Let $W_p(\cdot,\cdot)$ denote the Wasserstein distance and let $\mu$ and $\nu$ be probability measures on $\mathbb{R}^d$ and suppose we can decompose $\mu=\mu_1+\ldots+\mu_n$ and $\nu=\nu_1+\ldots+\nu_n$ where for all indices $i$, $\mu_i$ and $\nu_i$ are finite measures with the same support and also the support of $\mu_i$ and $\mu_j$ don't intersect if $i\neq j$ and the support of $\nu_i$ and $\nu_j$ don't intersect if $i\neq j$ (I am not sure if these condition are important). Then I would like to understand if we have the inequality
$W_p^p\left(\sum\limits_{k=1}^n \mu_k,\sum\limits_{k=1}^n\nu_k\right)\le \sum\limits_{k=1}^n W_p^p\left(\mu_k,\nu_k\right) \qquad(*)$
I could show that in general, $W_p^p(\frac{1}{2}\eta+\frac{1}{2}\xi,\chi)\le \frac{1}{2}W_p^p(\eta,\chi)+\frac{1}{2}W_p^p(\xi,\chi)$ because if $\gamma$ is a plan between $\eta$ and $\chi$ and $\hat{\gamma}$ is a plan between $\xi$ and $\chi$, then $\frac{1}{2}\gamma+\frac{1}{2}\hat{\gamma}$ is a plan between $\frac{1}{2}\eta+\frac{1}{2}\xi$ and $\chi$. Similarly I could show $W_p^p(\chi,\frac{1}{2}\eta+\frac{1}{2}\xi)\le \frac{1}{2}W_p^p(\chi,\eta)+\frac{1}{2}W_p^p(\chi,\xi)$. But then I don't get $(*)$ because I get cross terms as well. Maybe the inequality $(*)$ is wrong, but if this is the case, is it still wrong if $\mu=\frac{1}{n}\delta_1+\ldots+\frac{1}{n}\delta_n$?
Let $\varepsilon>0$ and $\pi_k$ be a coupling between $\mu_k$ and $\nu_k$ such that $\int_{\mathbb R^d\times\mathbb R^d}\vert x-y\vert^p\,\pi_k(dx,dy)\le W_p^p(\mu_k,\nu_k)+\varepsilon$. Then $\pi=\sum_{k=1}^n\pi_k$ is a coupling between $\sum_{k=1}^n\mu_k$ and $\sum_{k=1}^n\nu_k$. By definition of the Wasserstein distance we then have $$ \begin{align*} W_p^p\left(\sum_{k=1}^n\mu_k,\sum_{k=1}^n\nu_k\right)&\le\int_{\mathbb R^d\times\mathbb R^d}\vert x-y\vert^p\,\pi(dx,dy)\\ &=\sum_{k=1}^n\int_{\mathbb R^d\times\mathbb R^d}\vert x-y\vert^p\,\pi_k(dx,dy)\\ &\le\sum_{k=1}^nW_p^p(\mu_k,\nu_k)+n\varepsilon. \end{align*}$$
You then obtain the desired inequality by letting $\varepsilon$ go to $0$.