I'm trying to do the next problem, it seems easy but I can't do it!
Let G be a simple graph with $n\geq 2\delta$. Show that $\alpha' \geq \delta$
$\delta=$minimum degree in $G$
$\alpha'=$ the number of edges in a maximum matching of $G$
and $n$ the number of vertices of G.
I thought this: if $M$ is a maximum matching of $G$ and suppose $|M| < \delta$, then, somehow show that any two vertices not covered by M are connected by an M-augmenting path and derive a contradiction, but, even this, I can't see how to do it.
Any help would be appreciated! Thanks!