So I came across this proposition in the context of Inner Product Spaces
Let $y$ be a nonzero vector in $X$. For $x\in X$, let $d(x) := \inf_{α∈\mathbb{C}}||x+αy||$. Then $\forall x\in X$ $$|\langle x,y \rangle|^2 \leq ||y||^2 (||x||^2-d(x)^2)$$
I understand the proof involved but my difficulty is this: What is it actually telling us? It seems like a sort of optimisation concept but I'm not entirely sure.
On
It's a bit easier to see what's going on here if we relabel the dummy variable viz. $\alpha\mapsto-\alpha$ so $d(x)=\inf_{\alpha\in\Bbb C}\Vert x-\alpha y\Vert$ is the minimum distance from $x$ to the ray through $y$ – if you prefer, the length of the perpendicular from $x$ to that ray. In the special case $d(x)=0$, $x$ is parallel to $y$. The result you already know how to prove reduces in that case to the Cauchy–Schwarz inequality, and in this case it's saturated. More generally. In the general case, on the other hand, $\Vert x\Vert^2-d(x)^2$ is, by Pythagoras, the squared length of the multiple of $y$ closest to $x$. Thus we can restate the result as thus:
$|\langle x,\,y\rangle|^2\le\Vert y\Vert^2\Vert x_\parallel\Vert^2$, where $x_\parallel$ is the projection of $x$ onto the ray through $y$.
Indeed, $\langle x,\,y\rangle=\langle x_\parallel,\,y\rangle$, so we can restate this as the Cauchy-Schwarz inequality for $x_\parallel,\,y$.
Perhaps it tells you that if you have specific information regarding the distance of $x$ from the line spanned by $y$, then you can improve the general Cauchy-Schwartz inequality with respect to the inner product $\langle x,y\rangle$.