I wonder if there is a good to find the lower bound of the following term: \begin{equation} \min_{\boldsymbol{x}}\,a^2(\boldsymbol{x}^\textrm{T}\boldsymbol{A}\boldsymbol{y})^2-(\boldsymbol{x}^\textrm{T}\boldsymbol{A}\boldsymbol{x})^2 \end{equation} where $a>0$ is some constant; $\boldsymbol{x},\boldsymbol{y}\in\mathbb{R}^n$ have unit norm; and $\boldsymbol{A}\in\mathbb{R}^{n\times n}$ is a symmetric matrix. $\boldsymbol{A}$ and $\boldsymbol{y}$ are known. It can also be written as: \begin{equation} \min_{\boldsymbol{x}}\,(a\boldsymbol{y}-\boldsymbol{x})^\textrm{T}\boldsymbol{A}\boldsymbol{x}\boldsymbol{x}^\textrm{T}\boldsymbol{A}(a\boldsymbol{y}+\boldsymbol{x}) \end{equation} or \begin{equation} \min_{\boldsymbol{x}}\,\boldsymbol{x}^\textrm{T}\boldsymbol{A}(a^2\boldsymbol{y}\boldsymbol{y}^\textrm{T}-\boldsymbol{x}\boldsymbol{x}^\textrm{T})\boldsymbol{A}\boldsymbol{x} \end{equation} This is related to an earlier trivial question I asked, any help is greatly appreciated, thanks so much!
Edit: In other words, is there an $a$ that could make the above term greater than $0$?
You can't.
For any $y$, there is an $(n-1)$-dimensional space of vectors $x$ with $x^TAy =0$ (the orthogonal complement of $Ay$). For any of these vectors and for any $a$, $$a^2 (x^TAy)^2 - (x^TAx)^2 = -(x^TAx)^2 <= 0,$$ with equality only if $x$ is in the kernel of $A$. Therefore the only time that the minimum is guaranteed to be nonnegative is if $A=0$, or $A$ has rank one with $y$ in the column space of $A$ (in which case any $a\geq 1$ will do).