Exercise 1.21 of Villani Topics in Optimal Transportation I'm stuck on this exercise, can you please help me? 

1.2.2. Transshipment. The Kantorovich-Rubinstein theorem implies that
  the total cost only depends on the difference $\mu-\nu$ . Thus, when the cost
  function is a metric, the Kantorovich optimal transportation problem is
  equivalent to the Kantorovich-Rubinstein transshipment problem:
  $$\inf \{I[\pi] ; \quad \pi[A \times X]-\pi[X \times A]=(\mu-\nu)[A]\}$$
  The condition appearing above should be compared to the condition for
  $\pi \in \Pi(\mu, \nu),$ which is $\pi[A \times X]=\mu[A], \pi[X \times A]=\nu[A] .$ For a general cost function, the transshipment problem is a strongly relaxed version of
  the transportation problem. For instance, in the case of a quadratic cost
  in $\mathbb{R}^{n},$ the optimal transshipment cost between two given measures is in
  general 0. We shall not study the transhipment problem in this course, and
  refer to $[211]$ for motivations and detailed study.
Exercise 1.20. Give an interpretation of the Kantorovich-Rubinstein transshipment problem in (say) economics terms; contrast this interpretation with
  that of the Monge-Kantorovich problem.
Exercise 1.21 (Transshipment sometimes costs (almost) nothing).
  Let $c(x, y)=|x-y|^{2}$ in $\mathbb{R}^{n},$ and let $\mu, \nu$ be two probability measures
  on $\mathbb{R}^{n},$ such that $\mathcal T_{c}(\mu, \nu)<+\infty .$ Let $\pi \in \Pi(\mu, \nu)$ be any transference plan such that $I[\pi]<+\infty$ . Of course this transference plan can also be
  considered as a transhipment plan, with an associated transshiping cost.
  In order to lower this transshipment cost, you wish to improve this plan,
  and you come up with the following strategy. Whenever $x$ and $y$ are some
  initial and final points, respectively, instead of shipping $x$ to $y,$ you ship $x$ to $(x+y) / 2$ and simultaneously $(x+y) / 2$ to $y$. Show that this strategy
  can be implemented with an admissible transshipment plan, and express it
  in terms of image measures of $\pi .$ Make a schematic picture of how $\pi$ is modified by this transformation. Show that the cost has been lowered by a
  factor $2 .$ Deduce that the optimal transshipment cost is $0,$ which of course
  is not attained unless $\mu=\nu .$ Show that the conclusion is still valid for any power $|x-y|^{p}, p>1,$ or more generally as soon as $c(x, y)=\phi(|x-y|)$
  where $\phi$ is nondecreasing on $\mathbb{R}_{+}, \phi(0)=0, \phi^{\prime}(0)=0$.

(Original screenshot of text - Text Part 1
Text Part 2)
Following the notation of the book, $\Pi(\mu,\nu)$ is the set of probability measures $\pi$ on $\mathbb{R}^n\times \mathbb{R}^n $ such that
$$\pi(A\times \mathbb{R}^n )= \mu(A), \qquad \pi( \mathbb{R}^n \times A)= \nu(A)$$
and
$$I[\pi]:=\int_{\mathbb{R}^n\times \mathbb{R}^n} c(x,y)\,d\pi(x,y)$$
and
$$\mathcal{T}_{c}(\mu,\nu):=\inf_{\pi \in \Pi(\mu,\nu)}I[\pi]\,. $$
Following the hint in the exercise, my attempt was to consider the measure
$$\pi':= \frac{c(x,\frac{x+y}{2})}{c(x,y)}\cdot T_{1}\sharp \pi+ \frac{c(\frac{x+y}{2},y)}{c(x,y)}\cdot T_{2}\sharp \pi$$
where
$$T_{1}: (x,y)\mapsto (x, \frac{x+y}{2}) $$
$$T_{2}: (x,y)\mapsto (\frac{x+y}{2}, y ) $$
but I haven't been able to show that  $I[\pi']\leq\frac{1}{2}I[\pi]$.
 A: I know this is an old post, but I was just writing out the solution to myself independent of this post, so I thought I'd post it if you're still out there.
You should use $\pi' = T_1\#\pi + T_2\#\pi$.
You can see that $\pi'$ is still feasible for the transshipment problem. We can see this by noting: $$T_1^{-1}(A \times B) = \{ (x,y) : T_1(x,y) \in A \times B \} = \{(x,y) : (x,(x+y)/2) \in A \times B \}$$
Thus, $T_1^{-1}(A \times X) = A \times X$.
Similarly, $T_2^{-1}(X \times B) = X \times B$. Furthermore, $T_2^{-1}(A \times X) = T_1^{-1}(X \times A)$. 
Thus:
$$\pi'(A \times X) - \pi'(X \times A) = \pi(T_1^{-1}(A \times X)) + \pi(T_2^{-1}(A \times X)) - \pi(T_1^{-1}(X \times A)) -\pi(T_2^{-1}(X \times A)) = \pi(A \times X) - \pi(X \times A) = (\mu - \nu)(A)$$
Next, note that $|x-(x+y)/2|^2 = |x/2 - y/2|^2 = |(x+y)/2 - y|^2$, so:
$$\int c(x,y) d\pi'(x,y) = \int |x-y|^2 d\pi' = $$
$$\int |x-(x+y)/2|^2 d\pi + |(x+y)/2 - y|^2 d\pi$$
Thus, $$\int |x-y|^2 d\pi = 2 \int |x/2-y/2|^2 d\pi = (1/2) \int |x-y|^2 d\pi$$
