Graph Theory inequality I've been trying to prove the following inequality 
If G is r-regular graph and $\kappa (G)=1 $, then $\lambda (G)\leq \left \lfloor \frac{r}{2} \right \rfloor$
I've tried manipulating the Whitney inequality, but it doesn't seem to help. 

Let $G=(V,E)$ be the $r$-regular graph. Since $\kappa(G)=1$, there exist three vertices $u$, $v$ and $x$ such that, after removing $x$ from $G$, $u$ and $v$ are not connected. Call
$G'$ the graph obtained by removing $x$ from $G$, and define
$$ G_u=\{ y \in V \;|\; y \mbox{ is connected to } u \mbox{ in } G' \}$$
$$ G_v=\{ y \in V \;|\; y \mbox{ is connected to } v \mbox{ in } G' \}.$$
$G_u$ and $G_v$ are not connected in $G'$. Now define
$$ E_u=\{(x,y) \in E | y \in G_u \} $$
$$ E_v=\{(x,y) \in E | y \in G_v \}. $$
Observe that by removing one of these two sets the graph obtained is disconnected. Moreover, 
$$ |E_u|+|E_v|=d(x)=r.$$
Hence one of them has cardinality $\leq \lfloor\frac{r}{2} \rfloor$.
