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:

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

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

$\ds \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1/p} \ge \paren {\sum_{k \mathop = 1}^n a_k^p}^{1/p} + \paren {\sum_{k \mathop = 1}^n b_k^p}^{1/p}$

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:

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

Proof

Proof for $p = 2$

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

\(\ds \sum_{k \mathop = 1}^n \paren {a_k + b_k}^2\)

\(=\)

\(\ds \sum_{k \mathop = 1}^n \paren {a_k^2 + 2 a_k b_k + b_k^2}\)

\(\ds \)

\(=\)

\(\ds \sum_{k \mathop = 1}^n a_k^2 + 2 \sum_{k \mathop = 1}^n a_k b_k + \sum_{k \mathop = 1}^n b_k^2\)

\(\ds \)

\(\le\)

\(\ds \sum_{k \mathop = 1}^n a_k^2 + 2 \paren {\sum_{k \mathop = 1}^n a_k^2}^{1/2} \paren {\sum_{k \mathop = 1}^n b_k^2}^{1/2} + \sum_{k \mathop = 1}^n b_k^2\)

Cauchy's Inequality

\(\ds \)

\(=\)

\(\ds \paren {\paren {\sum_{k \mathop = 1}^n a_k^2}^{1/2} + \paren {\sum_{k \mathop = 1}^n b_k^2}^{1/2} }^2\)

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

Proof for $p > 1$

Without loss of generality, assume:

$\ds \sum_{k \mathop = 1}^n \paren {a_k + b_k}^p \ne 0$

Define:

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

Then:

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

It follows that:

\(\ds \sum_{k \mathop = 1}^n \paren {a_k + b_k}^p\)

\(=\)

\(\ds \sum_{k \mathop = 1}^n a_k \paren {a_k + b_k}^{p - 1} + \sum_{k \mathop = 1}^n b_k \paren {a_k + b_k}^{p - 1}\)

\(\ds \)

\(\le\)

\(\ds \paren {\sum_{k \mathop = 1}^n a_k^p}^{1/p} \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1/q}\)

Hölder's Inequality for Sums; by hypothesis $\paren {p - 1} q = p$

\(\ds \)

\(\)

\(\, \ds + \, \)

\(\ds \paren {\sum_{k \mathop = 1}^n b_k^p}^{1/p} \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1/q}\)

\(\ds \)

\(=\)

\(\ds \paren {\paren {\sum_{k \mathop = 1}^n a_k^p}^{1/p} + \paren {\sum_{k \mathop = 1}^n b_k^p}^{1/p} } \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1/q}\)

This article, or a section of it, needs explaining. In particular: Too many steps are being taken at the crucial point where it needs to be explained carefully, at "by Hölder's Inequality for Sums, and because $\paren {p - 1} q = p$, by hypothesis. Remedy needed: (a) separate out the manipulation of the indices according to the relation between $p$ and $q$ as one step, (b) use an instantiation of Hölder's Inequality for Sums which explains it in terms of summations (at the moment it is expressed in the compact and gnomic notation of $\norm x$, which requires one to go to a separate page again for the explanation of this notation) then (c) use as many steps of Hölder's Inequality for Sums (I'm not sure, but I believe there are two separate applications of it) as is necessary to provide the solution. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

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 $\ds 1 - \frac 1 q = \frac 1 p$. $\Box$

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.