Suppose that $A$, $B$ are deterministic $n\times n$ matrices and $G$ a Gaussian matrix of i.i.d. entries $N(0,1)$.

I'd like to establish an upper bound of the trace norm of $AGB$ as $$ \mathbb{E}\|AGB\|_\ast \leq \max\{\|A\|_\ast \|B\|_F, \|A\|_F\|B\|_\ast \}, $$ where $\|\cdot\|_\ast$ denotes the trace norm and $\|\cdot\|_F$ the Frobenius norm. I can show the inequality with an additional $\log n$ factor on the right-hand side but I do not think the $\log n$ factor is necessary...

(**First question**: is it true to replace $\max$ with $\min$?)

[Update] As answered by Mikael de la Salle below, it is possible to replace $\max$ with $\min$ for $p\in [1,2]$, to obtain that $$ \mathbb{E}\|AGB\|_p \leq \min\{\|A\|_p \|B\|_F, \|A\|_F \|B\|_p\}. $$

For $p>2$, it is necessary to have $\max$. Take $A$ to be identity matrix and $B = e_1e_1^T$ (zero matrix except the top-left entry being $1$), then $\|AGB\|_p \sim \sqrt{n}$ while $\|A\|_p \|B\|_F = n^{1/p} < \sqrt{n}$.

**Second question**: Does the following hold for $p>2$?
$$
\mathbb{E}\|AGB\|_p \leq C\max\{\|A\|_p \|B\|_F, \|A\|_F \|B\|_p\}.
$$
for some constant $C$ (which can depend on $p$)? I could get the inequality with an extra $\log n$ factor on the right-hand side, but I am not sure if it is necessary. For $p=\infty$, it is too easy to get $\sqrt{n}\|A\|_{op}\|B\|_{op}$, which is bigger than the bound I want.

**Third question**: How about the lower bound on $\mathbb{E}\|AGB\|_p$? Do we have
$$
\mathbb{E}\|AGB\|_{p} \geq c\max\{\|A\|_p \|B\|_F, \|A\|_F \|B\|_p\}
$$
for some constant $c$?

I can show that $$ \mathbb{E}\|AGB\|_{op} \geq \max\{\|A\|_{op} \|B\|_F, \|A\|_F \|B\|_{op}\} $$ So I think the lower bound inequality holds for $p\geq 2$. For $p = 1$, if we take $A$ to be identity matrix and $B=e_1e_1^T$ then $\mathbb{E}\|AGB\|_{p}\approx \sqrt{n}$, so the inequality cannot hold with $\max$. So the question is, will it hold $$ \mathbb{E}\|AGB\|_{p} \geq c\min\{\|A\|_p \|B\|_F, \|A\|_F \|B\|_p\} $$ for $p\in [1,2)$? Again I can show it with an extra $1/\log n$ factor on the right but I am not sure if it is necessary.