I want to find a condition on $\delta(G)$ (ex. $\delta(G) \geq an$) that guarantees $\kappa(G)=\delta(G)$ where $\kappa(G)$ is the vertex-connectivity of a bipartite graph $G$, and $\delta(G)$ is the minimum degree of $G$.<br>

In other words, I want to prove that the statement 

> If $\delta(G) \geq an$, then $\kappa(G)=\delta(G)$

holds for bipartite graphs.

I know that $\delta(G) \geq \frac{n+1}{4}$ implies $\kappa'(G)=\delta(G)$ where $\kappa'(G)$ is the 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 **vertex-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?