Let $A \in \mathbb{R}^{m\times p}$ and $B \in \mathbb{R}^{p \times n}$ where $m > n > p$ and $A, B$ are full rank.
For $x \in \mathbb{R}^n$ with $\|x\|_2=1$, let us consider $$ f(x) = x^T(A^TA)(BB^T)x. $$ Can we conclude that $$ f(x) \ge \sigma^2_{\min}(A^TABB^T)\|x\|^2 ? \quad \text{or} \quad \ge \sigma_{\min}^2(A)\sigma_{\min}^2(B)\|x\|^2 ? $$ or a lower bound which involves singular values of $A$ or $B$? Here, for a matrix $M$ of size $s \times k$, $\sigma_{\min}(M)$ is the $\min\{s,k\}$-th largest singular value of $M$.
Here is my attempt. Since $BB^T$ is symmetric positive definite, we can write it as $BB^T = R^{1/2}R^{1/2}$ where $R^{1/2}$ is a $p \times p$ symmeric matrix. Since $R^{1/2}A^TAR^{1/2}$ is symmetric, it is diagonalizable. Thus we have $$ A^TABB^T = R^{-1/2}(R^{1/2}A^TAR^{1/2})R^{1/2} = R^{-1/2}(Q\Sigma Q)R^{1/2}, $$ and $$ f(x) = x^TA^TABB^Tx = (x^TR^{-1/2}Q)\Sigma (QR^{1/2}x). $$ If $R^{-1/2} = (R^{1/2})^T$, $BB^T = I$ and the statement is obvious. If $R^{-1/2} \ne (R^{1/2})^T$, I am not sure what would be a valid lower bound in terms of singular values of $A$ or $B$.
Any suggestions/comments/answers will be very appreciated.