P-Product Metric on Real Vector Space is Metric/Proof 1

Theorem
Let $\R^n$ be an $n$-dimensional real vector space.

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

Let $d_p: \R^n \times \R^n \to \R$ be the $p$-product metric on $\R^n$:


 * $\displaystyle d_p \left({x, y}\right) := \left({\sum_{i \mathop = 1}^n \left \vert {x_i - y_i} \right \vert^p}\right)^{\frac 1 p}$

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

Then $d_p$ is a metric.

Proof of $M1$
So axiom $M1$ holds for $d_p$.

Proof of $M2$
It is required to be shown:
 * $d \left({x, y}\right) + d \left({y, z}\right) \ge d \left({x, z}\right)$

for all $x, y, z \in \R^n$.

Let:
 * $(1): \quad z = \left({z_1, z_2, \ldots, z_n}\right)$
 * $(2): \quad$ all summations be over $i = 1, 2, \ldots, n$
 * $(3): \quad \left\vert{x_i - y_i}\right\vert = r_i$
 * $(4): \quad \left\vert{y_i - z_i}\right\vert = s_i$.

Thus we need to show that:
 * $\displaystyle \left({\sum \left\vert{x_i - y_i}\right\vert^p}\right)^{\frac 1 p} + \left({\sum \left\vert{y_i - z_i}\right\vert^p}\right)^{\frac 1 p} \ge \left({\sum \left\vert{x_i - z_i}\right\vert^p}\right)^{\frac 1 p}$

We have:

So axiom $M2$ holds for $d_p$.

Proof of $M3$
So axiom $M3$ holds for $d_p$.

Proof of $M4$
So axiom $M4$ holds for $d_p$.