probability theory - If $X$, $Y$ are i.i.d. with mean $m$ then $E((X-Y)mathrm{sign}(X-m))>0$

It is obvious that if $X$ and $Y$ are independent and uniformly distributed on $[a,b]$ random variables, and $E[X]=E[Y]=m\equiv\frac{a+b}{2}$, then the random variable $Z=(X-Y)\mathrm{sign}(X-m)$ has the positive expectation, that is, $E[Z]>0$.

Indeed,
$$E[Z]=\frac{1}{(b-a)^2}\int_{a}^{b}\int_{a}^{b}(x-y)\mathrm{sign}(x-m)dxdy=$$
$$=\frac{1}{b-a}\int_{a}^{b}x\mathrm{sign}(x-m)dx-\frac{1}{(b-a)^2}\int_{a}^{b}\mathrm{sign}(x-m)dx\int_{a}^{b}ydy=$$
$$=\frac{1}{b-a}\Big(\int_{m}^{b}xdx-\int_{a}^{m}xdx\Big)-\frac{m}{b-a}\int_{a}^{b}\mathrm{sign}(x-m)dx=$$
$$=\frac{1}{b-a}\Big(\int_{m}^{b}xdx-\int_{a}^{m}xdx\Big)-\frac{m}{b-a}\Big(\int_{m}^{b}dx-\int_{a}^{m}dx\Big)=\frac{b-a}{4}>0.$$

Now, let $X$ and $Y$ be independent and identically distributed random variables with the probability density function $f(x)$, $E[X]=E[Y]=m$, and let $Z=(X-Y)\mathrm{sign}(X-m)$. Is it true that $E[Z]>0$?

My calculations:
$$E[Z]=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}(x-y)\mathrm{sign}(x-m)f(x)f(y)dxdy=$$
$$=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}x\mathrm{sign}(x-m)f(x)f(y)dxdy-\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}y\mathrm{sign}(x-m)f(x)f(y)dxdy=$$
$$=\int_{-\infty}^{\infty}xf(x)\mathrm{sign}(x-m)dx-m\int_{-\infty}^{\infty}f(x)\mathrm{sign}(x-m)dx=$$
$$=\int_{m}^{\infty}xf(x)dx-\int_{-\infty}^{m}xf(x)dx-m\Big(\int_{m}^{\infty}f(x)dx-\int_{-\infty}^{m}f(x)dx\Big).$$

And what will be if $X$ and $Y$ are not independent but identically distributed random variables with the joint probability density function $f(x,y)$, $E[X]=E[Y]=m$; $Z=(X-Y)\mathrm{sign}(X-m)$? Is it true that $E[Z]>0$ in this case too?

In this case we have
$$E[Z]=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}(x-y)\mathrm{sign}(x-m)f(x,y)dxdy=$$
$$=\int_{m}^{\infty}\int_{-\infty}^{\infty}(x-y)f(x,y)dxdy-\int_{-\infty}^{m}\int_{-\infty}^{\infty}(x-y)f(x,y)dxdy.$$