max degree and edge coloring 
Let G be a graph with max degree $\Delta$ , then there exists a valid $(\Delta+1)$-coloring of G's edges such that each color appears $\lceil{\frac{|E|}{\Delta+1}}\rceil$ or $\lfloor{\frac{|E|}{\Delta+1}}\rfloor$ times.

I have no clue what to do here...
 A: for any graph $\chi'(G) \ge \Delta(G)$
and by Vizing theorem $\chi'(G) \le \Delta (G) + 1$
this yields: $$\Delta \le \chi'(G) \le \Delta +1$$
the hardest part is to realize you don't need to prove that $\chi' = \Delta +1$
but that there exists some "legal" coloring that uses $\Delta + 1$ colors.
so if we can color it in $\Delta$ or $\Delta + 1$ different colors we can do it in $\Delta + 1$ for sure because if $\chi'(G) = \Delta$ we can recolor 1 edge in new color and get a new legal coloring with $\chi(G) = \Delta + 1$  (its just not a coloring that uses the minimal number of colors).
now to the second part . given a graph $G$ that have a maximal degree of $\Delta$
and have a legal coloring with $\Delta + 1$ colors we will think of a way to color it so all colors will will appear equal number of times since the number of edges in the graph  is not always divisible by $\Delta + 1$ we can only hope to achieve that the number of an edges of each color will differ at most by 1 for any two colors and that's exactly what we asked the question.  
as for the coloring itself one of the ways to do it is just to choose an vertex at random and start coloring all its edges 


*

*first edge in the first color

*second edge in the second color

*last edge in $i$'th color($i \le \Delta$)


now choose one of its neighbors and repeat this possess but start coloring from the color number $i+1$


*

*first edge in color $i+1$

*second edge in color $i+2$ and so on 


when you reach the color number $\Delta + 1$ just start over(color the next edge in first color).
when we complete this process we will get the required coloring
A: Hint :
Assume there is a color that appears more than $\lceil{\frac{|E|}{\Delta+1}}\rceil$ times , so there is also a color that appears less time than $\lfloor{\frac{|E|}{\Delta+1}}\rfloor$ times.
All that is left is to look at the connected components that that is induced by only edges that are colored by both those color and see how one may reduce the number of edges colored by the first color and increase the number of edges colored by the other color.
