Let $G$ be a simple, not complete graph. I'm asked to prove that:
$$\kappa(G)\geq2\delta(G)-|V(G)|+2$$
where $\kappa(G)$ is the vertex-connectivity of $G$, $\delta(G)$ is the minimum degree of $G$ and $V(G)$ is the set of all the vertices of $G$.
Any tips on how to start would be really appreciated.
So I have a solution, I guess I'd better update this.
Let $S$ be a vertex cut set of $G$ such that $|S| = \kappa(G)$ and let $G_{min}$ be the component of $G-S$ with the least number of vertices. Then, $$|V(G_{min})|\leq\frac{|V(G)| - |S|}{\omega(G-S)}\leq\frac{|V(G)| - |S|}{2}\tag{1}$$ Where $\omega(G-S)$ is the number of components of $G-S$.
Suppose a vertex $x\in V(G_{min})$. Then, $$|S|+|V(G_{min})|-1\geq d_{G}(x)\geq\delta(G)\tag{2}$$ From $(1)$ and $(2)$ we have $$|S|+\frac{|V(G)|-|S|}{2}-1\geq\delta(G)$$ And since $|S|=\kappa(G)$ $$\kappa(G)\geq2\delta(G)-|V(G)|+2$$