Theorem: Suppose $n \in \mathbb N$ and for each $i \in \mathbb N, (X_i, \tau_i)$ is a metric space . All conserving metrics on $\Pi_{i=1} ^{n}X_i$ are lipschitz equivalent. In particular, each is lipschitz equivalent to the euclidean product metric $\mu_2$ on $\Pi_{i=1} ^{n} X_i$
Proof. Suppose $d$ is a conserving metric on $\bigsqcup_{i=1}^n X_i$. Then $d$ and $\mu_2$ are Lipshitz stronger than $\mu_{\infty}$ and Lipshitz weaker than $\mu_1$. Moreover, we have $$\mu_{\infty}(a,b)\le\mu_2(a,b)\le\mu_1(a,b)\le n\mu_\infty(a,b),$$ for all $a,b\in\prod_{i=1}^n X_i$, so that $\mu_{\infty}$ and $\mu_1$ are Lipshitz constant.
Some Background:
Conserving metric: Suppose $(X,d) $ and $(X_i, \tau_i)$ are metric spaces for every $i$, where $X= \prod_{i=1} ^n X_i.~~~ d~$ is termed as a Conserving metric iff $\max \{\tau_i(a_i,b_i)~|~i \in (1,\cdots ,n)\} \le d(a,b) \le \sum_{i=1}^n \tau_i (a_i,b_i)$
- $\mu_1:(a,b)\mapsto\sum_{i=1}^n\tau_i(a_i,b_i);$
- $\mu_2:(a,b)\mapsto\sqrt{\sum_{i=1}^n\tau_i(a_i,b_i)^2};$
- $\mu_2:(a,b)\mapsto\max\{\tau_i(a_i,b_i):i\in\Bbb N\}$
Point of Confusion:
I understand that $d$ and $\mu_2$ are lipschitz stronger than $\mu_\infty$ as $d(a,b) \le n~\mu_\infty(a,b)$ and $\mu_2 \le n~\mu_\infty(a,b)$. But, I am unable to understand why they are lipschitz weaker than $\mu_1$ particularly when $d(a,b) \le \mu_1(a,b)$ as per definition of conserving metric.
By the definition of conserving metric you gave, we have
$$ \mu_\infty(a,b) \leq d(a,b) \leq \mu_1(a,b). $$
As $\mu_1(a,b) \leq n \mu_\infty (a,b)$, we have $\frac{1}{n}\mu_1(a,b) \leq \mu_\infty (a,b)$ and hence
$$\frac{1}{n} \mu_1(a,b) \leq \mu_\infty (a,b) \leq d(a,b). $$