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 $\R^n$:


 * $\ds \map {d_p} {x, y} := \paren {\sum_{i \mathop = 1}^n \size {x_i - y_i}^p}^{\frac 1 p}$

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

Then $d_p$ is a metric.

Proof of
So holds for $d_p$.

Proof of
It is required to be shown:
 * $\map {d_p} {x, y} + \map {d_p} {y, z} \ge \map {d_p} {x, z}$

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

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

Thus we need to show that:
 * $\ds \paren {\sum \size {x_i - y_i}^p}^{\frac 1 p} + \paren {\sum \size {y_i - z_i}^p}^{\frac 1 p} \ge \paren {\sum \size {x_i - z_i}^p}^{\frac 1 p}$

We have:

So holds for $d_p$.

Proof of
So holds for $d_p$.

Proof of
So holds for $d_p$.