Since $|C|=n$ is an even integer, the set of “odd” edges of $C$ gives a perfect matching for $G$. $\quad\Box$
I feel that using Dirac's theorem for this claim is an overkill. But after trying for a few days, couldn't come up with any other proofs. Can you give a more “direct” proof of the claim that avoids the machinery of Hamiltonian cycles?
Naturally, you might object to the vague requirement of “directness”. To clarify what I mean by it, I’ll give an example.
Claim 2. Suppose $G$ has a minimum degree at least $n/2$. Show that $G$ is connected.
Proof via Dirac's theorem. $G$ has a hamiltonian cycle as before, and hence is connected. $\Box$
A different proof. We'll show that any pair of vertices $u,v$ are connected by a path of length at most $2$. If $uv$ is an edge, we are done. Suppose not. Then $N(u) \cup N(v) \subseteq V \smallsetminus \{u,v\}$. Therefore $ |N(u) \cap N(v)| = |N(u)| + |N(v)| - |N(u) \cup N(v)| \geqslant \frac{n}{2}+\frac{n}{2}-(n-2) = 2 \gt 0. $ In particular there exists vertex $w$ such that $uw$ and $wv$ are both edges, and we are done. $\quad \Box$
I find the proof via Dirac's theorem much less illuminating in this example. In fact, as Qioachu points out below, the Claim 2 might even appear as an intermediate steps of Dirac’s theorem, making the above proof cyclic. My intention here is only to point out that Dirac’s theorem can be used as a powerful black box in killing much easier results.