TITLE: Prove that $\Bbb{E}(|X-Y|) \le \Bbb{E}(|X+Y|)$ for i.i.d $X$ and $Y$
QUESTION [25 upvotes]: Let $X$ and $Y$ be two independent identically distributed random
  variables with finite expectation $\Bbb{E}(X) = \Bbb{E}(Y) < \infty$. Prove that
$$\Bbb{E}(|X-Y|) \le \Bbb{E}(|X+Y|)$$

I think that this inequality may follow somehow from Jensen's inequality, but I failed to use it here. Or maybe it is worth considering an expression $|x+y|-|x-y|$ and making use of some of its properties?
I am interested to see a proof of this fact or some favorable ideas that may help here. Any suggestions would be greatly appreciated.

REPLY [8 votes]: Here's another argument. It doesn't seem to generalize to $p$ norms, but perhaps it is instructive anyway.
For independent variables $X,Y$ with finite expectation we can write
\begin{align*}
\mathbb E|X-Y|
&=\int_{-\infty}^\infty \mathbb P[X\leq t< Y]+\mathbb P[Y\leq t<X] dt\\
&=\int_{-\infty}^\infty F_X(t)(1-F_Y(t))+F_Y(t)(1-F_X(t)) dt\tag{1}
\end{align*}
where $F_Z$ denotes the cumulative distribution function $F_Z(t)=\mathbb P[Z\leq t]$ of a random variable $Z.$
In the case at hand, $X$ and $Y$ are i.i.d., so $F_{-Y}(t)=1-F_X(-t).$ Using $-Y$ instead of $Y$ in (1) gives
$$\mathbb E|X+Y|=\int_{-\infty}^\infty F_X(t)F_X(-t)+(1-F_X(-t))(1-F_X(t)) dt\tag{2}$$
We can get a comparable integrand for $\mathbb E|X-Y|$ by substituting $t$ for $-t$ in the final term in (1), and using $F_Y=F_X$:
$$
\mathbb E|X-Y|=\int_{-\infty}^\infty F_X(t)(1-F_X(t))+F_X(-t)(1-F_X(-t)) dt\tag{3}
$$
Writing $a=F_X(t)$ and $b=F_X(-t),$ clearly $((1-a)-b)((1-b)-a)=(1-a-b)^2\geq 0,$
so $a(1-a)+b(1-b)\leq ab+(1-a)(1-b).$ Integrating over $t$ and applying (3) and (2) gives $\mathbb E|X-Y|\leq \mathbb E|X+Y|.$ The equality case is when $F_X(-t)+F_X(t)=1$ a.e., which is when $X$ is symmetric about zero.