Assume that for matrices, $M,A$ where $M$ is positive definite and $0<\theta<1$ the inequality $x^T(A^TMA)x\leq \theta x^TMx$ holds for any vector $x$. I want to obtain a bound for $y^T(A^TMA\otimes A^TMA)y$ where $\otimes$ is the Kronecker product. My hope is that the bound is of the form $y^T(A^TMA\otimes A^TMA)y\leq \theta^2 y^T(M\otimes M)y$.
My attempt was to obtain a direct upper bound using matrix norms (induced by the Euclidean norm as $\|M\|=\max_{\|x\|=1} \|Mx\|$) and the Rayleigh inequality.
Concretelly, we know that $\|M^{1/2}A\|\leq \sqrt{\theta}\|M^{1/2}\|$. Thus $$ \begin{aligned} &y^T(A^TMA\otimes A^TMA)y = \| (M^{1/2}A\otimes M^{1/2}A)y \|^2\leq \|M^{1/2}A\|^4\|y\|^2 \\& \leq\theta^2\|M^{1/2}\|^4\|y\|^2 = \theta^2\|M\|^2\|y\|^2 \end{aligned} $$ Then, using the Rayleigh inequality I get $\|y\|^2\leq y^T(M\otimes M)y/\lambda_{\min}(M)^2$. Thus, instead of a multiplicative factor of $\theta^2$ I obtain $(\|M\|/\lambda_{\min}(M))^2\theta^2$.
However, $(\|M\|/\lambda_{\min}(M))^2\geq 1$. I wonder if its possible to obtain a tighter bound.
In general, for any two positive definite matrices $X$ and $Y$, we have $X\preceq Y$ if and only if $\rho(Y^{-1}X)\le1$. Therefore \begin{align} X\preceq Y &\Leftrightarrow \rho(Y^{-1}X)\le1\\ &\Leftrightarrow \rho(Y^{-1}X)^2\le1\\ &\Leftrightarrow \rho\left((Y^{-1}X)\otimes(Y^{-1}X)\right)\le1\\ &\Leftrightarrow \rho\left((Y\otimes Y)^{-1}(X\otimes X)\right)\le1\\ &\Leftrightarrow Y\otimes Y\preceq X\otimes X.\\ \end{align} Now put $X=A^TMA$ and $Y=\theta M$ and the result follows.