Minkowski's Inequality for Sums/Index Greater than 1/Proof 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:
- $\ds \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1 / p} \le \paren {\sum_{k \mathop = 1}^n {a_k}^p}^{1 / p} + \paren {\sum_{k \mathop = 1}^n {b_k}^p}^{1 / p}$
Proof
Define:
- $q = \dfrac p {p - 1}$
Then:
- $\dfrac 1 p + \dfrac 1 q = \dfrac 1 p + \dfrac {p - 1} p = 1$
It follows that:
| \(\ds \sum_{k \mathop = 1}^n \paren {a_k + b_k}^p\) | \(=\) | \(\ds \sum_{k \mathop = 1}^n a_k \paren {a_k + b_k}^{p - 1} + \sum_{k \mathop = 1}^n b_k \paren {a_k + b_k}^{p - 1}\) | ||||||||||||
| \(\ds \) | \(\le\) | \(\ds \paren {\sum_{k \mathop = 1}^n {a_k}^p}^{1 / p} \paren {\sum_{k \mathop = 1}^n \paren {\paren {a_k + b_k}^{p - 1} }^q}^{1 / q} + \paren {\sum_{k \mathop = 1}^n {b_k}^p}^{1 / p} \paren {\sum_{k \mathop = 1}^n \paren {\paren {a_k + b_k}^{p - 1} }^q}^{1 / q}\) | Hölder's Inequality for Sums (twice) | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {\sum_{k \mathop = 1}^n {a_k}^p}^{1 / p} \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1 / q} + \paren {\sum_{k \mathop = 1}^n {b_k}^p}^{1 / p} \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1 / q}\) | Power of Power, and by hypothesis: $\paren {p - 1} q = p$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {\paren {\sum_{k \mathop = 1}^n {a_k}^p}^{1 / p} + \paren {\sum_{k \mathop = 1}^n {b_k}^p}^{1 / p} } \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1 / q}\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1 - 1 / q}\) | \(\le\) | \(\ds \paren {\sum_{k \mathop = 1}^n {a_k}^p}^{1 / p} + \paren {\sum_{k \mathop = 1}^n {b_k}^p}^{1 / p}\) | dividing by $\ds \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1 / q}$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1 / p}\) | \(\le\) | \(\ds \paren {\sum_{k \mathop = 1}^n {a_k}^p}^{1 / p} + \paren {\sum_{k \mathop = 1}^n {b_k}^p}^{1 / p}\) | as $1 - \dfrac 1 q = p$ |
$\blacksquare$
Source of Name
This entry was named for Hermann Minkowski.