The comparison you want applies to any kind of tuples of numbers, whether vectors or arranged in blocks like a matrix. Namely, the Cauchy-Schwarz-Bunyakowsky inequality is $ \Big(\sum_n |a_nb_n|\Big)^2 \le \sum_n |a_n|^2\;\cdot \sum_n |b_n|^2 $
[Edit: additions] There are obviously many questions one could ask about the particulars... For example, for set-up, for $A$ self-adjoint, $A=UDU^{-1}$ with $U$ unitary and $D$ real diagonal, and the "absolute value" $|A|$ is $U|D|U^{-1}$ where $|D|$ has absolute values of $D$'s entries. Ok, and then the trace-norm of $A$ is the sum of the absolute values of eigenvalues, and the Hilbert-Schmidt norm of $A$ is the square root of the sum of squares of absolute values of eigenvalues. When $A,B$ are simultaneously diagonalizable (e.g., commute and are self-adjoint), then the Hadamard product is just the product, and $|A\circ B|=|A|\circ |B|$. In this simple case, Cauchy-Schwarz-B. implies that the HS-norm of $AB$ is at most the product of the HS norms. (This was the content of my earlier.) When $A,B$ do not necessarily commute, but at least one of them is self-adjoint (or normal), we can still take one (say $A$) to be diagonal, without loss of generality, and then the Hadamard product $A\circ B$ kills off all but the diagonal entries of $B$. Killing off entries reduces norms, etc.
Is this a relevant amplification?