Inverse of a metric under variation

Given a fixed metric $$g_{\mu\nu}$$, its variation by a small amount could be written as: $$g_{\mu\nu}+h_{\mu\nu}$$ or equivalently as: $$g_{\mu\nu}+\delta(g_{\mu\nu}).$$ The given metric has the property that: $$g_{\mu\alpha}g^{\alpha\nu}=\delta^\nu_\mu,$$ and that $$g_{\mu\nu}=g_{\nu\mu}$$, which applies also to the metric after variation $$g+h$$.

Question: What is the relationship between $$\delta g^{\mu\nu}$$ and $$h^{\mu\nu}$$? This is actually proving that for arbitrary tensors: $$A_{\mu\nu}B^{\mu\nu}=A^{\mu\nu}B_{\mu\nu}.$$ How to prove this?

Consider the disturbed metric $$\widetilde{g}_{\mu\nu}=g_{\mu\nu}+h_{\mu\nu}$$ with $$|h_{\mu\nu}|<1$$, meaning the pertubation is small and we can use $$g$$ to raise and lower indices, then $$\widetilde{g}^{\mu\nu}=g^{\mu\nu}-h^{\mu\nu}$$ with $$|h^{\mu\nu}|<1$$. The sign change arises since we want to have $$\widetilde{g}_{\lambda\mu}\widetilde{g}^{\mu\nu}=\delta_\lambda^\nu$$ as well as $$g_{\lambda\mu}g^{\mu\nu}=\delta_\lambda^\nu$$. Neglecting the pertubation in second order, we have: $$$$\widetilde{g}_{\lambda\mu}\widetilde{g}^{\mu\nu} =(g_{\lambda\mu}+h_{\lambda\mu})(g^{\mu\nu}-h^{\mu\nu}) =\delta_\lambda^\nu -\underbrace{g_{\lambda\mu}h^{\mu\nu}}_{=-h_\lambda^\nu} +\underbrace{h_{\lambda\mu}g^{\mu\nu}}_{=h_\lambda^\nu} +\mathcal{O}(h^2).$$$$ Therefore we need to put $$\delta g^{\mu\nu}=-h^{\mu\nu}$$. Furthermore, we have: $$$$A_{\mu\nu}B^{\mu\nu} =A^{\kappa\lambda}g_{\kappa\mu}g_{\lambda\nu}B^{\mu\nu} =A^{\kappa\lambda}B_{\kappa\lambda}.$$$$