The author defined the Bessel kernel that \begin{align*} G_{\alpha}(x)=\int_{0}^{\infty}t^{(\alpha-N)/2}e^{-\pi(|x|^{2}/t)-(t/4\pi)}\dfrac{dt}{t},~~~~x\in\mathbb{R}^{N} \end{align*} for $\alpha>0$. He noted that \begin{align*} G_{\alpha}(x)\leq AG_{\alpha}(x+y),~~~~|x|\geq 2,~~~~|y|\leq 1. \end{align*} for some constant $A>0$. I fail to see the reasoning right here.
First off, $|x+y|\leq|x|+1\leq |x|+(1/2)|x|=(3/2)|x|$, and so \begin{align*} G_{\alpha}(x+y)\geq\int_{0}^{\infty}t^{(\alpha-N)/2}e^{-\pi(|x|^{2}/t)-(t/4\pi)}e^{-(5\pi/4)(|x|^{2}/t)}\dfrac{dt}{t}, \end{align*} this wouldn't help in getting the inequality greater than $G_{\alpha}(x)$ since $e^{-(5\pi/4)(|x|^{2}/t)}\rightarrow 0$ as $t\rightarrow 0^{+}$.
But the author claimed that it is an easy exercise. So I wonder that there must be some trick to go on but I have totally missed it.
The bound $(3/2)|x|$ is simply too crude for the estimate.
First off, we also denote that \begin{align*} G_{\alpha}(r)=\int_{0}^{\infty}t^{(\alpha-N)/2}e^{-\pi(r^{2}/t)-(t/4\pi)}\dfrac{dt}{t},~~~~r\geq 0,~~~~\alpha>0, \end{align*} so that $G_{\alpha}(\cdot)$ is radial, decreasing, and continuous on $(0,\infty)$.
Since $G_{\alpha}(|x|+1)\leq G_{\alpha}(x+y)$, it suffices to show that $G_{\alpha}(x)\leq A G_{\alpha}(|x|+1)$, or simply that \begin{align*} \varphi(r)&=\dfrac{G_{\alpha}(r+1)}{G_{\alpha}(r)},\\ \inf_{r\geq 2}\varphi(r)&=\text{constant}>0. \end{align*}
We appeal to the asymptotic expansion that \begin{align*} G_{\alpha}(r)\approx r^{(\alpha-N-1)/2}e^{-r},~~~~r\rightarrow\infty,~~~~\alpha>0. \end{align*} Then it is easy to see that such a constant is strictly positive. Such an asymptotic expansion also shows that the bound $(3/2)|x|$ will not work for the estimate since $e^{-(3/2)r}/e^{-r}=e^{-(1/2)r}\rightarrow 0$ as $r\rightarrow\infty$.
The point is that, the deduction of such an asymptotic expansion is by no means easy, a reference is given by Aronszajin, N., Smith, K. T. Theory of Bessel potentials, I.
An easy one-sided estimate is given in many harmonic analysis books that \begin{align*} G_{\alpha}(x)=\mathcal{O}(e^{-|x|/2}),~~~~|x|\rightarrow\infty. \end{align*} But this one-sided estimate is simply not enough to showing that the strictly positivity of such a constant.
An easier argument should be ready for such an estimate.