Proof that $\lambda(G) = \kappa(G)$ Let $\lambda(G)$ be edge connectivity of graph and $\kappa(G)$ vertex connectivity.
How can I proof that $\lambda(G) = \kappa(G)$ for every graph where every vertex of this graph has degree not greater than 3.
 A: Since $0\leq\kappa(G)\leq\lambda(G)\leq\delta(G)\leq 3$, we shall show that $\kappa<\lambda$ is impossible. We consider the following cases.
If $\kappa(G)=0$, then $G$ is not connected, and hence $\lambda(G)=0$.
If $\kappa(G)=1$, then $G$ is connected but $G-v$ is not connected for some $v\in V(G)$. $G-v$ can have two or three connected components. Since $|\delta(v)|\leq 3$, there is a connected component $C=G[X]$ ($X\subset V(G)$) in $G-v$ such that $|\delta(X)|=1$, hence $\lambda(G)=1$.
If $\kappa(G)=2$, then $G-v$ is connected for all $v\in V(G)$ but $G-v_1-v_2$ is not connected for some $v_1,v_2\in V(G)$. Note that every connected components in $G-v_1-v_2$ is adjacent to $v_1$ and $v_2$, respectively. Suppose, towards contradiction, $\lambda(G)=3$, then, for each connected component $C$ in $G-v_1-v_2$, we have $|\delta(V(C))|=3$. It means that we have exactly two connected components $C_1,C_2$ in $G-v_1-v_2$, with say there are two edges between $C_i$ and $v_i$ ($i=1,2$). Hence, $|\delta(C_1\cup\{v_1\})|=2$, contradiction.
