As suggested by Robin, we would need to prove that if $X$ and $Y$ are independent random variables with the same distribution, the inequality
$E|X|\leq E|X-Y|$
holds.

This is certainly NOT true in general; for example suppose $X$ and $Y$ have some constant non-zero value; then the RHS is 0 and the LHS is positive.

However, if you add the condition that $E(X)=0$ then the result is true. (In your notation,
$\int_{-\infty}^{\infty}xf(x)=0$ ).

In that case write $p=P(X\geq 0)$, and write $X_+$ for the positive part of $X$ and $X_-\leq 0$ for the negative part of $X$.

We have $0=EX=E|X_+|-E|X_-|$, so $E|X_+|=E|X_-|$.

Hence also $E|X|=E|X_+|+E|X_-|=2E|X_+|$.

Now $E|X-Y| \geq E(|X-Y|I(Y<0)I(X\geq 0)) + E(|X-Y|I(X<0)I(Y \geq 0))$

$=2 E(|X-Y|I(Y<0)I(X\geq0))$

$=2 E(|X_+|I(Y<0)I(X\geq0)) + 2E(|Y_-|I(Y<0) I(X\geq 0))$

$=2(1-p)E(|X_+|I(X\geq 0)) + 2p E(|Y_-|I(Y<0))$

$=2(1-p)E|X_+| + 2pE|Y_-|$

$=2E|X_+|$

$=E|X|$

(using at various times the facts that $X$ and $Y$ are independent and
that $X$ and $Y$ have the same distribution)