Minkowski's Inequality for Sums/Index Less 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.
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 $p < 1$, $p \ne 0$, then:
- $\ds \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1/p} \ge \paren {\sum_{k \mathop = 1}^n a_k^p}^{1/p} + \paren {\sum_{k \mathop = 1}^n b_k^p}^{1/p}$
Proof
Define:
- $q = \dfrac p {p - 1}$
Then:
- $\dfrac 1 p + \dfrac 1 q = \dfrac 1 p + \dfrac {p - 1} p = 1$
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.
$\blacksquare$
Also known as
This result is also known as the Reverse Minkowski's Inequality (for Sums).
Source of Name
This entry was named for Hermann Minkowski.