yes it is correct.
in general, if there are |V| vertices and 'd' is the minimum degree then, it is possible that there will be vertices with degree 'd' or more. Now if you consider that all the vertices have degree 'd', then total degree will be less than or equal to |V|.d (as there could be vertices with higher degree)
also we know that in a graph, total degree = 2|e| (e being no. of edges), as each edge contributes 2 in degree.
equting the above results => |V|.d <= 2|e|