Question about matches (Graph theory) :
Let $\alpha_{1}$ denote the maximum number of edges in a matching. Prove that $\alpha_{1}=\min\{\delta(G),\left\lfloor\frac{|G|}2\right\rfloor\}$, if $G$ is a complete multiparty graph.
+++ I know that if the graph is complete multipartite then the vertices of minor degree are exactly those that are in the set with more vertices and it had to see cases and sea that the degree is greater, less than or equal to the floor of $\frac{|G|}{2}$ but I don't know how to prove that the greatest number of edges of a mating is the minimum degree or the floor of $\frac{|G|}{2}$.
Someone could help me? Thank you so much. +++
We assume that $|G|$ is the size of the set $V$ of vertices of $G$ and $\delta(G)$ is the smallest degree of a vertex of $G$. Let $V_1$ be the largest multipartite part of vertices of $G$. The multipartiteness of $G$ implies that $\delta(G)=|V\setminus V_1|$
To show the upper bound for $\alpha_1(G)$ note that each mathching $M$ for $G$ has no more than $|G|/2$ edges. Also each edge of $M$ has at least one endvertex in $V\setminus V_1$ and so $|M|\le |V\setminus V_1|=\delta(G)$.
Let’s show the lower bound for $\alpha_1(G)$. If $\delta(G)\le |G|/2$ then since each vertex of $V\setminus V_1$ is adjacent to any vertex of $V_1$ and s $$|V\setminus V_1|=\delta(G) \le |G|/2\le |V|-\delta(G) =|V_1|,$$ there exists a matching of size $|V\setminus V_1|$ consisting of edges with one endvertex in $|V\setminus V_1|$ an the other endvertex in $V_1$. If $\delta(G)\ge |G|/2$ then we can construct a matching $M$ of size $\lfloor|G|/2\rfloor$ consecutively adding to $M$ an arbitrary edge $e$ with endvertices in two largests multipartite part of vertices of $G$ and removing from $G$ the endvertices of $e$. It is easy to check that this operation keeps both multipartiteness of $G$ and the inequality $\delta(G)\ge |G|/2$, so we can proceed till $M$ there remains at most one vertex, assuring the other vertices of the initial graph were removed and so they are endvertices of edges of $M$.