$\newcommand\R{\mathbb R}$Let $\|\cdot\|$ be any norm on $\R^n$. Take any real $t$. Let $Z$ be a random vector in $\R^n$ such that (i) $Z$ is independent of $X$ and (ii) $Z\sim N(0,\Sigma_Y-\Sigma_X)$. Then $X+Z$ equals $Y$ in distribution.

So, it suffices to show that \begin{equation*} P(\|X\|\le t)\ge P(\|X+Z\|\le t). \tag{1} \end{equation*}

Note that \begin{equation*} P(\|X+Z\|\le t)=Eg(Z), \tag{2} \end{equation*} where \begin{equation*} g(z):=P(\|X+z\|\le t)=\int_{\R^n}dx f(x)1(\|x+z\|\le t) \end{equation*} and $f$ is the pdf of $X$. So, by the [Prékopa–Leindler][1] theorem, $g$ is a log-concave function. Also, the function $g$ is even. So, $g(z)\le g(0)=P(\|X\|\le t)$ for all $z\in\R^n$, and hence (1) follows from (2).

$\newcommand\R{\mathbb R}$Let $\|\cdot\|$ be any norm on $\R^n$. Take any real $t$. Let $Z$ be a random vector in $\R^n$ such that (i) $Z$ is independent of $X$ and (ii) $Z\sim N(0,\Sigma_Y-\Sigma_X)$. Then $X+Z$ equals $Y$ in distribution.

So, it suffices to show that \begin{equation*} P(\|X\|\le t)\ge P(\|X+Z\|\le t). \tag{1} \end{equation*}

Note that \begin{equation*} P(\|X+Z\|\le t)=Eg(Z), \tag{2} \end{equation*} where \begin{equation*} g(z):=P(\|X+z\|\le t)=\int_{\R^n}dx\, f(x)1(\|x+z\|\le t) \end{equation*} and $f$ is the pdf of $X$. The functions $f$ and $x\mapsto1(\|x+z\|\le t)$ are log concave, and hence the function $x\mapsto f(x)1(\|x+z\|\le t)$ is log concave. So, by the Prékopa–Leindler theorem, $g$ is a log-concave function. Also, the function $g$ is even. So, $g(z)\le g(0)=P(\|X\|\le t)$ for all $z\in\R^n$, and hence (1) follows from (2).

$\newcommand\R{\mathbb R}$Let $\|\cdot\|$ be any norm om $\R^n$. Take any real $t$. Let $Z$ be a random vector in $\R^n$ such that (i) $Z$ is independent of $X$ and (ii) $Z\sim N(0,\Sigma_Y-\Sigma_X)$. Then $X+Z$ equals $Y$ in distribution.

So, it suffices to show that \begin{equation} P(\|X\|\le t)\le P(\|X+Z\|\le t). \end{equation}

But this follows because (i) $P(\|X+Z\|\le t)=Eg(Z)$ with $g(z):=P(\|X+z\|\le t)$ and (ii) $g$ is an even log-concave function and hence $g(z)\le g(0)=P(\|X\|\le t)$ for all $z\in\R^n$.

$\newcommand\R{\mathbb R}$Let $\|\cdot\|$ be any norm on $\R^n$. Take any real $t$. Let $Z$ be a random vector in $\R^n$ such that (i) $Z$ is independent of $X$ and (ii) $Z\sim N(0,\Sigma_Y-\Sigma_X)$. Then $X+Z$ equals $Y$ in distribution.

So, it suffices to show that \begin{equation*} P(\|X\|\le t)\ge P(\|X+Z\|\le t). \tag{1} \end{equation*}

Note that \begin{equation*} P(\|X+Z\|\le t)=Eg(Z), \tag{2} \end{equation*} where \begin{equation*} g(z):=P(\|X+z\|\le t)=\int_{\R^n}dx f(x)1(\|x+z\|\le t) \end{equation*} and $f$ is the pdf of $X$. So, by the [Prékopa–Leindler][1] theorem, $g$ is a log-concave function. Also, the function $g$ is even. So, $g(z)\le g(0)=P(\|X\|\le t)$ for all $z\in\R^n$, and hence (1) follows from (2).