For every graph $G$, prove that $\beta(G)\leq2\alpha'(G)$. ($\beta(G)$ is the minimum cardinality of a vertex cover, $\alpha'(G)$ is the number of edges in a maximum matching of $G$.)
I went in circles:
Consider a (connected) graph with $n$ vertices. Gallai's Theorem tells us $\alpha(G)+\beta(G)=n=\alpha'(G)+\beta'(G)$. So write $\beta(G)=\alpha'(G)-\alpha(G)+\beta'(G)\Rightarrow-\beta(G)=-\alpha'(G)+\alpha(G)-\beta'(G)$. Theorem 10.12 (just some theorem in our book) tells us $\beta(G)\geq\alpha'(G)$, so $-\alpha'(G)\geq-\beta(G)=-\alpha'(G)+\alpha(G)+-\beta'(G)$. $-\alpha'(G)\geq-\alpha'(G)+\alpha(G)-\beta'(G)$, and...
obviously I'm just going in circles, because the other half of Theorem 10.12 says $\beta'(G)\geq\alpha(G)$, which is what would fall out of my argument above...not helpful. So how do I go about doing this? The fact we are proving this for any graph leads me to believe I need to use some previously established inequalities here, and this "Theorem 10.12" is the only inequality that we've really done in this chapter.
The point here is that if you take a maximum matching $M$ and look at the set $N$ of all vertices incident to edges in $M$, then $N$ is a vertex cover (as otherwise there is an edge with neither of its incident vertices in $N$. But then we could have added that edge to $M$ to form a bigger matching, contradicting the maximality of $M$).
So $N$ is a vertex cover of size $2\alpha^\prime(G)$, and so certainly $\beta(G)\leq 2\alpha^\prime(G)$ since $\beta(G)$ is the size of a smallest possible vertex cover.