Prove that no bipartite graph of order $3$ or more is Hamiltonian connected
A graph $G$ is Hamiltonian connected if for every pair $u,v$ of vertices of $G$, there is a Hamiltonian $u-v$ path in $G$
Theorem 3.23: Let $G$ be a graph of order $n$ such that $deg(v) \geq \frac{n+1}{2}$ then $G$ is Hamiltonian connected.
this is what I got so far
Let $G$ be a bipartite graph with 2 partites $U$ and $V$. Assume that $G$ is connected, otherwise, we have nothing to prove. We will prove this by induction
Base: $n=3$, then either $|U|=1$ and $|V|=2$ or the other way around, there is a vertex of degree $1$, which is less than $\frac{n+1}{2}$, so $K_{1,2}$ isn’t Hamiltonian connected.
Inductive: Assume that for $n=k$, G isn’t Hamiltonian connected. That mean there is at least one vertex $v$ of degree less than $\frac{k+1}{2}$. Add a new vertex $w$ the connect $w$ to every vertex on the opposite partite
Case 1: $w$ is on the same partite of $v$ then $deg(v)$ is unchange, and $G$ still isn't Hamitonian connected.
Case 2: $w$ is on the opposite partite of $v$, Can this change my result?
It suffices to show that no complete bipartite graph is Hamiltonian connected, as any bipartite graph can be obtained from a complete one by removing edges. And removing edges cannot make a graph Hamiltonian connected.
So let $G$ be a complete bipartite graph with parts $X$ and $Y$.
If $|X| = |Y|$, take $u, v \in X$ (why must $u,v$ exist, by the way ?). No Hamiltonian path can start at $u$ and finish at $v$, since any Hamiltonian path starting in $X$ must finish in $Y$.
If $|X| < |Y|$, take $u \in Y, v \in X$. No Hamiltonian path can start at $u$ and finish at $v$, since any Hamiltonian path starting in $Y$ must finish in $Y$.
Similar when $|X| > |Y|$.
In fact I didn't even need to suppose $G$ was a complete bipartite graph...