I'm struggling with this question, and I'll be happy for a hint or a direction.
Let G=($V_1\cup V_2$,E) with no isolated Vertices s.t. the average degree on $V_1$ is large than the average degree on $V_2$. Prove that there exists an edge $\{v_1,v_2\}\in E$ s.t. $v_i\in V_i$ and $deg(v_1)>deg(v_2)$.
Thank you.
Things I tried: Since the sum of degrees is the same on every side, the assumptions lead to the fact that $|V_1|<|V_2|$.
I defining a function $f:V_1\rightarrow N$ by $f(v_1)$ being the minimal degree of the vertex's neighbors. we see that $$\sum_{v\in V_1} f(v) \geq \sum_{v\in V_1} deg(v)=|E|$$ This could solve the question since I'm counting degree's in $V_2$ but I could be double counting.
Another thing I noticed is that if I assume the contrary, I get $$\sum_{v\in V_1} deg^2(v)\leq \sum_{v\in V_2} deg^2(v)$$ but I'm not sure if this helps in any way.
Define $f : E \to \mathbb{R}$ as follows $$f\big((v_1,v_2)\big) = \frac{1}{\deg(v_1)} - \frac{1}{\deg(v_2)}$$ and observe that we need to prove that there exists an edge $e$ such that $f(e) < 0$. However, assuming that there are no isolated vertices, $$ \sum_{e \in E}f(e) = \sum_{v_1 \in V_1}\deg(v_1)\cdot\frac{1}{\deg(v_1)} - \sum_{v_2 \in V_2}\deg(v_2)\cdot\frac{1}{\deg(v_2)} = |V_1|-|V_2| < 0 $$
because $|V_2| > |V_1|$, so there has to be some edge $e$ with $f(e) < 0$.
I hope this helps $\ddot\smile$