A strict inequality for the $L^1$-norm of a symmetrized zero-mean random variable

Question

Suppose that $Y$ is an independent copy of a random variable (r.v.) $X$ with a zero-mean nondegenerate distribution. Is it then always true that $E|X-Y|>E|X|$?

To get the non-strict version of this inequality, condition on $X$ and then apply Jensen's inequality to the zero-mean random variable $Y$.

If $p\in(1,\infty)$ and $E|X|^p<\infty$, then $E|X-Y|^p>E|X|^p$ -- because then the function $|\cdot|^p$ is strictly convex.

In view of the well-known expression of the absolute moments in terms of the characteristic function (c.f.), the highlighted question can be restated as follows:

Suppose that $X$ is a r.v with a zero-mean nondegenerate distribution and c.f. $f$. Is it then always true that $$\int_0^\infty\frac{dt}{t^2}\,(\Re f(t)-|f(t)|^2)>0\,\text{?}$$

Answer 1

No, a symmetric random sign gives equality ($X$ is $\pm 1$ with probability $1/2$).