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I am asked to prove that for a graph $G$ with vertex set $V$:

  1. $\exists \text{ a partition } V_1 \cup V_2 \text { of } V \text{ such that } e(G[V_1]) + e(G[V_2]) \leq \frac{1}{2}e(G)$
  2. We may also require that $e(G[V_1]), \;e(G[V_2]) \leq \frac{1}{3}e(G)$

Using an inductive argument I have proven the first part, but am unsure of how to continue to prove the second. I want to use an inductive argument on the number of vertices by first removing a vertex, finding a partition of the remaining subgraph, and then appropriately adding the original vertex and its edges back in.

However, my problem is that I don't know how I could justify that no matter how many edges the removed vertex has, there is a way to place it in either $V_1$ or $V_2$ such that the conditions are still met. In fact I suspect it might not actually be always possible, however I am unsure how else I may approach this question.

user366818
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  • The first statement is sometimes proved using a probabilistic argument: Assume we flip a coin for each vertex, the outcome determining whether it goes into $V_1$ or $V_2$. Then, an edge $e={v,w}$ has a probability of exactly one half to be inside one of the two sets. Hence, there must be at least one way to distribute the vertices such that at most $\frac12$ of all edges are completely inside one of the two sets. I have no idea how to apply it to #2 though. – Jesko Hüttenhain Dec 16 '17 at 18:24

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