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

Theorem
The general Euclidean metrics are metrics.

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

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

The general Euclidean metric is defined as follows:

Proof for p = 1
This is the taxicab metric.

This is proved by Taxicab Metric on Real Vector Space is Metric.

Proof for p = 2, 3, ...
It is easy to see that conditions M1, M3 and M4 of the conditions for being a metric are satisfied. So all we need to do is check M2.

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

Then:


 * $\displaystyle \left({\sum \left({x_i - y_i}\right)^p}\right)^{\frac 1 p} + \left({\sum \left({y_i - z_i}\right)^p}\right)^{\frac 1 p} = \left({\sum r_i^p}\right)^{\frac 1 p} + \left({\sum s_i^p}\right)^{\frac 1 p}$

So we have to prove:


 * $\displaystyle \left({\sum \left|{r_i}\right|^p}\right)^{\frac 1 p} + \left({\sum \left|{s_i}\right|^p}\right)^{\frac 1 p} \ge \left({\sum \left|{r_i + s_i}\right|^p}\right)^{\frac 1 p}$

This is Minkowski's Inequality.

Proof for Infinite Case
This is proved in Chebyshev Distance on Real Vector Space is Metric.