The $\|\cdot\|_{\infty}$-norm on $\mathbb{R}^n$ for $n\in \mathbb{Z}^+$ is not a smooth function. However, I came across this post which essentially says that a pointwise approximation to the maximum function, and therefore $\|\cdot\|_{\infty}$ is $$ m_{\lambda}(x) = \sum_{i=1}^n\,\frac{e^{\lambda\, x_i}x_i}{\sum_{j=1}^n\,e^{\lambda\, x_j}} $$ where $x=(x_i)_{i=1}^n\in \mathbb{R}^n$ is arbitrary.

Fix $\lambda>0$ and consider the map $d_{\lambda}:\mathbb{R}^n\times \mathbb{R}^n\rightarrow [0,\infty)$ defined by $$ d_{\lambda}(x,y) = m_{\lambda}(|x-y|) $$ where $|z|:=(|z_i|)_{i=1}^n$ for any $z\in \mathbb{R}^n$.

Does $d_{\lambda}$ define a quasi-metric on $\mathbb{R}^n$?

It certainly satisfies positivity, symmetry and $d_{\lambda}(x,y)=0$ if and only if $x=y$ by its not obvious that it should satisfy a relaxed triangle inequality: ie $$ d_{\lambda}(x,y) \le C\big(d_{\lambda}(x,z)+d_{\lambda}(z,y)\big) $$ for every $x,y,z\in \mathbb{R}^n$ and some $C\ge 1$ independent of $x,y,z$.