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, $$ \mathbb{E}|\tilde X-\tilde Y|^2 \le \mathbb{E}|X-Y|^2 + |\mathbb{E}XY| , $$ is trivially true.

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.

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 $c>0$, $$ \mathbb{E}|\tilde X-\tilde Y| \le c\left( \mathbb{E}|X-Y| + \sqrt{|\mathbb{E}XY|} \right). $$

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, $$ \mathbb{E}|\tilde X-\tilde Y|^2 \le \mathbb{E}|X-Y|^2 + |\mathbb{E}XY| , $$ is trivially true.