Problem:
Suppose $d$ and $e$ are metrics on a set $X$. Let $g$ be the function $(x,y) \to min\{d(x,y),e(x,y)\}$ defined on $X × X$. Find a condition under which $g$ is a metric.
My solution: If $d(x,y) = ke(x,y)$, where $k$ is a positive constant, then $g$ will be a metric. Seems to be trivial. Is there any other possibility?
Related question here: Is the minimum of two metrics is again a metric?
On
Consider $\|\ \|_1$ on $\mathbb{R}^2$. Its unit ball is a convex hull of $$ (\pm 1,0),\ (0,\pm 1) $$ We will find two metrics $d_i$ whose infimum is $\| \ \|_1$.
Step 1 : This convex set is a union of two convex sets $A,\ B$ : $A$ is a convex hull of $$ (\pm 1,0),\ (\pm 1/2,1/2),\ (\pm 1/2,-1/2) $$
and $B$ is a convex hull of $$ (0,\pm 1) ,\ (\pm 1/2,1/2),\ (\pm 1/2,-1/2) $$
Here we have norms $d_i$ whose unit balls are $A,\ B$.
Step 2 : Then $\|\ \|_1$ is infimum of $d_i$.
You found an (obviously very strong) sufficient condition. Can you make it weaker? Can you find a necessary and sufficient one?
HINT
To be a metric, $g$ has to map $X \times X \to [0,\infty)$ (which it obviously is) and satisfy some properties. Most of the properties, except one, will carry over directly from the definition.
On the last one, you have to think which are the least restrictive condition(s) to impose on $d$ and $e$ for $g=\min\{d,e\}$ to remain a metric still.