I am stuck on the following problem. I have a discrete distribution $\mu_0$ (it is actually an empirical distribution). I have some $\mu_i$ (again discrete, some empirical distribution). I have some bound on the Wasserstein distance $W_2(\mu_0, \mu_i).$ I now want to consider a simple mixture of $\mu_i,$ that is, $\nu=\sum\limits_{i=1}^{m}\lambda_i\mu_i$ where $\sum\lambda_i=1, \lambda_i>0.$

My goal is to bound $W_2^2(\mu_0, \nu).$ I felt that it would be easy to get a bound on $W_2^2(\mu_0, \nu)$ in terms of $W_2^2(\mu_0, \mu_i),$ but I am unable to prove anything. I want something like $$W_2^2(\mu_0, \nu)\le \sum \lambda_i^2 W_2^2(\mu_0, \mu_i).$$

This does not look terribly hard, but I am stuck. Can anyone please say if it is true or not? If anyone can give a simple demonstration of why this is true, it would be great.