Let $X,Y$ be iid random variables with mean $\mu$ and having finite moments of, say, all orders.
It is an easy exercise to show that $$\operatorname{Var}(X) = E[|X-\mu|^2] = \frac{1}{2} E[|X-Y|^2].\tag{*}$$ If we want an $L^p$ version of this statement, $1 \le p < \infty$, we can write $$\begin{align*} E\left[|X-\mu|^p\right] &= E\left[\left|E[X-Y \mid X]\right|^p\right] \\ &\le E\left[E[|X-Y|^p \mid X]\right] &&\text{(conditional Jensen)} \\ &= E[|X-Y|^p]. \end{align*}$$ However, this does not reduce to (*) when $p=2$, because the factor of $1/2$ is missing. Is there a "better" version of this $L^p$ inequality that contains the equality for $p=2$?
It is only a partial answer, for $p\geqslant 3$. Assume that $\mu=0$. Taylor's formula gives for any $x$ and $y$ that $$\lvert x+y\rvert^p=\lvert x\rvert^p+p\lvert x\rvert^{ p-2}xy+p(p-1)\int_0^1(1-s)y^2 \lvert x+sy\rvert^{p-2}\mathrm ds . $$ Using this with $x=X$, $y=-Y$ and integrating, one gets $$\mathbb E\lvert X-Y\rvert^p= \mathbb E\lvert X\rvert^p+p(p-1)\int_0^1(1-s)\mathbb E\left[Y^2\lvert X+sY\rvert^{p-2} \right] \mathrm ds. $$ Now, since $p-2\geqslant 1$, we get by Jensen's inequality \begin{align} \mathbb E\left[Y^2\lvert X+sY\rvert^{p-2} \right]&= \mathbb E\left[\mathbb E\left[ Y^2\lvert X+sY\rvert^{p-2}\right] \mid \sigma(Y) \right] \\ &=\mathbb E\left[Y^2\mathbb E\left[ \lvert X+sY\rvert^{p-2}\right] \mid \sigma(Y) \right]\\ &\geqslant \mathbb E\left[Y^2\lvert\mathbb E\left[ X+sY\mid \sigma(Y) \right]\rvert^{p-2} \right]\\ &=s^{p-2}\mathbb E\lvert Y\rvert^p \end{align}
hence $$\mathbb E\lvert X-Y\rvert^p\geqslant \mathbb E\lvert X\rvert^p\left( 1+p(p-1)\int_0^1(1-s)s^{p-2} \mathrm ds\right) . $$
We end with $$\mathbb E\lvert X\rvert^p\leqslant \frac 12\mathbb E\lvert X-Y\rvert^p.$$