Minkowski's Inequality for Sums/Index 2

From ProofWiki
Jump to navigation Jump to search


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


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


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


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.