Theorem: P-Product Metric on Real Vector Space is Metric

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:
=== Proof of {{Metric-space-axiom|1|nolink}} ===
{{begin-eqn}}
{{eqn | l = \map {d_p} {x, x}
      | r = \paren {\sum_{i \mathop = 1}^n \size {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 = 
}}
{{eqn | r = 0
      | c = 
}}
{{end-eqn}}
So {{Metric-space-axiom|1}} holds for $d_p$.
{{qed|lemma}}
=== Proof of {{Metric-space-axiom|2|nolink}} ===
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:
{{begin-eqn}}
{{eqn | l = \map {d_p} {x, y} + \map {d_p} {y, z}
      | r = \paren {\sum \size {x_i - y_i}^p}^{\frac 1 p} + \paren {\sum \size {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 | o = \ge
      | r = \paren {\sum \paren {\size {x_i - y_i} + \size {y_i - z_i} }^p}^{\frac 1 p}
      | c = Definition of $r_i$ and $s_i$
}}
{{eqn | r = \paren {\sum \size {x_i - z_i}^p}^{\frac 1 p}
      | c = Triangle Inequality for Real Numbers
}}
{{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 \size {x_i - y_i}^p}^{\frac 1 p}
      | c = Definition of $d_p$
}}
{{eqn | r = \paren {\sum \size {y_i - x_i}^p}^{\frac 1 p}
      | c = 
}}
{{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 = x_k - y_k
      | o = >
      | r = 0
      | c = as $d_k$ fulfils {{Metric-space-axiom|4}}
}}
{{eqn | ll= \leadsto
      | l = \paren {\sum \size {x_k - y_k}^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}}