If $A, B$ are positive $n \times n$ complex matrices, $n$ some integer, then obviously \begin{equation\*} \|ABA\|\_\text{tr} = tr(ABA) = tr(A^2 B). \end{equation\*} 


But can we say there is a constant $C\_n > 0$ depending only on $n$ where $\|ABA\|\_\text{tr} \geq C\_n \| A^2 B\|\_\text{tr}$? 


Note that it's easy to get the reverse: 


$\|ABA\|\_\text{tr} \leq \|A^2B\| \_\text{tr} $ 


so it's the above inequality I really need. 


More generally though, I'm guessing $\|AB\|\_\text{tr}$ 


and $\|BA\|\_\text{tr}$ are not equivalent (modulo a constant depending only on $n$) 



trace norm inequality for positive matrices 
If $A$ are $B$ are projections which are not orthogonal, but are close to being orthogonal so that $ABA \not= 0$ but has only small eigenvalues then we have $\| A B A \|\_\text{tr} \ll \| (A B A)^{1/2} \|\_\text{tr} = \| A^2 B \|\_\text{tr}$. Hence, no such constant $C\_n$ exists.


For a specific example, if $A = 
\left( \begin{array}{cc} 
1 & 0 \\ 
0 & 0
\end{array} \right)$ and $B =
\left( \begin{array}{cc}
\sin^2 t & - \sin t \cos t \\ 
 - \sin t\cos t & \cos^2 t 
\end{array} \right)$, then as $t \to 0$ we have $\| ABA \|\_\text{tr} / \| A^2 B \|\_\text{tr} = | \sin t | \to 0$.


