I am embarrassed to be stuck on this seemingly simple question.


Suppose that $X,Y$ are mean-zero real-valued random variables and $\tilde X,\tilde Y$ are their "independent copies": $\tilde X,\tilde Y$ are mutually independent and independent of $(X,Y)$, and $\tilde X$ (resp., $\tilde Y$) is distributed identically to $X$ (resp., $Y$).


Here is the inequality I am trying to prove/disprove:
for some universal constant $c>0$,
$$
\mathbb{E}|\tilde X-\tilde Y|
\le
c\left(
\mathbb{E}|X-Y|
+
\sqrt{|\mathbb{E}XY|}
\right).
$$


Update. Note that the related inequality,
$$
\mathbb{E}|\tilde X-\tilde Y|^2
\le
\mathbb{E}|X-Y|^2
+
2|\mathbb{E}XY|
,
$$
is trivially true.



Decoupling inequality/counterexample
[**EDITED** *to ensure that the random variables have expectation 0*]
I think the answer is no.


Let $Z$ be a random variable taking the values $\pm 1$ with probability $p$ each and 0 with probability $1-2p$; and let $N$ be a standard normal independent of $Z$.


Set $X=ZM+(1-|Z|)N$ and $Y=-ZM+(1-|Z|)N$, where $M=\sqrt{(1-2p)/(2p)}$.
Now $\mathbb E|X-Y|=2pM=O(\sqrt p)$ and $\mathbb EXY=-2pM^2+(1-2p)=0$.
On the other hand, $\mathbb E|\tilde X-\tilde Y|=\Omega(1)$.


