I think "$\forall \kappa[dv(\kappa)=\kappa]$" is a consequence of GCH:

For successor $\kappa$, by GCH at the predecessor of $\kappa$ we have that the set of bounded subsets of $\kappa$ is just $\kappa^{\kappa^-}=({2^{\kappa^-}})^{\kappa^-}=2^{\kappa^-}=\kappa$ (indulging in a bit of awful notation :P).

For limit $\kappa$: Write $\kappa=\sup\{\lambda_0<\lambda_1< . . . <\lambda_{\alpha}< . . .: \alpha\in \eta\}$ with each $\lambda_\eta$ a cardinal. Now, given a bounded subset $B$ of $\kappa$, let $\alpha_B$ be the least $\alpha$ such that $B\subseteq\lambda_{\alpha}$. Let $\mathcal{B}_\alpha=\{B: \alpha_B=\alpha\}$, and note that $2^{<\kappa}=\bigcup_{\alpha\in\eta} \mathcal{B}_\alpha$. By GCH below $\kappa$, each $\mathcal{B}_\alpha$ has size $\lambda_\alpha^+$, so $2^{<\kappa}=\kappa$.

Note that this is all a very coarse argument - all we're doing is counting the set $B(\kappa)$ and showing that your inequality collapses.

Your statement "$\forall \kappa[dv(\kappa)=\kappa]$" is a consequence of GCH:

For successor $\kappa$, by GCH at the predecessor of $\kappa$ we have that the set of bounded subsets of $\kappa$ is just $\kappa^{\kappa^-}=({2^{\kappa^-}})^{\kappa^-}=2^{\kappa^-}=\kappa$ (indulging in a bit of awful notation :P).

For limit $\kappa$: Write $\kappa=\sup\{\lambda_0<\lambda_1< . . . <\lambda_{\alpha}< . . .: \alpha\in \eta\}$ with each $\lambda_\eta$ a cardinal. Now, given a bounded subset $B$ of $\kappa$, let $\alpha_B$ be the least $\alpha$ such that $B\subseteq\lambda_{\alpha}$. Let $\mathcal{B}_\alpha=\{B: \alpha_B=\alpha\}$, and note that $2^{<\kappa}=\bigcup_{\alpha\in\eta} \mathcal{B}_\alpha$. By GCH below $\kappa$, each $\mathcal{B}_\alpha$ has size $\lambda_\alpha^+$, so $2^{<\kappa}=\kappa$.

Note that this is all a very coarse argument - all we're doing is counting the set $B(\kappa)$ and showing that your inequality collapses.