I am searching for a proof or a reference of a proof of $$ (\mathbf{v}^\prime\ \mathbf{Sv})^{-1}\le\mathbf{v}^\prime\ \mathbf{S}^{-1}\ \mathbf{v} $$ for every symmetric positive definite matrix $\mathbf{S}$ and every normed vector $\mathbf{v}$ (i.e. $\mathbf{v^\prime\ \mathbf{v}=1}$) with both sides of the inequality become equal if $\mathbf{v}$ is an eigenvector of $\mathbf{S}$.
In my application $\mathbf{S}$ is the Fisher information, $\mathbf{v}^\prime\ \mathbf{Sv}$ is the directional information (i.e. the negative of the curvature of the likelihood in the direction of $\mathbf{v}$). The substantive meaning is that the negative inverse of the curvature of the likelihood is smaller than the asymptotic variance in the respective direction of the parameter space (or equal in directions that correspond to an eigenvector).
This is a consequence of the Cauchy Schwarz inequality. We have \begin{align} 1 &= |v'v|^2 = |v'S^{1/2}S^{-1/2}v|^2 \\ &= |(S^{1/2}v)'(S^{-1/2}v)|^2 \\ & \leq \|S^{1/2}v\|^2 \cdot \|S^{-1/2}v\|^2 = (v'Sv) \cdot (v'S^{-1}v), \end{align} which we can rearrange to get $(v'Sv)^{-1} \leq (v'S^{-1}v)$. In the above, $S^{1/2}$ denotes the unique positive definite square root of $S$, and $S^{-1/2}$ denotes its inverse.
As for the case of equality, we know that the Cauchy-Schwarz inequality $|p'q| \leq \|p\|\cdot\|q\|$ is an equality if and only if $p$ is a positive multiple of $q$. In this case, we see that $p = S^{1/2}v = Sq$ is a (positive) multiple of $q = S^{-1/2}v$ if and only if $q = S^{-1/2}v$ is an eigenvector of $S$.
Because $S^{1/2}$ and $S$ have the same eigenvectors, we can actually conclude that $q$ is an eigenvector of $S$ iff $S^{1/2}q = v$ is an eigenvector of $s$, hence the equality becomes equality iff $v$ is an eigenvector of $S$.