Holder inequality for matrices

Recall that $|U|=(U^*U)^{1/2}$. If $D=I$, then, in general, the proposed inequality does not work; the correct inequality is $|tr(A^*B)|\leq (tr(|A|^p)^{1/p}(tr(|B|^q)^{1/q}$.

Of course, we consider $tr(DA^*B)$ where $D$ is symmetric $\geq 0$ (the condition $tr(D)=1$ is useless). Using a reasoning by continuity, we may assume that $D>0$ and , moreover, that $D$ is diagonal $>0$. One has $|tr(DA^*B)|=|tr(D^{1/p}A^*BD^{1/q})|=|<AD^{1/p},BD^{1/q}>|\leq (tr(|AD^{1/p}|)^p)^{1/p}(tr(|BD^{1/q}|)^q)^{1/q}$.

The question: is $tr((|AD^{1/p}|)^p)=tr(|A|^pD)$ true ?

If $p=2$, then the answer is yes; indeed the LHS is $tr(D^{1/2}A^*AD^{1/2})$ and the RHS is $tr(A^*AD)$. If $p\not= 2$, then the equality does not work (at least , I think so).