Seeking a More Elegant Proof to an Expectation Inequality Let $X$ and $Y$ be i.i.d. random variables, and $\mathbb E[|X|]<\infty$, prove that $$\mathbb E[|X+Y|]\geq\mathbb E[|X-Y|].$$

This question is a re-posting of An expectation inequality. I can prove this with integration. But there must be a more elegant proof via perhaps a symmetry argument. Can someone come up with such a one?
 A: A first observation is that $|x+y|\geqslant|x-y|$ if and only if $xy\geqslant 0$, so defining 
$$f(x,y):=|x+y|-|x-y|,$$
the positivity of $f$ is linked to that of $xy$. We would like to find a more tractable expression for $f$.


*

*Assume that $x\gt 0$ and $ y\gt 0$. Then $f(x,y)=x+y-|x-y|=2\min\{x,y\}
=\min\{|x|,|y|\}$. 

*Since $f(-x,-y)=f(x,y)$ we get $f(x,y)=2\min\{|x|,|y|\}$ if $xy\gt 0$. 

*Assume that $x\gt 0$ and $y\lt 0$. Then $x-y\gt 0$, hence $|x-y|=x-y$
$$f(x,y)=|x+y|+y-x=-(x-y-|x-(-y)|)=-2\min\{x,-y\}=-2\min\{|x|,|y| \}.$$

*By symmetry of $f$, we obtain this expression if $x\lt 0$ and $y\gt 0$. 


To sum up: for each $(x,y)\in\mathbf R^2$, 
$$f(x,y)= 2\min\{|x|,|y|\}\left(\mathbf 1\{xy\gt 0\}-\mathbf 1\{xy\lt 0\}\right).$$
Now we compute the expectation: using the fact that the random variable are i.i.d., we have 
$$\mathbb E[\min\{|X|,|Y|\}\mathbf 1\{XY\gt 0\}]=\int_0^{+\infty}\mu\{X\gt t\}^2+ \mu\{-X\gt t\}^2\mathrm dt  \mbox{ and }   $$
$$\mathbb E[\min\{|X|,|Y|\}\mathbf 1\{XY\lt 0\}]=2\int_0^{+\infty}\mu\{X\gt t\}\cdot \mu\{-X\gt t\} \mathrm dt. $$
This follows from the equality
$$\mathbb E[Y]=\int_0^{+\infty}\mu\{Y\gt t\}\mathrm dt $$
and the fact that 
$$\mu\left(\{ \min\{|X|,|Y|\}\gt t \}\cap\{X\gt 0\}\cap\{Y\gt 0\}\right) 
=\mu\{X\gt t\}\mu\{Y\gt t\} ,$$
$$\mu\left(\{ \min\{|X|,|Y|\}\gt t \}\cap\{X\lt 0\}\cap\{Y\lt 0\}\right) 
=\mu\{-X\gt t\}\mu\{-Y\gt t\}\mbox{ and }   $$
$$\mu\left(\{ \min\{|X|,|Y|\}\gt t \}\cap\{X\lt 0\}\cap\{Y\gt 0\}\right) 
=\mu\{X\gt t\}\mu\{-Y\gt t\}.   $$
We thus infer that 
$$\mathbb E|X+Y|-\mathbb E|X-Y|=2
\int_0^{\infty}\left(\mu\{X\gt t\}-\mu\{-X\gt t\} \right)^2\mathrm dt. $$
This gives the wanted lower bound, and it shows that the equality is achieved if and only if $X$ is symmetric.
