# Minkowski's Inequality for Sums

## 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 $p \in \R$, $p \ne 0$ be a real number.

If $p < 0$, then we require that $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n$ be strictly positive.

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 $p < 1$, $p \ne 0$, then:

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

### Corollary

Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n \in \R$ be real numbers.

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

If $p > 1$, then:

$\displaystyle \paren {\sum_{k \mathop = 1}^n \size {a_k + b_k}^p}^{1/p} \le \paren {\sum_{k \mathop = 1}^n \size {a_k}^p}^{1/p} + \paren {\sum_{k \mathop = 1}^n \size {b_k}^p}^{1/p}$

## Proof

### Proof for $p = 2$

$p = 2$ is an easily proved special case:

 $\displaystyle \sum_{k \mathop = 1}^n \paren {a_k + b_k}^2$ $=$ $\displaystyle \sum_{k \mathop = 1}^n \paren {a_k^2 + 2 a_k b_k + b_k^2}$ $\displaystyle$ $=$ $\displaystyle \sum_{k \mathop = 1}^n a_k^2 + 2 \sum_{k \mathop = 1}^n a_k b_k + \sum_{k \mathop = 1}^n b_k^2$ $\displaystyle$ $\le$ $\displaystyle \sum_{k \mathop = 1}^n a_k^2 + 2 \paren {\sum_{k \mathop = 1}^n a_k^2}^{1/2} \paren {\sum_{k \mathop = 1}^n b_k^2}^{1/2} + \sum_{k \mathop = 1}^n b_k^2$ Cauchy's Inequality $\displaystyle$ $=$ $\displaystyle \paren {\paren {\sum_{k \mathop = 1}^n a_k^2}^{1/2} + \paren {\sum_{k \mathop = 1}^n b_k^2}^{1/2} }^2$

The result follows from Order is Preserved on Positive Reals by Squaring.

$\Box$

### 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:

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

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

$\Box$

### Proof for $p < 1$, $p \ne 0$

In this case, $p$ and $q$ have opposite sign.

The proof then follows the same lines as the proof for $p > 1$, except that the Reverse Hölder's Inequality for Sums is applied instead.

$\blacksquare$

## Source of Name

This entry was named for Hermann Minkowski.