I want to find a condition of $\delta(G)$(ex. $\delta(G) \geq an$) that guarantees $\kappa(G)=\delta(G)$ where $\kappa(G)$ is a connectivity of bipartite graph $G$, and $\delta(G)$ is the minimum degree of $G$.<br>
In other words, I want to prove the statement 'If $\delta(G) \geq an$, then $\kappa(G)=\delta(G)$.'<br>
I know that $\delta(G) \geq \frac{n+1}{4}$ implies $\kappa'(G)=\delta(G)$ where $\kappa'(G)$ is an edge-connectivity of $G$, and it is sharp.<br>
Also, I proved that bipartite $G$ is connected if $\delta(G) \geq \frac{n+1}{4}$.<br>
But I have no idea with the **connectivity**.<br>
Here are my ideas to find such boundary $an$:
1. (Contrapositive) Prove the statement 'If $\kappa(G)<\delta(G)$, then $\delta(G)<an$ . 
2. Use the statement $\delta(G) \geq \frac{n+1}{4}\Rightarrow \kappa'(G)=\delta(G)$. Find the boundary of $\delta(G)$ that implies $\kappa(G)=\kappa'(G)$, and then show this boundary also satisfies $\delta(G) \geq \frac{n+1}{4}$. 

Would you help me?