# Minkowski's Inequality for Sums/Index Greater than 1

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## 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:

- $\displaystyle \left({\sum_{k \mathop = 1}^n \left({a_k + b_k}\right)^p}\right)^{1/p} \le \left({\sum_{k \mathop = 1}^n a_k^p}\right)^{1/p} + \left({\sum_{k \mathop = 1}^n b_k^p}\right)^{1/p}$

## Proof

Without loss of generality, assume that:

- $\displaystyle \sum_{k \mathop = 1}^n \left({a_k + b_k}\right)^p \ne 0$

Define:

- $q = \dfrac p {p-1}$

Then:

- $\dfrac 1 p + \dfrac 1 q = \dfrac 1 p + \dfrac {p-1} p = 1$

It follows that:

\(\displaystyle \sum_{k \mathop = 1}^n \left({a_k + b_k}\right)^p\) | \(=\) | \(\displaystyle \sum_{k \mathop = 1}^n a_k \left({a_k + b_k}\right)^{p-1} + \sum_{k \mathop = 1}^n b_k \left({a_k + b_k}\right)^{p-1}\) | |||||||||||

\(\displaystyle \) | \(\le\) | \(\displaystyle \left({\sum_{k \mathop = 1}^n a_k^p}\right)^{1/p} \left({\sum_{k \mathop = 1}^n \left({a_k + b_k}\right)^p}\right)^{1/q}\) | by Hölder's Inequality for Sums, and because $\left({p - 1}\right) q = p$, by hypothesis | ||||||||||

\(\displaystyle \) | \(\) | \(\, \displaystyle + \, \) | \(\displaystyle \left({\sum_{k \mathop = 1}^n b_k^p}\right)^{1/p} \left({\sum_{k \mathop = 1}^n \left({a_k + b_k}\right)^p}\right)^{1/q}\) | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left({\left({\sum_{k \mathop = 1}^n a_k^p}\right)^{1/p} + \left({\sum_{k \mathop = 1}^n b_k^p}\right)^{1/p} }\right) \left({\sum_{k \mathop = 1}^n \left({a_k + b_k}\right)^p}\right)^{1/q}\) |

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 $\displaystyle 1 - \frac 1 q = \frac 1 p$.

$\blacksquare$

## Source of Name

This entry was named for Hermann Minkowski.