Are there fairly general conditions on $f\colon\mathbb{R}\to\mathbb{R}$ such that $$ E[f(X^2)] \geq E[f(X\tilde{X})] $$ for all random variables $X$, where $\tilde{X}\sim X$ can be arbitrarily dependent on $X$ but has the same distribution?
For $f=Id$ and $\tilde{X}$ independent of $X$ the inequality becomes $E[X^2]\geq E[X]^2$, which can be proven either via Var X>0 or via Jensen's inequality, neither of which generalizes to the conjectured generalization.
For $f=\log$, the conjectured inequality holds sharply (and trivially) for all $X$.
Unfortunately, I didn't make more progress than that.
Here is a counterexample. Let $X$ be uniformly distributed among $1,2,3,4,5$. Let $\tilde{X}$ have the values $1,5,4,3,2$ in the corresponding situations. Then $$ f(y)= \begin{cases} 1 & \text{if } y\ge 10\\ 0 & \text{otherwise} \end{cases}\ \ \implies\ \ E[f(X\tilde{X})]=\frac45>\frac25=E[f(X^2)]$$
As a similar but smooth variant, $$f(y)=\arctan(y-9) \ \ \implies\ E[f(X\tilde{X})]=.524>.023=E[f(X^2)]$$