Minkowski's Inequality for Sums/Index Greater than 1

Theorem
Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n \in \R_{\ge 0}$ be non-negative real numbers.

Let $p \in \R$ be a real number such that $p > 1$.

Then:
 * $\ds \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1 / p} \le \paren {\sum_{k \mathop = 1}^n {a_k}^p}^{1 / p} + \paren {\sum_{k \mathop = 1}^n {b_k}^p}^{1 / p}$

Proof
, assume:
 * $\ds \sum_{k \mathop = 1}^n \paren {a_k + b_k}^p \ne 0$

Define:
 * $q = \dfrac p {p - 1}$

Then:
 * $\dfrac 1 p + \dfrac 1 q = \dfrac 1 p + \dfrac {p - 1} p = 1$

It follows that:

The result follows by dividing both sides of the above inequality by $\ds \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1 / q}$, and using the equation $1 - \dfrac 1 q = \dfrac 1 p$.