Let $G=(V,E)$ be a finite graph. We write $\nu(G)$ for the matching number of $G$. Is there $\varepsilon > 0$ such that we have $$\frac{\nu(G)+\Delta(G)}{V(G)} \geq \varepsilon$$ for all finite graphs $G=(V,E)$?



Maximum degree and matching number
The answer is no (even for connected graphs: if you consider general graph, simply take $E$ empty for a counter-example).


Let $n,d>2$ be any positive integers, define $G\_1$ as a path of length $\ell$, and let $G$ be the tree obtained by adding $d-2$ leaves adjacent to each vertex of $G\_1$. Then we have $\Delta(G)=d$, $V(G)=(d-1)n$ and $\nu(G)=n$ (indeed, any edge of a matching of $G$ must have one vertex of $G\_1$ has an endpoint). We get
$$\frac{\nu(G)+\Delta(G)}{V(G)} = \frac{n+d}{n(d-1)}$$
and this can be made arbitrarily small. 


