Minkowski's Inequality for Sums

From ProofWiki
Jump to navigation Jump to search

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:

\(\ds \sum_{k \mathop = 1}^n \paren {a_k + b_k}^2\) \(=\) \(\ds \sum_{k \mathop = 1}^n \paren {a_k^2 + 2 a_k b_k + b_k^2}\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^n a_k^2 + 2 \sum_{k \mathop = 1}^n a_k b_k + \sum_{k \mathop = 1}^n b_k^2\)
\(\ds \) \(\le\) \(\ds \sum_{k \mathop = 1}^n a_k^2 + 2 \paren {\sum_{k \mathop = 1}^n a_k^2}^{1/2} \paren {\sum_{k \mathop = 1}^n b_k^2}^{1/2} + \sum_{k \mathop = 1}^n b_k^2\) 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 $\paren {p - 1} q = p$
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {\sum_{k \mathop = 1}^n b_k^p}^{1/p} \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1/q}\)
\(\ds \) \(=\) \(\ds \paren {\paren {\sum_{k \mathop = 1}^n a_k^p}^{1/p} + \paren {\sum_{k \mathop = 1}^n b_k^p}^{1/p} } \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1/q}\)



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.