The Prokhorov (or Levy-Prokhorov metric) $d_P$ on the space of probability measures on $\Bbb R$ with respect to the standard metric $d(x,y) = \vert x-y\vert $ is defined as $$d_P (\mu , \nu ) := \inf \{ \varepsilon > 0 : \mu (B) \leq \nu (B^\varepsilon) + \varepsilon \text{ for all Borel sets } B \}$$ where $B^\varepsilon := \{ x\in \Bbb R : \inf_{y\in B} \vert x - y \vert \leq \varepsilon \}$.
Question: Does it hold that $d_P ( \mathcal N (0,1) * \mu , \mathcal N (0,1) * \nu ) \leq d_P (\mu , \nu ) $?
(Here $\mathcal N (0,1)$ shall denote the normal distribution with mean $0$ and variance $1$ and $*$ is the usual convolution of measures.)
Motivation: For "stronger" metrics on (sub)sets of probability measures, or more precisely, for the 1-Wasserstein (or Kantorovich) distance with respect to $d(x,y) = \vert x-y\vert $, namely $$d_W (\mu , \nu) := \inf \{ \Bbb E [\vert X -Y \vert] : X \sim \mu , Y\sim \nu\}$$ and the total variation distance $$d_{TV} (\mu , \nu) := \inf \{ \Bbb P (X\neq Y) : X\sim \mu , Y\sim \nu\}$$ this inequality suggested in the question above is true. This can be easily seen by the chosen coupling representations for the definition of the distances and the fact that $X + Z \sim \mathcal N (0,1)* \mu$ and $Y+Z\sim \mathcal N (0,1)* \nu$, whenever $X\sim \mu$ and $Y\sim\nu$ and $Z\sim \mathcal N (0,1)$ independently from $X$ and $Y$. This could be, of course, extended to an more general convolution statement, since the argument does not depend on $\mathcal N (0,1)$ being the normal distribution.