Minkowski's Inequality for Sums

Theorem
Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n \ge 0$ be real numbers.

Let $p \in \R$ be a real number.

If $p > 1$, then:
 * $\displaystyle \left({\sum_{k \mathop = 1}^n \left({a_k + b_k}\right)^p}\right)^{1/p} \le \left({\sum_{k \mathop = 1}^n a_k^p}\right)^{1/p} + \left({\sum_{k \mathop = 1}^n b_k^p}\right)^{1/p}$

If $p < 1$ and $p \ne 0$, then the reverse inequality holds.

Proof for $p = 2$
The result follows from Order Preserved on Positive Reals by Squaring.

Proof for $p > 1$
Without loss of generality, assume that:
 * $\displaystyle \sum_{k \mathop = 1}^n \left({a_k + b_k}\right)^p \ne 0$

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

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

It follows that:

The result follows by dividing both sides of the above inequality by $\displaystyle \left({\sum_{k \mathop = 1}^n \left({a_k + b_k}\right)^p}\right)^{1/q}$, and using the equation $\displaystyle 1 - \frac 1 q = \frac 1 p$.

Proof for $p < 1$, $p \ne 0$
In this case, $p$ and $q$ have opposite sign.

Therefore, the proof is identical to the proof for $p > 1$, except that the reverse Hölder's inequality is applied.