Euclidean Metric on Real Vector Space is Metric/Proof 2

Theorem
The Euclidean metric is a metric.

Proof
Consider the Euclidean space $M = \left({\R^n, d}\right)$ where $d$ is the distance function given by:
 * $\displaystyle d \left({x, y}\right) = \left({\sum_{i \mathop = 1}^n \left({x_i - y_i}\right)^2}\right)^{\frac 1 2}$

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

Axioms $M1$, $M3$ and $M4$ follow immediately from the above definition.

It remains to show:
 * $d \left({x, y}\right) + d \left({y, z}\right) \ge d \left({x, z}\right)$

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 $i = 1, 2, \ldots, n$
 * $(3): \quad x_i - y_i = r_i$
 * $(4): \quad y_i - z_i = s_i$.

Thus we need to show that:
 * $\displaystyle \left({\sum \left({x_i - y_i}\right)^2}\right)^{\frac 1 2} + \left({\sum \left({y_i - z_i}\right)^2}\right)^{\frac 1 2} \ge \left({\sum \left({x_i - z_i}\right)^2}\right)^{\frac 1 2}$

We have:


 * $\displaystyle \left({\sum \left({x_i - y_i}\right)^2}\right)^{\frac 1 2} + \left({\sum \left({y_i - z_i}\right)^2}\right)^{\frac 1 2} = \left({\sum r_i^2}\right)^{\frac 1 2} + \left({\sum s_i^2}\right)^{\frac 1 2}$

So we have to prove:


 * $\displaystyle \left({\sum \left({r_i + s_i}\right)^2}\right)^{\frac 1 2} \le \left({\sum r_i^2}\right)^{\frac 1 2} + \left({\sum s_i^2}\right)^{\frac 1 2}$

This is Minkowski's Inequality for Sums:

The result follows: