Decoupling inequality/counterexample 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.
 A: [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, $\mathbb E|\tilde X-\tilde Y|=\Omega(1)$.
A: Let $\varepsilon \to 0$ and $M= \sqrt{(1-\varepsilon)/\varepsilon}$. Define $X$ and $Y$ as follows:

*

*w.p. $\frac{1-\varepsilon}2$: $X=Y=1$

*w.p. $\frac{1-\varepsilon}2$: $X=Y=-1$

*w.p. $\frac{\varepsilon}2$: $X=M$ and $Y=-M$

*w.p. $\frac{\varepsilon}2$: $X=-M$ and $Y=M$
Then $\mathbf{E}[XY] = (1-\varepsilon) - \varepsilon M^2 = 0$. Also, $\mathbf{E}|X-Y| = \varepsilon \cdot (2M) = 2 \sqrt{\varepsilon(1-\varepsilon)}$. Finally, $\mathbf{E}[X] = \mathbf{E}[Y] = 0$.
However, $\mathbf{E}|\tilde X- \tilde Y| \geq \mathbf{Pr}(\tilde X = - \tilde Y) \cdot 2\geq  \frac{(1-\varepsilon)^2}{2} \cdot 2 = (1-\varepsilon)^2$.

On a separate note, observe that $\mathbf{E}|\tilde X - \tilde Y| = \Theta(\mathbf{E}|\tilde X| + \mathbf{E}|\tilde Y|)$.
