Show that a graph with exactly one vertex of degree $i$ : $2\le i \le m$ and $k$ other vertices (which are monovalent) satisfies $\left\lfloor\frac{m+3}2\right\rfloor \le k\;.$
Here is my current approach. I want to a construct a graph of minimal $k$ - to do this I need to make sure that all the all the other vertices of degree $i$ connect to the vertex of degree $m$, and that the other vertices of degree $m-1$, $m-2$, ... share the largest number of vertices that won't result in a double. Here is my current construction.
Start with a vertex of degree $m$ $v_0$. choose some $v$ adjacent to that vertex, $v_1$. Connect it to $m-2$ of the other vertices giving it degree $m-1$ and creating $m-2$ vertices of degree $2$. At the next step, choose another of the vertices of degree $2$, call it $v_3$. connect it to $m-4$ vertices of degree $2$, leaving $1$ vertex of degree $2$, and creating $m-4$ vertices of degree $3$.
This produces a sequence of graphs $G_n$, where each graph contains exactly one vertex of degree $1,\dots ,n+1$ and degree $m,\dots ,(m-n)+1$.
Now at some point, we will reach an $i$ s.t $m-i = i+1$, at which point we will have two duplicate vertices. We then only have to move at most $i$ edges in order to get a graph with the "minimal number" of monovalent vertices satisfying this property.
Problem is I can't prove this graph is minimal, and I can't figure out how to get the bound to arise from this construction.