I have a set $X = X_1 \times X_2 \times \dots \times X_n$.
I have metrics that satisfy triangular inequality "per component" $$ \rho_i: X_i \times X_i \to \mathbb{R} $$
for each $i=1,2,\dots,n$.
I want to know what are possible ways to combine $\rho_i$ for $i=1,2,\dots,n$ into $$\rho:X\times X \to \mathbb{R}$$
Specifically:
- Would weighted $\max$ work?
- Would weighted sum work?
- Would rooted weighed sum of squares of components work?
All 3 would work.
First, you reduce the weighted problems to the unweighted ones by pre-scaling the initial metrics. That means to prove that e.g. $$\rho ((x_1,x_2), (y_1,y_2)) = \max (\alpha_1\rho_1(x_1,y1), \alpha_2\rho_2(x_2,y_2))$$ is a metric for any $\rho_1,\rho_2$ it is enough to prove that
$$\rho^* ((x_1,x_2), (y_1,y_2)) = \max (\rho_1^*(x_1,y1), \rho_2^*(x_2,y_2))$$
is a metric for any $\rho_1^*,\rho_2^*$ and then set
$$\rho_1^*(x_1,y_1)=\alpha_1\rho(x_1,y_1),\quad \rho_2^*(x_2,y_2)=\alpha_2\rho(x_2,y_2),$$
which are metrics on $X_1$ and $X_2$ resp., if $\alpha_i > 0, i=1,2$.
So the 3 'combination functions' you considered are
$$f_m(d_1,\ldots,d_n) = \max(d_1,\ldots,d_n)$$ $$f_s(d_1,\ldots,d_n) = d_1+\ldots+d_n$$ $$f_e(d_1,\ldots,d_n) = \sqrt{d_1^2+\ldots+d_n^2}$$
Those are the well known maximum, sum and euclidean norms on $\mathbb R^n$. I'll show that any such norm that is also increasing in each component under the assumption that all components are non-negative will work.
Let $f_r$ be such a norm. Then you define
$$\rho((x_1,\ldots,x_n), (y_1,\ldots,y_n)) = f_r(\rho_1(x_1,y_1),\ldots,\rho_n(x_,y_n))$$
This $\rho$ is $0$ iff $(x_1,\ldots,x_n)=(y_1,\ldots,y_n)$ because that is true for each component's $\rho_i$ and also $f_r$ is zero only for the null vector.
This $\rho$ is symmtric because your component $\rho_i$'s are symmetric (nothing needed from $f_r$).
Of course, the triangle inequality is the most problematic thing to prove usually. Define
$$r_i=\rho_i(x_i,y_i),\quad s_i=\rho_i(y_i,z_i),\quad t_i=\rho_i(x_i,z_i), \forall i=1,\ldots,n$$ We then have
$$\rho((x_1,\ldots,x_n), (z_1,\ldots,z_n))=f_r(t_1,\ldots,t_n)$$ Because I assumed that $f_r$ was increasing in each component for non-negative values, we can replace each $t_i$ with $r_i+s_i$ and we have $r_i+s_i \ge t_i$ because of the triangle inequality of $\rho_i$ between $x_i,y_i$ and $z_i$:
$$\rho((x_1,\ldots,x_n), (z_1,\ldots,z_n)) \le f_r(r_1+s_1,\ldots,r_n+s_n)$$ Because $f_r$ is a norm, the norm-version of the triangle inequality applies here:
$$f_r(r_1+s_1,\ldots,r_n+s_n) \le f_r(r_1,\ldots,r_n) + f_r(s_1,\ldots,s_n) = \rho((x_1,\ldots,x_n), (y_1,\ldots,y_n)) + \rho((y_1,\ldots,y_n), (z_1,\ldots,z_n)),$$
which finally proves that $\rho$ is indeed a metric.
That means that any such norms can be used to combine the component's metrics into a new one for the product set.