For square matrices $A,B,C\in\mathbb C^{n\times n}$ is it correct that
$$\lvert\mathrm{Tr}(ABC)\rvert \le \sqrt{\mathrm{Tr}(A^\ast A \lvert B\rvert)\mathrm{Tr}(CC^\ast \lvert B\rvert)},$$
where $\lvert B\rvert$ is the positive-semidefinite square-root $\lvert B\rvert=(B^\ast B)^{1/2}$? Note that the rhs. is non-negative since the product of positive-semidefinite matrices has a non-negative trace.
The answer is trivially yes if $B=\lvert B\rvert$ since then
$$\lvert\mathrm{Tr}(ABC)\rvert=\lvert\mathrm{Tr}(A\lvert B\rvert^{1/2}\lvert B\rvert^{1/2} C)\rvert\le \sqrt{\mathrm{Tr}(A \lvert B\rvert A^\ast)\mathrm{Tr}( \lvert B\rvert C C^\ast )}$$
due to Cauchy-Schwarz.
Best Answer
The proposed inequality is valid as long as we are allowed to tweak it a bit: $$ |\mathrm{Tr}(abc)| \le \sqrt{\mathrm{Tr}(a^\ast a | b^*|)\ \mathrm{Tr}(cc^\ast | b|)}. $$ The difference is the term $| b^*|$ instead of $| b|$ in the RHS.
To prove it let $b=u|b|$ be the polar decomposition of $b$ and notice that by Cauchy-Schwarz $$ |\mathrm{Tr}(abc)|^2 = |\mathrm{Tr}(au|b|^{1/2}|b|^{1/2}c)|^2 \leq $$$$ \leq \mathrm{Tr}(au|b|^{1/2}|b|^{1/2}u^*a^*) \ \mathrm{Tr}(c^*|b|^{1/2}|b|^{1/2}c)|^2 = $$$$ = \mathrm{Tr}(au|b|u^*a^*) \ \mathrm{Tr}(c^*|b|c)|^2. \tag {1} $$
Next notice that $$ (u|b|u^*)^2 = u|b|u^*u|b|u^* = u|b||b|u^* = bb^*, $$ which means that $u|b|u^*$ is a positive matrix whose square coincides with $bb^*$, whence $$ u|b|u^* = |b^*|. \tag{2} $$ Plugging (2) in $(1)$ then completes the proof.
Given the nature of Math, in which the pieces fit together in an incredibly elegant way, I dare say that the problem that led the OP to formulate this question might be better served by the above inequality, as compared to the originally proposed one. I would be very thankful if they could confirm it or deny it in a comment!