Relation between P-Product Metric and Chebyshev Distance on Real Vector Space

Theorem
For $n \in \N$, let $\R^n$ be a Euclidean space.

Let $p \in \R_{\ge 1}$.

Let $d_p$ be the $p$-product metric on $\R^n$.

Let $d_\infty$ be the Chebyshev distance on $\R^n$.

Then
 * $\forall x, y \in \R^n: d_\infty \left({x, y}\right) \le d_p \left({x, y}\right) \le n^{1/p} d_\infty \left({x, y}\right)$

It follows that $d_p$ and $d_\infty$ are Lipschitz equivalent.

Proof
By definition of the Chebyshev distance on $\R^n$, we have:
 * $\displaystyle d_\infty \left({x, y}\right) = \max_{i \mathop = 1}^n {\left\vert{x_i - y_i}\right\vert}$

where $x = \left({x_1, x_2, \ldots, x_n}\right)$ and $y = \left({y_1, y_2, \ldots, y_n}\right)$.

Let $j$ be chosen so that:
 * $\displaystyle \left\vert{x_j - y_j}\right\vert = \max_{i \mathop = 1}^n {\left\vert{x_i - y_i}\right\vert}$

Then: