2

Sum:

$$ s =\frac{\left(\sum_{i=1}^n a_i^{p-1} v_i\right)\left(\sum_{i=1}^{n} a_i^{p-1} v_i\right)}{\sum_{i=1}^{n} a_i^p} - \sum_{i=1}^{n} a_i^{p-2} v_i^2 $$

Where $0<p<1$ and $a_i, v_i$ are real numbers.

EDIT - also $a_i$ > 0

JDS
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  • No it's not. Sorry I will edit to make more clear. – JDS Feb 28 '14 at 15:57
  • Just to clarify, is that what you want to calculate? $$ s = \frac{\big(\sum_1^n a_i^{p-1} v_i\big)\big(\sum_1^n a_i^{p-1} v_i\big)}{\sum_1^n a_i^p} - \sum_1^n a_i^{p-2} v_i^2 = \frac{\big(\sum_1^n a_i^{p-1} v_i\big)^2}{\sum_1^n a_i^p} - \sum_1^n a_i^{p-2} v_i^2 $$ – Integral Feb 28 '14 at 16:02
  • @YoungMoney You get stacked fractions with \frac {top}{bottom} – Ross Millikan Feb 28 '14 at 16:04
  • What about the possibility $=0$? That can certainly happen if, for example, $a_i=v_i=1$ for all $i$. – Barry Cipra Feb 28 '14 at 16:05
  • @Integral - yes exactly – JDS Feb 28 '14 at 16:05

1 Answers1

7

This is an application of The Cauchy-Schwarz Inequality: $$ \left|\,\sum_{i=1}^na_i^{p-2}a_iv_i\,\right| \le\left(\sum_{i=1}^na_i^{p-2}a_i^2\right)^{1/2} \left(\sum_{i=1}^na_i^{p-2}v_i^2\right)^{1/2}\tag{1} $$ Squaring $(1)$ and dividing by $\sum\limits_{i=1}^na_i^{p-2}a_i^2$ and subtracting $\sum\limits_{i=1}^na_i^{p-2}v_i^2$ yields $$ \frac{\left(\sum\limits_{i=1}^na_i^{p-2}a_iv_i\right)\left(\sum\limits_{i=1}^na_i^{p-2}a_iv_i\right)} {\sum\limits_{i=1}^na_i^{p-2}a_i^2}-\sum_{i=1}^na_i^{p-2}v_i^2 \le0\tag{2} $$

robjohn
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