Suppose $d = f(d_1, d_2)$, i.e. $d(x,y) = f(d_1(x,y), d_2(x,y))$. The only requirements for a metric are:
- $d \ge 0$,
- $d(x,y)=0$ iff $x=y$,
- $d(x,y) = d(y,x)$, and
- $d(x,z) \le d(x,y) + d(y,z)$.
Suppose these are metrics on a set $M$.
Let $S$ be the set of pairs $\{(d_1(x,y), d_2(x,y)): x,y \in M\}$.
This is a subset of $[0,\infty)^2$.
- For (1) you need $f(a,b) \ge 0$ for all $(a,b) \in S$.
- For (2) (noting that $(0,0) \in S$ but $(0,x) \notin S$ for $x \ne 0$), you need $f(0,0) = 0$ and $f(a,b) > 0$ for all other $(a,b) \in S$.
- (3) is automatically true.
- (4) is more complicated, because in principle it
requires knowing the set of sextuples $T = \{ (d_1(x,y), d_1(y,z), d_1(x,z), d_2(x,y), d_2(y,z), d_2(x,z)): x,y,z \in M\}$. You need
$$ f(t_3, t_6) \le f(t_1, t_4) + f(t_2, t_5)\ \text{for all}\ (t_1,t_2,t_3,t_4,t_5,t_6) \in T$$
A sufficient condition would be
$$ f(t_3, t_6) \le f(t_1, t_4) + f(t_2, t_5)\ \text{whenever}\ t_3 \le t_1 + t_2 \ \text{and}\ t_6 \le t_4 + t_5 $$