Question: G is a graph with n$\ge$2 vertices an min degree $\delta$. Prove that G contains a spanning sub graph of a min degree $\delta$ with at most $(n-1)\delta$ edges.
Thoughts: For the induction step I've tried 4 different approaches:
Removing the vertex with min degree, using the induction hypothesis and then bringing back the vertex and it's edges- that implied either min degree became lower or equal (and that was okay, as I could have bounded the new min degree with the old one), but If it increased I didn't know what to do.
Removing the vertex with the max degree- That made sure (?) that the min degree won't increase, but then I couldn't bound the number of edges in the spanning sub graph.
Removing a vertex which is a neighbor of a vertex with the min degree. This implies the n vertices sub graph will have a spanning subgraph with (n-1)($\delta$-1) edges. I can bring back at most n edges if the graph is simple, but this (n-1)($\delta$-1)+n is still less then the needed $n\delta$.
Removing a vertex which is not a neighbor of a vertex with the min degree. I'm not sure I can do this, cause I can't prove that such vertex always exists. Also, to such a vertex I think I can add back more than $\delta$ edges which may turn to give me more than $n\delta$ edges in total.