Balanced component cut ($\max (|V_1|,|V_2|)\le (k+1)n/(2k+1)$)

If every graph component has at most $(k+1)n/[k(2k+1)]$ vertices, assigning whole graph components to two sides gives an empty graph cut with the displayed bound. To prove it, maximize the smaller side's weight $S\le n/2$. If $S<kn/(2k+1)$, every component on the larger side has weight at least $n-2S$, because moving a smaller one would improve the smaller side. There are at least $k+1$ such components, since their total exceeds $k$ times the permitted maximum weight. Thus $n-S\ge (k+1)(n-2S)$, a contradiction. The proof also works for arbitrary positive weights.