Minkowski's Inequality for Sums/Index 2

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n \ge 0$ be non-negative real numbers.

Then:

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


Proof

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

$\blacksquare$


Also see

This result is a special case of Minkowski's Inequality for Sums, where $p = 2$.


Source of Name

This entry was named for Hermann Minkowski.


Sources