Let $G = (X \cup Y, E)$ be a bipartite graph with no isolated vertices and let $S \subseteq X$. Suppose $|S| = |N(S)|$ (neighboors of $S$) and that $\forall \{x,y\} \in E$, where $x\in X$ and $y\in Y$, $\deg(x) \ge \deg(y)$.
It is not obvious to me why $\sum_{x\in S}{\deg(x)} \ge \sum_{y\in N(S)}{\deg(y)}$. It seems obvious for my colleagues but they can't explain it to me. Can anyone provide a formal proof or clear intuition ?
The stated result is wrong, at least in its current form form. Consider for instance $S=\{x_1,x_2\}$, $X=S\cup\{x_3\}$, $Y=\{y_1,y_2\}$ and edges $(x_1,y_1)$, $(x_1,y_2)$ $(x_3,y_1)$, $(x_3,y_2)$. Remark that $x_2$ is isolated.
Then all assumptions are verified, but the conclusion is not: sum of degrees is $2$ in $S$ and $4$ in $N(S)$.