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The problem at hand is the following, out of Bondy & Murty's graph theory text.

"Let $G[X,Y]$ be a bipartite graph, each vertex of which is joined to at least one, but not all, vertices in the other part. Suppose that $d(x) \geq d(y)$ for all $xy \not\in E$. Show that $|Y| \geq |X|$, with equality if and only if $d(x) = d(y)$ for all $xy \not\in E$ with $x \in X$ and $y \in Y$."

I'm trying to proceed by assuming $|Y| \leq |X|$ and proving that it then must be the case that $|Y| = |X|$. I've tried looking at the adjacency matrix, and dividing by either $d(x)$ or $d(y)$ depending on if I was summing over the columns or not, but it hasn't gotten me anywhere. Any hints would be much appreciated.

Update: I have found this hint, but am not sure how to apply it/how it's useful. I've proved that $\frac{1} {|X|(|Y| - d(x))} \geq \frac{1} {|Y|(|X|-d(y))}$

http://mindseye.fr/GT/wp-content/uploads/2008/10/hints-bmv1.pdf

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Clearly we have $\displaystyle\sum_{x\in X}\frac{1}{|X|}\displaystyle\sum_{y\in Y, xy \notin E}\frac{1}{|Y|-d(x)}=\displaystyle\sum_{y\in Y}\frac{1}{|Y|}\displaystyle\sum_{x\in X, xy \notin E}\frac{1}{|X|-d(y)}=1$.

Thus if $|X|>|Y|$, we have $|X|d(x)>|Y|d(y) \> \forall xy \notin E$; thus $1=\displaystyle\sum_{x\in X}\displaystyle\sum_{y\in Y, xy \notin E}\frac{1}{|X|(|Y|-d(x))}>\displaystyle\sum_{y\in Y}\displaystyle\sum_{x\in X, xy \notin E}\frac{1}{|Y|(|X|-d(y))}=1$, a contradiction. Thus $|X|\leq|Y|$.

By the equation on the first line, by either conditioning on $|X|=|Y|$ or $d(x)=d(y) \> \forall xy \notin E$, by the two inequalities $|X|\leq|Y|$ and $d(x)\geq d(y) \> \forall xy \notin E$ we obtain an inequality which actually leads to an equality case (LHS and RHS being $1$). Thus the summands should be equal for all $x \in X, y\in Y, xy \notin E$; this suffices to prove the necessary and sufficient conditions.