Wasserstein Metric, a triangle type inequality.

Denote

$$K:=\Big\{ \rho :\mathbb{R}^d\to [0,\infty)~ \text{measurable}, : \int_{\mathbb{R}^d} \rho(x) dx=1, ~M(\rho)<\infty \Big\},$$

with

$$M(\rho):=\int_{\mathbb{R}^d}|x|^2\rho(x)dx.$$

Think of $$\rho$$ as densities of probability distributions. We define the Wasserstein distance $$d$$ between two Borel probability measures on $$\mathbb{R}^d$$ as usual :

$$d(\mu_1,\mu_2)^2=\inf_{\mu} \int_{\mathbb{R}^d\times \mathbb{R}^d} |x-y|^2 \mu(dx,dy),$$ with the infimum taken over all couplings with marginals $$\mu_1,\mu_2$$. Denote $$d(\rho_1,\rho_2)=d(\mu_1,\mu_2)$$ for distributions $$\mu_1,\mu_2$$ with densities $$\rho_1,\rho_2$$.

$$\textbf{Question :}$$

I am reading some text that says clearly : $$M(\rho_1) \leq 2M(\rho_0)+2d(\rho_1,\rho_0)^2$$

I don't see how this works? it seems like $$|x|^2\leq 2|x|^2+2|x-y|^2$$ could be used? Any help? My main difficulty is starting on the LHS how to introduce $$\rho_0$$, adding and subtracting in some way?

Let $$\mu$$ with marginals $$\mu_1,\mu_0$$, then $$2M(\rho_0)+2\int_{\mathbb{R}^d\times\mathbb{R}^d}|x-y|^2\mu(dx,dy)=2\int_{\mathbb{R}^d\times\mathbb{R}^d}\left(|y|^2+|x-y|^2\right)\mu(dx,dy)$$ because $$\int_{\mathbb{R}^d\times\mathbb{R}^d}|y|^2\mu(dx,dy)=M(\rho_0)\int_{\mathbb{R}^d}\rho_1(x)dx=M(\rho_0)$$ Moreover $$2|y|^2+2|x-y|^2=|x|^2+|x-2y|^2\geqslant |x|^2$$ so that \begin{aligned} 2M(\rho_0)+2\int_{\mathbb{R}^d\times\mathbb{R}^d}|x-y|^2\mu(dx,dy)&\geqslant\int_{\mathbb{R}^d\times\mathbb{R}^d}|x|^2\mu(dx,dy) \\ &=M(\rho_1)\int_{\mathbb{R}^d}\rho_0(y)dy\\ &=M(\rho_1) \end{aligned} Taking the infimum on $$\mu$$ leads to $$2M(\rho_0)+2d(\rho_1,\rho_0)\geqslant M(\rho_1)$$.