TITLE: $\alpha^{'}(G)\geq\frac{n}{1+\Delta(G)}$
QUESTION [6 upvotes]: Let $G$ be a connected graph of order $n$ and $\alpha^{'}(G)$ be the edge independence number(or matching number). Can you help me prove the inequality $\alpha^{'}(G)\geq\frac{n}{1+\Delta(G)}$ where $\Delta(G)$ is the maximum vertex degree?

REPLY [5 votes]: The answer by Adam Lowrance proves the following two theorems:
Tree Theorem. If $G$ is a tree with $n$ vertices, $n\ge2,$ then
$$\alpha'(G)\ge\frac n{\Delta(G)+1}.$$
General Theorem. If $G$ be a connected graph with $n$ vertices, $n\ge2,$ then
$$\alpha'(G)\ge\frac n{\Delta(G)+1}.$$
Here is a short proof of the Tree Theorem. Let $e(G)=|E(G)|$ be the size of $G,$ and let $\chi'(G)$ be the edge chromatic number of $G.$ If $G$ is a tree, then we have $e(G)=n-1\ge\Delta(G)$ and $\chi'(G)=\Delta(G),$ whence
$$\alpha'(G)\ge\frac{e(G)}{\chi'(G)}=\frac{n-1}{\Delta(G)}\ge\frac n{\Delta(G)+1}.$$
Here is a short proof of the General Theorem using Vizing's theorem. Without loss of generality, we may assume that $G$ is a simple graph; moreover, we may assume that $G$ is not a tree. By Vizing's theorem, $\chi'(G)\le\Delta(G)+1.$ Therefore, since $G$ is not a tree, $e(G)\ge n$ and
$$\alpha'(G)\ge\frac{e(G)}{\chi'(G)}\ge\frac n{\Delta(G)+1}.$$
Alternatively (without Vizing's theorem), let $T$ be a spanning tree in $G;$ then
$$\alpha'(G)\ge\alpha'(T)\ge\frac n{\Delta(T)+1}\ge\frac n{\Delta(G)+1}.$$
Corollary. If $G$ is a graph with $n$ vertices, and if each connected compenent of $G$ has at least $2$ vertices, then
$$\alpha'(G)\ge\frac n{\Delta(G)+1}.$$
Proof. Let $G_1,G_2,\dots,G_k$ be the components of $G$; suppose $G_i$ has $n_i$ vertices, $\alpha'(G_i)=\alpha'_i$, and $\Delta(G_i)=\Delta_i$. Then
$$\alpha'(G)=\sum_{i=1}^k\alpha'_i\ge\sum_{i=1}^k\frac{n_i}{1+\Delta_i}\ge\sum_{i=1}^k\frac{n_i}{1+\Delta(G)}=\frac n{1+\Delta(G)}$$