# If $G$ is a graph of order $n \ge 5$ and size $m\ge 3n - 5$ then $G$ contains a subgraph with minimum degree 4.

Since $m \ge 3n-5$ and $n \ge 5$, I have a theorem which tells me that $G$ is not planar. It seems like a natural approach to do a proof by contradiction and arrive at the fact that $G$ must be planar. It seems easiest to do this with Kuratowksi's theorem. However I'm having troubles making any progress like this since I don't really know how many edges or vertices would be in the subgraph which is a subdivision of $K_5$ or $K_{3,3}$ (although I'm not sure I necessarily need to know that).

Any help or tips on a different approach would be greatly appreciated.

Assume the converse. Then the graph $G$ contains a vertex $v_1$ with degree at most $3$. Removing $v_1$ and edges incident to it from $G$ we obtain a graph with $n-1$ vertices and at least $3(n-1)-5$ edges which contains a vertex $v_2$ with degree at most $3$. Remove the vertex $v_2$ edges incident to it from the graph ... and so forth. We stop when there remains $4$ vertices, so we have removed $n-4$ vertices and at most $3(n-4)$ edges and the final graph has at most ${4 \choose 2}=6$ edges. Thus initially we had at most $6+3(n-4)<3n-5$ edges, a contradiction.