Theorem: P-Product Metric is Metric

Let $M_{1'} = \struct {A_{1'}, d_{1'} }, M_{2'} = \struct {A_{2'}, d_{2'} }, \ldots, M_{n'} = \struct {A_{n'}, d_{n'} }$ be metric spaces.
Let $\ds \AA = \prod_{i \mathop = 1}^n A_{i'}$ be the cartesian product of $A_{1'}, A_{2'}, \ldots, A_{n'}$.
Let $p \in \R_{\ge 1}$.
Let $d_p: \AA \times \AA \to \R$ be the $p$-product metric on $\AA$:
:$\ds \map {d_p} {x, y} := \paren {\sum_{i \mathop = 1}^n \paren {\map {d_{i'} } {x_i, y_i} }^p}^{\frac 1 p}$
where $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in \AA$.
Then $d_p$ is a metric.

Proof:
=== Proof of {{Metric-space-axiom|1|nolink}} ===
{{begin-eqn}}
{{eqn | l = \map {d_p} {x, x}
      | r = \paren {\sum_{i \mathop = 1}^n \paren {\map {d_{i'} } {x_i, x_i} }^p}^{\frac 1 p}
      | c = Definition of $d_p$
}}
{{eqn | r = \paren {\sum_{i \mathop = 1}^n 0^p}^{\frac 1 p}
      | c = as $d_{i'}$ fulfills {{Metric-space-axiom|1}}
}}
{{eqn | r = 0
      | c = 
}}
{{end-eqn}}
So {{Metric-space-axiom|1}} holds for $d_p$.
{{qed|lemma}}
=== Proof of {{Metric-space-axiom|2|nolink}} ===
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:
:$\ds \paren {\sum \paren {\map {d_{i'} } {x_i, y_i} }^p}^{\frac 1 p} + \paren {\sum \paren {\map {d_{i'} } {y_i, z_i} }^p}^{\frac 1 p} \ge \paren {\sum \paren {\map {d_{i'} } {x_i, z_i} }^p}^{\frac 1 p}$
We have:
{{begin-eqn}}
{{eqn | l = \map {d_p} {x, y} + \map {d_p} {y, z}
      | r = \paren {\sum \paren {\map {d_{i'} } {x_i, y_i} }^p}^{\frac 1 p} + \paren {\sum \paren {\map {d_{i'} } {y_i, z_i} }^p}^{\frac 1 p}
      | c = Definition of $d_p$
}}
{{eqn | r = \paren {\sum r_i^p}^{\frac 1 p} + \paren {\sum s_i^p}^{\frac 1 p}
      | c = 
}}
{{eqn | o = \ge
      | r = \paren {\sum \paren {r_i + s_i}^p}^{\frac 1 p}
      | c = Minkowski's Inequality for Sums
}}
{{eqn | r = \paren {\sum \paren {\map {d_{i'} } {x_i, y_i} + \map {d_{i'} } {y_i, z_i} }^p}^{\frac 1 p}
      | c = Definition of $r_i$ and $s_i$
}}
{{eqn | o = \ge
      | r = \paren {\sum \paren {\map {d_{i'} } {x_i, z_i} }^p}^{\frac 1 p}
      | c = as $d_{i'}$ fulfills {{Metric-space-axiom|2}}
}}
{{eqn | r = \map {d_p} {x, z}
      | c = Definition of $d_p$
}}
{{end-eqn}}
So {{Metric-space-axiom|2}} holds for $d_p$.
{{qed|lemma}}
=== Proof of {{Metric-space-axiom|3|nolink}} ===
{{begin-eqn}}
{{eqn | l = \map {d_p} {x, y}
      | r = \paren {\sum \paren {\map {d_{i'} } {x_i, y_i} }^p}^{\frac 1 p}
      | c = Definition of $d_p$
}}
{{eqn | r = \paren {\sum \paren {\map {d_{i'} } {y_i, x_i} }^p}^{\frac 1 p}
      | c = as $d_i$ fulfills {{Metric-space-axiom|3}}
}}
{{eqn | r = \map {d_p} {y, x}
      | c = Definition of $d_p$
}}
{{end-eqn}}
So {{Metric-space-axiom|3}} holds for $d_p$.
{{qed|lemma}}
=== Proof of {{Metric-space-axiom|4|nolink}} ===
{{begin-eqn}}
{{eqn | l = x
      | o = \ne
      | r = y
      | c = 
}}
{{eqn | ll= \leadsto
      | q = \exists k \in \closedint 1 n
      | l = x_k
      | o = \ne
      | r = y_k
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map {d_k} {x_k, y_k}
      | o = >
      | r = 0
      | c = as $d_k$ fulfills {{Metric-space-axiom|4}}
}}
{{eqn | ll= \leadsto
      | l = \paren {\sum \paren {\map {d_{i'} } {y_i, x_i} }^p}^{\frac 1 p}
      | o = >
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map {d_p} {x, y}
      | o = >
      | r = 0
      | c = Definition of $d_p$
}}
{{end-eqn}}
So {{Metric-space-axiom|4}} holds for $d_p$.
{{qed}}