Prove that $\Bbb{E}(|X-Y|) \le \Bbb{E}(|X+Y|)$ for i.i.d $X$ and $Y$

Taking integration by parts to the Dirichlet integral, it is easy to check that

$$ \int_{-\infty}^{\infty} \frac{1-\cos(at)}{t^2} \, dt = \pi|a|. \tag{1}$$

Taking advantage of the fact that the integrand of $\text{(1)}$ is non-negative, by the Tonelli's theorem, for any real-valued random variable $Z$ we have

$$ \pi \Bbb{E}[|Z|] = \Bbb{E}\left[ \int_{-\infty}^{\infty} \frac{1-\cos(Zt)}{t^2} \, dt \right] = \int_{-\infty}^{\infty} \frac{1-\Bbb{E}[\cos(Zt)]}{t^2} \, dt. $$

Therefore

\begin{align*} \pi \Bbb{E}[|X+Y| - |X-Y|] &= \int_{-\infty}^{\infty} \frac{\Bbb{E}[\cos((X-Y)t)-\cos((X+Y)t)]}{t^2} \, dt \\ &= \int_{-\infty}^{\infty} \frac{\Bbb{E}[2\sin(Xt)\sin(Yt)]}{t^2} \, dt \\ &= \int_{-\infty}^{\infty} \frac{2\Bbb{E}[\sin(Xt)]^2}{t^2} \, dt \\ &\geq 0. \end{align*}

Moreover, notice that the equality holds if and only if $\Bbb{E}[\sin(Xt)] = 0$ for all $t$. This means that the c.f. $\varphi_X(t) = \Bbb{E}[e^{itX}]$ is real-valued, which is equivalent to the symmetry condition: $X \stackrel{d}{=} -X$.