How to prove $ E(|X-Y|) \le E(|X+Y|)$ when $X,Y$ are i.i.d variables? Let $X$ and $Y$ be independent random variables having the same distribution and the finite mathematical expectation. How to prove the inequality $$ E(|X-Y|) \le E(|X+Y|)?$$
After a little inspection, we see that
$$
E(|X+Y|-|X-Y|) = 2E[Z(1_{XY\geq 0}-1_{XY&lt;0})]
$$
where $Z = \min(|X|,|Y|)$.

Remember that for any non-negative random variable $T$,
$$
E(T) = \int_0^\infty P(T&gt;t)\,dt.
$$

We apply this with $T=Z\,1_{X \geq 0, Y\geq 0}$, $T=Z\,1_{X &lt; 0, Y&lt; 0}$ and $T=Z\,1_{X \geq 0, Y&lt; 0}$. Since $\{Z &gt; t\} = \{|X| &gt; t\}\cap\{|Y| &gt; t\}$, we obtain

$$
E(Z \,1_{X \geq 0,Y\geq 0}) = \int_0^\infty P(X &gt; t)P(Y &gt; t)\,dt = \int_0^\infty P(X &gt; t)^2\,dt
$$

$$
E(Z\, 1_{X &lt; 0, Y &lt; 0}) = \int_0^\infty P(X &lt; -t)P(Y &lt; - t)\,dt = \int_0^\infty P(X &lt; -t)^2\,dt
$$

$$
E(Z\,1_{X \geq 0, Y&lt; 0}) = E(Z\,1_{\{X &lt; 0, Y \geq 0\}}) = \int_0^\infty P(X &gt; t)P(X &lt; -t)\,dt
$$

So finally,
$$
E(|X+Y|-|X-Y|) = 2\int_0^\infty (P(X&gt;t)-P(X&lt;-t))^2\,dt \geq 0
$$



Remark 1. The inequality is an equality if and only if the distribution of $X$ is symetric, that is $P(X &gt; t) = P(X &lt; -t)$ for any $t \geq 0$.



Remark 2. When $|X|=1$ a.s. the inequality is nothing but the semi-trivial fact that if $X$ and $Y$ are independent with same distribution, then $P(XY \geq 0) \geq \dfrac{1}{2}$.



Remark 3. It is worthwile to mention a nice corollary : $E(|X+Y|) \geq E(|X|)$. The function $x \mapsto |x|$ is convex hence $|X| \leq \frac{1}{2}(|X+Y|+|X-Y|)$. Taking expectations we find
$$
\Bbb E(|X+Y|-|X|) \geq \frac{1}{2}\Bbb E(|X+Y|-|X-Y|) \geq 0.
$$
Furthermore, there is an equality if and only if $X=0$ a.s.
E[x] is a linear operator. 

This means E[X + Y] = E[X] + E[Y]

Also, E[X - Y]  = E[X] - E[Y]

The statement will be true when $E[Y] \ge 0$