Decide, whether it's true, that:
$\forall $ graph $G$ : if $K(G) \neq \lambda(G) $ then $\exists $ vertex $v$ : $deg(v)\geq 4$ (where $K(G)$ is the vertex-connectivity and $\lambda (G)$ is the edge-connectivity)
So, I have 2 solutions:
1, if I negate it right away I get:
$ \exists G : K(v) = \lambda(G) : \forall v : deg(v) \leq 3 $ - which is absolutely true because of $K_4$ - so if the negation is true, the original problem should be false.
2, by contradiction I say that $\forall v : deg(v) \leq 3 $ while $K(G) \neq \lambda(G)$ still holds
because of the degrees we can say, that $K(G) \leq \lambda(G) \leq 3$ but I know, that $K(G) \neq \lambda(G)$ so I get $K(G) < \lambda(G) \leq 3$ which seems like a contradiction, but I don't know how to prove it. (of course that would mean that the original sentence is true)
Any help is appreciated.
Your negation of the statement is incorrect. The correct negation would be: There exists a graph $G$ so that $K(G) \neq \lambda(G)$ and the maximum degree of $G$ is at most 3. Remember that the negation of ($p \Rightarrow q$) is ($p$ and not $q$).
Your second strategy is a good start. You know $K(G) < \lambda(G) \leq 3$. Therefore $K(G) \in \{0,1,2\}$. If $K(G) = 0$, then $\lambda(G) = 0$, the graph is already disconnected. If $K(G) = 1$, there is a vertex $v \in V(G)$ so that deleting $v$ disconnects the graph. But $\deg_G(v) \leq 3$, so for some component of $G - v$, it must be the case that $v$ sends at most 1 edge $e$ to that component. But then $e$ is a cut edge, and $\lambda=1$. If $K(G) = 2$, you can do something similar - look at the cut vertices, each must send at most 1 edge to one of the components, and those edges together would be an edge cut.
In fact you can use this same strategy without doing the case analysis - just look at any set of cut vertices $V$ in the graph, as long as the maximum degree is at most 3, each sends at most one edge to one of the components of $G-V$, and that set of edges would be an edge cut.