Let ${\xi}$ and ${\eta}$ have equal distribution on the interval [0;1]. Show that despite the distribution of vector (${\xi}$, ${\eta}$), $E|{\xi}-{\eta|}\le{\frac{1}{2}}$. I was thinking about the fact that the mathematical expectation of absolute value of a random variable is an integral from 0 to + infinity of 1 minus the distribution function of the foretold random variable. But that’s all I can say right now. All suggestions are welcome
 A: The inequality does not hold when $X$ and $Y$ are not independent. Consider indeed $X$ such that $P(X=0)=P(X=1)=\frac 12$ and let $Y=1-X$, so that $X$ and $Y$ have the same distribution. 
Then $E(|X-Y|) = E(|2X-1|)$ and $|2X-1|=1$ a.s, hence  $E(|X-Y|)=1>\frac 12$.
If $X$ and $Y$ are independent, the identity $E(|X-Y|)=2\int_0^1 F(t)(1-F(t)) dt$ and the bound $x(1-x)\leq \frac 14$ for $x\in [0,1]$ yield $E(|X-Y|)\leq \frac 12$.
A: This follows from a coupling argument: by Kantorovich's duality, 
$$\sup_{\xi,\eta\sim\text{Unif}([0,1])}\mathsf E|\xi-\eta|=\inf_{\phi\in L^1([0,1])}\left(\int_0^1\sup_{x\in[0,1]}(\phi(x)+|y-x|)dy-\int_0^1\phi(x)dx\right).$$
Set $\phi(x)=\frac12-\left|x-\frac12\right|$, it is straightforward to check that
$$\int\sup_{y\in[0,1]}(\phi(y)+|x-y|)dx-\int\phi(x)dx=\frac34-\frac14=\frac12,$$
as desired.

Addendum. We do not need the full strength of Kantorovich's duality here, what is actually needed is only
$$\mathsf E|\xi-\eta|\le\int_0^1\sup_{x\in[0,1]}(\phi(x)+|y-x|)dy-\int_0^1\phi(x)dx$$
for any $\phi\in L^1([0,1])$. In this case the above inequality can be easily obtained by
\begin{align}
&\int_0^1\sup_{x\in[0,1]}(\phi(x)+|y-x|)dy-\int_0^1\phi(x)dx\\
=&\mathsf E\sup_{x\in[0,1]}(\phi(x)+|\eta-x|)-\mathsf E\phi(\xi)\\
\ge&\mathsf E(\phi(\xi)+|\eta-\xi|)-\mathsf E\phi(\xi)\\
=&\mathsf E|\eta-\xi|.
\end{align}
