How to reconcile these $L^p$ and $L^2$ (in)equalities? 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 $p=2$?
 A: 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 
$$\mathbb E\lvert X\rvert^p\leqslant \frac 12\mathbb E\lvert X-Y\rvert^p.$$
