Let the two metrics on $X$ be defined as
$$d_2(x,y) = \sqrt{(x_1 - y_1)^2 + \dots + (x_n - y_n)^2}$$ $$d_{\infty}(x,y) = \text{max}\left\{|x_1 - y_1|, \dots, |x_n - y_n|\right\}.$$
In the general case, for some $\alpha, \beta \in \mathbb{N}$ we have that all metrics $d_p$ and metric $d_{\infty}$ are strongly/lipschitz equivalent since we can find some $\alpha$, $\beta$ such that
$$\alpha d_{\infty}(x,y) \leq d_p(x,y) \leq \beta d_{\infty}(x,y).$$
However, in my particular case,
$$\alpha d_{\infty}(x,y) \leq d_2(x,y) \leq \beta d_{\infty}(x,y).$$
Obviously, $\alpha = 1$, works. What would be a suitable $\beta$. Could we also simply say $\beta = \text{sup}~d_2(x,y)$?
$$d_2(x,y)=\sqrt{(x_1-y_1)^2+...+(x_n-y_n)^2}\leq \sqrt{(x_1-x_2)^2}+...+\sqrt{(x_n-y_n)^2}$$ $$=|x_1-x_2|+...|x_n-y_n| \leq nd_{\infty}(x,y)$$