An expectation inequality

Let $X$ and $Y$ be iid random variables, and $\mathbb E[|X|]<\infty$, prove that $$\mathbb E[|X+Y|]\geq\mathbb E[|X-Y|]$$

Let $F(x)$ denote the distribution, after calculation, I need to prove $$\int_{-\infty}^{+\infty}x[1-F(x)-F(-x)]F(dx)\geq0$$ but I stuck with it. Any help, thanks!

Take $G(x)=F(x)+F(-x)$, $F'(x) = f(x)$

$2\int_{-\infty}^{\infty} x(1 - F(x) - F(-x)) f(x) dx = 2\int_0^{\infty} x (1 - G(x)) f(x) - 2\int_0^{\infty} x(1 - G(-x))f(-x)dx = 2\int_0^{\infty} x(1-G(x)) (f(x) - f(-x)) = 2\int_0^{\infty} x(1-G(x))G'(x) dx \\= \displaystyle -x(1- G(x))^2\Big|_0^{\infty} + \int_0^{\infty} (1 -G(x))^2 dx =\int_0^{\infty} (1 -G(x))^2 dx \ge 0$

It is easy to show that $x(1 - G(x))^2 \to 0$, when $x\to \infty$, using the fact the expectation is finte.