Why is the shift the optimal plan between $[0,1)$ and $[1,2)$ (with distance-squared cost function)? Example 1.3 of Optimal and Better Transport Plans
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Consider the task to transport points on the real line (equipped with the Lebesgue measure) from the interval [0, 1) to [1, 2) where the cost of moving one point to another is the squared distance between these points ($X = [0, 1)$, $Y = [1, 2)$, $c(x, y) = (x − y)^2$, $\mu = \nu = \lambda$). The simplest way to achieve this transport is to shift every point by 1. This results in transport costs of 1 and one easily checks that all other transport plans are more expensive.

Why are all other transport plans more expensive? I.e., given a transport plan $P$ (which is a measure on $[0,1) \times [1,2)$ such that $P \circ \pi_{1}^{-1} = \mu$ and $P \circ \pi_{2}^{-1} = \nu$, where $\pi_i$ is the projection onto the $i$-th coordinate) why is it true that
$$
\int_{[0,1) \times [1,2)} (x-y)^2 \, dP \geq 1 ?
$$
What I tried initially. By writing $(x-y)^2 = x^2 -2xy + y^2$ and using the fact that integrating a function of $x$ with respect to $P$ is the same as integrating with respect to $\mu$ (and similarly for $y$ and $\nu$) gives
$$
\begin{align*}
\int_{[0,1) \times [1,2)} (x-y)^2 \,dP &= \int_{0}^{1}x^{2} \,dx - 2\int_{[0,1) \times [1,2)} xy \,dP + \int_{1}^{2} y^{2}\, dy \\
&= 8/3 - 2\int_{[0,1) \times [1,2)} xy \,dP,
\end{align*}
$$
so it suffices to show that $\int_{[0,1) \times [1,2)} xy \,dP \leq 5/6$. I don't see how to show this.
Addendum. I'm not sure if what I tried above was leading to a solution, but I think I have the answer, based on @Niels Diepeveen's hint.
As noted in my comment to Niels' answer, it suffices to show that the 'diagonal' transport plan $P_{o}$ on $[0,1)\times [0,1)$ is optimal for the cost function $c(x,y+1) = (x-y-1)^{2}$ ('diagonal' here means that $P_{o}$ is supported by $\{(x,y):x=y\}$. As noted by Niels, it is clear that $P_{o}$ is optimal for the cost function $c(x,y)=(x-y)^{2}$. But $c(x,y+1) = c(x,y) -2(x-y)+ 1$, so if $P$ is any transport plan,
$$
\begin{align*}
\int_{[0,1)^{2}} c(x,y+1)\, dP &= \int_{[0,1)^{2}} (c(x,y) - 2(x-y) + 1) \, dP \\
&= \int_{[0,1)^{2}} c(x,y) \, dP - 2\int_{[0,1)^{2}} x \, dP + 2\int_{[0,1)^{2}} y \, dP + 1  \\
&= \int_{[0,1)^{2}} c(x,y) \, dP - 2\int^{1}_{0} x \, d\lambda + 2\int^{1}_{0} y \, d\lambda + 1  \\
&= \int_{[0,1)^{2}} c(x,y) \, dP +1 \\
&\geq \int_{[0,1)^{2}} c(x,y) \, dP_{o} +1 \\
&= \int_{[0,1)^{2}} c(x,y+1)\, dP_{o}, \qquad \text{(By the computation just done.)}
\end{align*}
$$
where we've used the fact that integrating a function of just one variable with respect to a transport measure (plan) is the same as integrating that same function with respect to the marginal measure.
 A: Hint:
There is an obvious correspondence between plans for moving $[0, 1)$ to $[1, 2)$ and plans for moving $[0, 1)$ to itself, and the latter is clearly optimized by
a diagonal measure. All you need to do is show that the costs are related by
a strictly increasing function.

Clarification:
Essentially, what you have done is what I was hinting at inasmuch as you
have proved that translating the target simply adds a constant to the cost.
What I would do at that point is sit back and observe that the two problems
are isomorphic in the sense that there is a bijection $T$ between the feasible
sets which induces an order isomorphism between the images of the objectives,
that is, you have
$$
J_1(T(x)) < J_1(T(y)) \iff J_0(x) < J_0(y)
$$
This implies that $T$ and $T^{-1}$ take minimizers to minimizers.
At this point, you might be content, or you might use the isomorphism you
have found to transform the minimality proof for one problem into  an
elementary proof for the other. You can do that by rewriting
$$
\begin{eqnarray*}
\int_{[0,1) \times [1,2)} (x-y)^2 dP 
  &=& \int_{[0,1) \times [1,2)} ((x - y +1) - 1)^2 dP \\
  &=& \int_{[0,1) \times [1,2)} (x - y + 1)^2 dP + \int_{[0,1) \times [1,2)} -2x + 2y -1 \,  dP \\
\end{eqnarray*}
$$
then arguing that the second integral is constant under the constraints and
the first can be minimized by concentrating $P$ where the integrand is minimal
i.e. where $y = x + 1$.
