Is Riemannian distance function equivalent to Euclidean one?

Consider the Riemannian manifold $\mathbb{R}^n$ and a smooth Riemannian metric $G:\mathbb{R}^n\rightarrow\mathbb{R}^{n\times{n}}$. We know that if $w_1I_n\leq{G}(x)\leq{w_2}I_n$ for some $w_1,w_2\in\mathbb{R}^+$, any $x\in\mathbb{R}^n$, and identity matrix $I_n$, then we have $\sqrt{w_1}\Vert{x_1}-x_2\Vert_2\leq\mathbf{d}_G(x_1,x_2)\leq\sqrt{w_2}\Vert{x_1}-x_2\Vert_2$, where $\mathbf{d}_G$ is the Riemannian distance function with respect to the metric $G$. Now, I am interested in the following equivalency between $\mathbf{d}_G$ and the usual Euclidean distance: $\alpha_1(\Vert{x_1}-x_2\Vert_2)\leq\mathbf{d}_G(x_1,x_2)\leq\alpha_2(\Vert{x_1}-x_2\Vert_2)$, where $\alpha_i:\mathbb{R}^+_0\rightarrow\mathbb{R}^+_0$ are continuous, strictly increasing functions, $\alpha_i(0)=0$ and $\alpha_i(r)\rightarrow\infty$ as $r\rightarrow\infty$, for $i=1,2$. I tried to find a characterization on the metric $G$ to reach to this version of equivalency, but I could not find anything. I am just wondering if there is any nice result for this kind of equivalency.