Question: prove that every graph with $n\ge7$ vertices and at least 5n-14 edges contains a sub graph with minimum degree at least 6.
My proof: By induction. For n=7, the number of edges is 21=$7 \choose 2$ which implies this is the full graph, so each degree is 6, and therefore the min degree as well. Now using the induction hypothesis, proof for n: A graph G with n vertices and 5n-14 edges, can actually be thought of as a graph G' with n-1 vertices with 5(n-1)-14=5n-21 edges, with an added vertex and 7 added edges. Since the hypothesis is true for n-1, then adding this vertex and the 7 edges won't decrease the min degree.
I want to know if this proof suffices or am I missing something?
For the induction step: Consider $G$ with $n$ vertices. If all vertices of $G$ have degree $\ge 6$, $G$ itself is a subgraph as required. Otherwise, $G$ has a vertex $v$ of degree $\le 5$, and then $G':=G-v$ has $n-1$ vertices and $5n-14-\rho(v)\ge 5(n-1)-14$ edges, hence by induction hypotheses $G$' contains a subgraph with minimum degree $\ge 6$. As this is also a subgraph of $G$, we are done.