Suppose $A, B, X$ are all real, symmetric and positive definite matrices. I want to upper bound ${\bf tr}(ABXBA)$ in terms of ${\bf tr}(AXA)$? Is it possble? My guess would be ${\bf tr}(ABXBA) \le \lambda_{\max}(BB) {\bf tr}(AXA)$? This is obviously true if $A, B$ commute. I am wondering whether it still holds if $A, B$ does not commute.
I don't have any insight to prove this. But I tried a numerical simulation and it seems to be true. I would not trust numerical simulations as often they cannot capture good counterexamples.
The bound I guessed is wrong as pointed out by user293121. But actually there is a very crude bound, i.e. \begin{align*} {\bf tr}(ABXBA) \le \lambda_{\max}(A^2) {\bf tr}(BXB) \le \lambda_{\max}^2 (A) \lambda_{\max}^2 (B) {\bf tr}(X) \le \frac{\lambda_{\max}^2 (A) \lambda_{\max}^2 (B)}{\lambda_{\min}^2(A)} {\bf tr}(AXA). \end{align*}
I think the answer is no. The reason is that the eigenvectors of $A$ and $X$ might be misaligned, leading to small trace on the right side, but $B$ can help align them. Take for instance (putting in the $\epsilon$ just to make things positive definite): $$A = \begin{pmatrix} 1 & 0\\ 0 & \epsilon \end{pmatrix}, B = \begin{pmatrix} .5 +\epsilon & .5\\ .5& .5+\epsilon \end{pmatrix}, X = \begin{pmatrix} \epsilon & 0\\ 0 & 1 \end{pmatrix}.$$
With $\epsilon=.01$, we find that $$ ABXBA\approx\begin{pmatrix} .2526 & .0026\\ .0026 & 0 \end{pmatrix}, AXA = \begin{pmatrix} .01 & 0\\ 0 & .0001 \end{pmatrix}, $$ while $\lambda_{max}(B)^2\approx 1.02$ (one can also do this without the $\epsilon$'s, see the left hand side is $.25$ and the right hand side is $0$, so this will be true with small $\epsilon$ perturbations). So with these values, and assuming I didn't type wrong in MatLab, this fails.