Trace inequality for product of three matrices 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.
 A: 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!
A: For reasons of continuity we need only consider this purported inequality when $B$ is invertible.
The answer to OP's question is No and I construct an explicit counter example below using Polar Form and Cauchy-Schwarz.
using Polar Decomposition $B = UP$
and for concreteness suppose that $U$ is the discrete Fourier transform and $P\succ \mathbf 0$ is a diagonal matrix with distinct diagonal entries
$\Big\vert\text{trace}\big(ABC\big)\Big\vert $
$=\Big\vert\text{trace}\big(AUP^\frac{1}{2}P^\frac{1}{2}C\big)\Big\vert $
$=\Big\vert\text{trace}\big((P^\frac{1}{2}C)(AUP^\frac{1}{2})\big)\Big\vert $
$=\Big\vert\text{trace}\big((C^*P^\frac{1}{2})^*(AUP^\frac{1}{2})\big)\Big\vert $
$\leq  \Big\Vert AUP^\frac{1}{2}\Big \Vert_F \Big \Vert C^*P^\frac{1}{2}\Big \Vert_F $
by Cauchy Scwharz
now I select $A$ such that $U^*A^*AU = P$
(note that $A^*A = UPU^* \neq P$)
finally select $C^*:=  AU$. Then the above Cauchy-Schwarz is met with equality, so
$\Big\vert\text{trace}\big(ABC\big)\Big\vert $
$=\Big\Vert AUP^\frac{1}{2}\Big \Vert_F \Big \Vert C^*P^\frac{1}{2}\Big \Vert_F $
$=\Big\Vert AUP^\frac{1}{2}\Big \Vert_F^2$
$=\text{trace}\Big(  U^*A^*AUP\Big)$
$=\Big \Vert U^*A^*AU\Big \Vert_F \Big \Vert P\Big \Vert_F$
$=\Big \Vert A^*A\Big \Vert_F \Big \Vert P\Big \Vert_F$
$\gt \text{trace}\Big(  A^*AP\Big)$
by Cauchy Schwarz
putting this all together, we have found a case where
$ \sqrt{\mathrm{Tr}(A \lvert B\rvert A^\ast)\cdot \mathrm{Tr}( \lvert B\rvert C C^\ast )} =  \sqrt{\mathrm{Tr}(A \lvert B\rvert A^\ast)\cdot \lvert\mathrm{Tr}(ABC)\rvert} \lt \lvert\mathrm{Tr}(ABC)\rvert$
A: As the other two answers show, the inequality as it stands is false. It can be corrected by changing the first $|B|$ on the RHS to $|B^\ast|$. It is easier to prove the corrected inequality by singular value decomposition $B=USV^\ast$ than by polar decomposition. Denote by $\langle X,Y\rangle$ the Frobenius inner product $\operatorname{tr}(XY^\ast)$. By Cauchy-Schwarz inequality,
\begin{aligned}
|\operatorname{tr}(ABC)|
&=|\operatorname{tr}(AUSV^\ast C)|\\
&=|\langle AUS^{1/2},C^\ast VS^{1/2}\rangle|\\
&\le\sqrt{\langle AUS^{1/2},AUS^{1/2}\rangle \langle C^\ast VS^{1/2},C^\ast VS^{1/2}\rangle}\\
&=\sqrt{\operatorname{tr}(AUSU^\ast A^\ast) \operatorname{tr}(C^\ast VSV^\ast C)}\\
&=\sqrt{\operatorname{tr}(A|B^\ast|A^\ast) \operatorname{tr}(C^\ast|B|C)}\\
&=\sqrt{\operatorname{tr}(A^\ast A|B^\ast|) \operatorname{tr}(CC^\ast|B|)}.\\
\end{aligned}
