Graph Theory Proof Problem(Bipartites graphs) I'm struggling to proof something the Professor gave us. I asked several friends and no one knows what to do. Basically is this:

Prove for every simple bipartite Graph with $n \ge 1$ vertex: $$\delta(G)+\Delta(G)\le n.$$

Professor's Tip: Split you answer in two. First prove the follow: 
If $G$ is simple and bipartite with at least $3$ vertices, $G$ has a vertex $v \in V(G)$ that satisfies at least one of these properties:
• $\Delta(G−v)=\Delta(G)$
• $\delta(G−v)= \delta(G)$
Now, prove by induction on $n$ using these properties.
Guys, I don't know where to start and how to use these information! I checked several books ([West], [Chartrand-Zhang] and [Diestel]) and none of them proved to be useful.
Sorry if my English was bad. Thanks!

You want to use that result in your induction step. Set up your base case (true for say, $n\le 3$) and let $G$ be a bipartite graph with $n > 3$ vertices. By the result in the tip given to you, there exists a vertex $v$ such that $\delta(G-v) = \delta(G)$ or $\Delta(G-v) = \Delta(G)$. Remove $v$ to obtain $G-v$. By the induction hypothesis we have $$\delta(G-v) + \Delta(G-v) \le n-1.$$ If $\delta(G-v) = \delta(G)$ and $\Delta(G-v) \ne \Delta(G)$ then we have $$\delta(G) + (\Delta(G)-1) \le n-1 \implies \delta(G) + \Delta(G) \le n$$ and similarly if $\Delta(G-v) = \Delta(G)$ and $\delta(G-v) \ne \delta(G)$. If both conditions are satisfied then we have $$\delta(G') + \Delta(G') \le n-1 \implies \delta(G) + \Delta(G) \le n-1\le n.$$
