Solution
= Solution
Repeatedly delete a vertex whose current degree is less than $d/2$. If every vertex were deleted, charge each edge to the endpoint deleted first. At each deletion fewer than $d/2$ remaining edges are charged, so the original graph would have fewer than $|V|d/2$ edges, contradicting average degree at least $d$. The nonempty graph left by the process has minimum degree at least $d/2$.