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Feb 28, 2018 at 6:36 comment added Fedor Petrov if $d_G^{\max}=2k+1$ is odd, we get at least $\frac{k}{2k+1}|E|=\frac12|E|-\frac1{2(2k+1)}|E|$, and since $|E|\leqslant (2k+1)\min(|V_1|,|V_2|)$ this yields at least $\frac12(|E|-\min(|V_1|,|V_2|))$ edges.
Feb 27, 2018 at 16:33 vote accept Penelope Benenati
Feb 27, 2018 at 16:32 comment added Penelope Benenati Great! It is correct to say that if $d^{\max}_G$ is not necessarily even, using this approach would yield to replace condition (1) with the inequality $|E'| \ge \frac{1}{2}|E|-|V_1|-|V_2|$?
Feb 27, 2018 at 8:13 comment added Fedor Petrov No, I mean $k$ most presented (or popular, or how do you call it) colors.
Feb 27, 2018 at 5:27 history answered Fedor Petrov CC BY-SA 3.0