Consider the $M \times M$ complex positive semidefinite matrices ${\bf A}, {\bf B}, {\bf Z}$. We have the relation $\mu_{\text{max}}{\bf I} \succeq {\bf A} \succeq {\bf B} \succ \mu_{\text{min}}{\bf I} $, i.e. both are nonsingular and the eigenvalues are bounded by $\mu_{\text{max}}$ and $\mu_{\text{min}}$. Further assume $\bf Z = {\bf U} {\bf U}^H$ for some complex $M \times N$ matrix $\bf U$, so $\bf Z$ might be singular. I need to find under what conditions it holds that
$$\left\| {\bf A} \bf{U} \right\|_F^2 \geq \left\| {\bf B} \bf{U} \right\|_F^2$$
or equivalently
$$\text{Tr}\left( \left( {\bf A}^2 - {\bf B}^2 \right) {\bf Z}\right) \geq 0 . \tag{1}$$
Intuitively one might say that (1) is fulfilled if ${\bf A} \succeq {\bf B} \succ {\bf 0}$ holds. This intuition will (usually) also be confirmed by numerical tests with random positive semidefinite matrices ${\bf A}, {\bf B}, {\bf Z}$. But inequality (1) is not true in general. Since the matrix square is not operator monotone we have
$$ {\bf A} \succeq {\bf B} \quad \rightarrow \quad {\bf A}^2 \nsucceq {\bf B}^2.
$$
For a counter example, assume that ${\bf A}^2 - {\bf B}^2$ has some negative eigenvalue $\lambda({\bf A}^2 - {\bf B}^2) < 0$ with corresponding eigenvector $\bf u$. If matrix $\bf Z$ is chosen as ${\bf Z} = {\bf u} \bf{u}^H$ then we have $$\text{Tr}\left( \left( {\bf A}^2 - {\bf B}^2 \right) {\bf Z}\right) < 0$$
i.e., inequality (1) is not fulfilled.
So I wonder if we can make any statements on the trace inequality if we only know the bounds $\mu_{\text{max}}$ and $\mu_{\text{min}}$ on the eigenvalues of ${\bf A}$ and ${\bf B}$ and the eigenvalues of ${\bf Z}$.
My current approach is as follows: Define ${\bf C} = {\bf A}^2 - {\bf B}^2 \nsucceq {\bf 0}$ which is a hermitian matrix with real eigenvalues. Further define the sorted eigenvalues of $\bf C$ and $\bf Z$ as
$$\lambda_1({\bf C}) \geq \ldots \geq \lambda_M({\bf C}) \quad \text{ and } \quad \lambda_1({\bf Z}) \geq \ldots \geq \lambda_M({\bf Z})\geq 0.\tag{2}$$ Then we can bound the trace in (1) as follows
$$
\sum_{m=1}^M \lambda_m ({\bf C}) \; \lambda_{M-m+1} ({\bf Z}) \leq
\text{Tr} ( {\bf C} {\bf Z} ) \leq
\sum_{m=1}^M \lambda_m ({\bf C}) \; \lambda_{m} ({\bf Z}) \tag{3}
$$
(see "A Trace Inequality for Matrix Product" by Lasserre).
Also, since
$$ \text{Tr} \left( {\bf A}^2 - {\bf B}^2 \right) =
\text{Tr} \left( \left( {\bf A} + {\bf B} \right) \left( {\bf A} - {\bf B} \right) \right) \geq 0 \tag{4}
$$
we know that the sum of eigenvalues is nonnegative.
Further, by Weyl's theorem we can bound the eigenvalues of $\bf{C}$ as
$$
\mu_{\text{min}}^2 - \mu_{\text{max}}^2
\leq \lambda_M^2( {\bf A} ) - \lambda_1^2({\bf B})
\leq \lambda_M( {\bf A}^2 - {\bf B}^2 )
\leq \; ... \quad\quad\quad \\ \quad\quad\quad ... \;
\leq \lambda_1( {\bf A}^2 - {\bf B}^2 )
\leq \lambda_1^2( {\bf A} ) - \lambda_M^2({\bf B})
\leq \mu_{\text{max}}^2 - \mu_{\text{min}}^2 \tag{5} $$
where the lower (more interesting) bound is quite loose.
From equations (2)-(5) we can form a set of linear inequalities. For better illustration assume that $M=4$. Then we have
$$
\left[
\begin{matrix}
-\lambda_4({\bf Z}) & -\lambda_3({\bf Z}) & -\lambda_2({\bf Z}) & -\lambda_1({\bf Z}) \\
1 & 1 & 1 & 1 \\
1 & -1 & 0 & 0 \\
0 & 1 & -1 & 0 \\
0 & 0 & 1 & -1 \\
-1 & 0 & 0 & 0\\
0 & 0 & 0 & 1 \\
\end{matrix}
\right]
\left[
\begin{matrix}
\lambda_1({\bf C}) \\ \lambda_2({\bf C}) \\ \lambda_3({\bf C}) \\ \lambda_4({\bf C})
\end{matrix}
\right]
\geq
\left[
\begin{matrix}
0 \\ 0 \\ 0 \\ 0 \\ 0 \\ \mu_{\text{min}}^2 - \mu_{\text{max}}^2 \\ \mu_{\text{min}}^2 - \mu_{\text{max}}^2
\end{matrix}
\right] \tag{6}
$$
where the 1st ineqaulity stems from (3), the 2nd stems from (4), 3rd to 5th stem from (2) and 6th and 7th stem from (5). If the ineqaulities in (6) describe an empty set for eigenvalues $\lambda_m({\bf C})$, then inequality (1) must be true. Farka's lemma can be used to see if (6) is infeasible. I get good results if I have a good lower bound for (5), but the lower bound by Weyl's theorem is too loose. Are there any better lower bounds than Weyl's theorem?