https://en.wikipedia.org/wiki/Edge_coloring#Vizing's_theorem

So, here, minimum number of colors can be 7 or 8. So, we have to convince ourselves that minimum number of colors required to proper color the edges of the given graph can't be 8 by making atleast one graph which required only $7$ colors for proper edge coloring.

https://gateoverflow.in/?qa=blob&qa_blobid=10132893589318938572

-- Suppose, maximum degree of a graph $G$ is $\Delta(G).$

if $|E| > \Delta(G)*\left \lfloor \frac{|V|}{2} \right \rfloor $ then graph $G$ requires $\Delta(G) + 1 $ minimum colors for edge coloring of graph. But if $|E| \ngtr \Delta(G)*\left \lfloor \frac{|V|}{2} \right \rfloor $ then graph $G$ requires $\Delta(G) $ . It is wrong statement. Counter-example is Petersen graph with 10 vertices, 15 edges and maximum degree=3. Petersen graph requires 4 colors with maximum degree=3 for proper edge-coloring.

-- All regular graphs $G$ with odd no. of vertices requires $\Delta(G) + 1$ minimum no. of colors for proper edge-coloring. For example-Triangle.