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How can we prove the following conjecture?

Given any simple unweighted bipartite graph $G(V_1, V_2, E)$, there always exists a subgraph $G'(V_1, V_2, E')$ of $G$ such that the two following conditions are simultaneously satisfied:

1) $|E'| \ge \frac{1}{2}|E|$ .

2) The degree of each vertex in $G'$ is at most half of the maximum vertex degree $d^{\max}_G$ in $G$ (for the sake of simplicity assume $d^{\max}_G$ is an even integer).

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If $d_G^{\max}$ is odd and the graph is regular, this clearly is not possible. But if $d_G^{\max}=2k$ is even, this is possible: the edges may be properly colored with $2k$ colors [this is well-known bipartite variant of Vizing's theorem, which may be proved, for example, by induction in number of edges: color all edges except the edge $uv$, there are free colors for vertices $u,v$ --- call them 1,2. There can not exist a 1-2 chain from $v$ to $u$ since it would produce an odd cycle in $G$. Thus interchange 1 and 2 in the $u$-component of the graph formed by edges of colors 1,2, after that $uv$ may be colored with 2.] Now take $k$ most popular colors.

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  • $\begingroup$ No, I mean $k$ most presented (or popular, or how do you call it) colors. $\endgroup$ Commented Feb 27, 2018 at 8:13
  • $\begingroup$ 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 $|E'| \ge \frac{1}{2}|E|-|V_1|-|V_2|$? $\endgroup$ Commented Feb 27, 2018 at 16:32
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    $\begingroup$ 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. $\endgroup$ Commented Feb 28, 2018 at 6:36

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