I don't know how to prove this:
If $X$ and $Y$ are i.i.d. then $\mathbb{E}(|X|) \le \mathbb{E}(|X+Y|)$
I first want to use conditional expectation. When $\mathbb{E}(X)=0$, I know it is right. $\mathbb{E}(X+Y|X)=X$, so $\mathbb{E}(|X|) \le \mathbb{E}(|X+Y|)$.
But with the general situation, how can I solve it? How can I use the same distribution?
It is known that
$$ \mathbb E|X-Y| \le \mathbb E|X+Y| \tag{1}$$
for $X$ and $Y$ that are independent random variables with the same finite expectation [1].
As
$$|x|= \left | \frac12 (x-y)+ \frac12 (x+y)\right | \le \frac12|x-y|+\frac12|x+y|$$
and from (1) we have
$$ \mathbb E|X| \le \frac12 \mathbb E \left |X-Y\right |+\frac12 \mathbb E \left |X+Y\right |\le \mathbb E \left |X+Y\right |.$$
PS: The inequality holds if $X$ and $Y$ are independent with the same expectation, which is much weaker than having the same distribution.
