Euclidean Metric is Metric/Proof 2

Proof
We have that the Euclidean metric on $\mathcal A$ is defined as:


 * $\displaystyle \map {d_2} {x, y} = \paren {\sum_{i \mathop = 1}^n \paren {\map {d_{i'} } {x_i, y_i} }^2}^{\frac 1 2}$

where $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in \mathcal A$.

Proof of $M1$
So axiom $M1$ holds for $d_2$.

Proof of $M2$
Let:
 * $(1): \quad z = \tuple {z_1, z_2, \ldots, z_n}$
 * $(2): \quad$ all summations be over $i = 1, 2, \ldots, n$
 * $(3): \quad \map {d_{i'} } {x_i, y_i} = r_i$
 * $(4): \quad \map {d_{i'} } {y_i, z_i} = s_i$.

Thus we need to show that:
 * $\displaystyle \paren {\sum \paren {\map {d_{i'} } {x_i, y_i} }^2}^{\frac 1 2} + \paren {\sum \paren {\map {d_{i'} } {y_i, z_i} }^2}^{\frac 1 2} \ge \paren {\sum \paren {\map {d_{i'} } {x_i, z_i} }^2}^{\frac 1 2}$

We have:

So axiom $M2$ holds for $d_2$.

Proof of $M3$
So axiom $M3$ holds for $d_2$.

Proof of $M4$
So axiom $M4$ holds for $d_2$.