TITLE: Is it possible to upper bound ${\bf tr}(ABXBA)$ in terms of ${\bf tr}(AXA)$ for positive definite $A, B, X$?
QUESTION [6 upvotes]: 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*}

REPLY [2 votes]: 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.