I have two (real-valued) independent random variables $X,Y$, and a function $f\colon \mathbb{R}\times\mathbb{R}\to\mathbb{R}$. Is it true that $$ \mathbb{E}[\lvert f(X,Y) \rvert]+\lvert\mathbb{E}[ f(X,Y) ] \rvert \geq \mathbb{E}[\lvert \mathbb{E}[f(X,Y) \mid X]\rvert]+\mathbb{E}[\lvert \mathbb{E}[f(X,Y) \mid Y]\rvert] \tag{$\dagger$} $$ ? (If not, what is a counter-example?)
It looks eerily simple and intuitive, but at that point I wouldn't believe in my intuition, nor in eeriness in general. Note that if $f$ only depends on its first variable (for instance), i.e. if $f(X,Y) = g(X)$ for some function $g$, then setting $Z=g(X)$ we have $$ \mathbb{E}[\lvert f(X,Y) \rvert]+\lvert\mathbb{E}[ f(X,Y) ] \rvert = \mathbb{E}[\lvert Z \rvert]+\lvert\mathbb{E}[ Z ] \rvert $$ and $$ \mathbb{E}[\lvert \mathbb{E}[f(X,Y) \mid X]\rvert]+\mathbb{E}[\lvert \mathbb{E}[f(X,Y) \mid Y]\rvert]= \lvert \mathbb{E}[Z]\rvert+\mathbb{E}[\lvert Z\rvert] $$ and we actually have equality in ($\dagger$). If $X,Y$ are two i.i.d. random variables, then ($\dagger$) seems to be equivalent to $$ \mathbb{E}[\lvert X\rvert]^2 + \mathbb{E}[X]^2 \geq 2\mathbb{E}[\lvert X\rvert]\lvert\mathbb{E}[X]\rvert $$ which again is true (this time from e.g. the AM-GM inequality).
The inequality is not true in general; here is a counter example. If $X$ is uniformly distributed on $\{1,2\}$ and $Y$ is uniformly distributed over $\{-1,-2\}$. Then if $f(X,Y)=X+Y$: $$ P(X+Y=-1)=P(X+Y=1)=\frac 14. $$ We have: $$ E[|f(X;Y)|]=\frac 12\\ E[f(X;Y)]=0\\ E[f(X;Y)|X]=X-\frac 32\in\{-\frac 12,\frac 12\}\\ E|E[f(X;Y)|X]|=E[\frac 12]=\frac 12\\ E[f(X;Y)|Y]=Y+\frac 32\in\{-\frac 12,\frac 12\}\\ E[|E[f(X;Y)|Y]|]=\frac 12. $$