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.
By a simple $u$-substitution, we find that
$$ \int_{-\infty}^{\infty} \frac{1-\cos(at)}{t^2} \, \mathrm{d}t = C|a|, \tag{1}$$
where $C = \int_{-\infty}^{\infty} \frac{1-\cos t}{t^2} \, \mathrm{d}t$ is a positive, finite constant. (It can be shown that $C = \pi$, although the exact value of $C$ plays no role in this solution.)
Note that the integrand of $\text{(1)}$ is non-negative. Taking advantage of this fact, we can apply Tonelli's theorem to find that, for any real-valued random variable $Z$, the following identity holds:
$$ \Bbb{E}[|Z|] = \frac{1}{C} \Bbb{E}\left[ \int_{-\infty}^{\infty} \frac{1-\cos(Zt)}{t^2} \, \mathrm{d}t \right] = \frac{1}{C} \int_{-\infty}^{\infty} \frac{1-\Bbb{E}[\cos(Zt)]}{t^2} \, \mathrm{d}t $$
Therefore
\begin{align*} \Bbb{E}[|X+Y| - |X-Y|] &= \frac{1}{C} \int_{-\infty}^{\infty} \frac{\Bbb{E}[\cos((X-Y)t)-\cos((X+Y)t)]}{t^2} \, \mathrm{d}t \\ &= \frac{1}{C} \int_{-\infty}^{\infty} \frac{\Bbb{E}[2\sin(Xt)\sin(Yt)]}{t^2} \, \mathrm{d}t \\ &= \frac{1}{C} \int_{-\infty}^{\infty} \frac{2\Bbb{E}[\sin(Xt)]^2}{t^2} \, \mathrm{d}t \\ &\geq 0. \tag{2} \end{align*}
Moreover, the equality holds for $\text{(2)}$ if and only if $\Bbb{E}[\sin(Xt)] = 0$ for all $t$. This implies that the characteristic function $\varphi_X(t) = \Bbb{E}[e^{itX}]$ is real-valued, which in turn is equivalent to the symmetry condition: $X \stackrel{d}{=} -X$.
Remark. Using a similar argument, we can show that: