Prove a graph $(V,E)$ with $d$-maximal degree let $k=d/2+1$ can be decomposed as $V=V_1 \cup\cdots \cup V_k$ where each $V_i$ is a loopless graph I tried looking at a vertex v with the maximal degree, that is $d(v)=d$ and started looking at it's neighbours
$$N(v):=\{u\mid (u,v) \in E\}$$
therefore $|N(v)|=d$, now between every two vertice $z,w\in N(v)$ 
Where are no cycle so you can devide the nieghbour group into $d/2$ subgroups, 
now I get stuck with the other $V-\{v\}-N(v)$ vertices and I was trying to somehow put then into these groups but got stuck.
Help anyone?
 A: You can use the Nash-Williams arboricity theorem (see here), in particular, for any subgraph $H \subseteq G$ we have
\begin{align*}
  \left\lceil\frac{|E_H|}{|V_H|-1}\right\rceil
  &= \left\lceil \frac{\frac{1}{2}\sum_{v \in V_H}\deg_H(v)}{|V_H|-1}\right\rceil \\
  &\leq \left\lceil \frac{\frac{1}{2}d_H\big(|V_H|-1\big)+\frac{1}{2}d_H}{|V_H|-1}\right\rceil \\
  &\leq  \left\lceil \frac{1}{2}d_H + \frac{1}{2}\right\rceil
  &\leq k 
\end{align*}
where $\displaystyle d_H = \max_{v \in V_H}\deg_H(v)$.
I hope this helps $\ddot\smile$
A: I was thinking long and hard about it and I realised that induction can be applied:
for n=3 It's obvious we have three vertice $ v_1 , v_2, v_3 $
Highest possible degree here could be 2 so:
$ V_1=v_1 $ $V_2$={$v_2, v_3$} 
Suppose that the claim is true for all m where m < n:
Let'a look at our G=(V,E) where |V|=n.
Take a vertex $ v\in V $ such v has the minimal degree so we know that d(v)<=d.
If we now look at V-{v} this graph has m=n-1 vertices and still has maximal degree of d (we didn't delete the vertex with the maximal degree) therefore by induction where's a decomposition into k=d/2+1 disjoint vertice group. Now if the v we removed closes a cycle in each of $ V_1, V_2,..., V_k $ it must be that it has at least 2 edges going into each component whereby it has to have at least 2k edges meaning d(v)>2k-1=d+1 but we don't have that many edges (bouded by d edges max degree) so where exists a component $V_j$ such that v joined with it doesn't close a cycle, so none of the groups have a cycle and where is exactly k disjoint vertice groups as such.
A: This shows what happens when we removed the minimal vertex and it had an edge touching the maximal degree:

