Welcome to MathOverflow

Question

-1

1

Given a finite list $x_i$ of $N$ positive reals, it seems that $\sum_{i=1}^N x_i = \sum_{i=1}^N x_i {}^{-1} \Rightarrow \sum_{i=1}^N x_i \geq N$. Can anyone give me a proof?

edited Sep 22 2010 at 22:56

Gerry Myerson
15.7k●1●47●87

asked Sep 22 2010 at 14:52

Jamie Vicary
597●2●10

3

Cauchy-Schwarz yields $\left(\sum x_i\right)\left(\sum x_i^{-1}\right) \geq \left(\sum \sqrt{x_ix_i^{-1}}\right)^2 = N^2$. – darij grinberg Sep 22 2010 at 14:54

1

Use the Cauchy-Schwarz inequality. – Byron Schmuland Sep 22 2010 at 14:55

darij beat me to it! – Byron Schmuland Sep 22 2010 at 14:55

Cauchy-Schwarz is an overkill. This is just the inequality between the arithmetic and harmonic means. – Sergei Ivanov Sep 22 2010 at 19:15

5

$a+1/a\ge 2$. Add and divide by 2. – Andres Caicedo Sep 23 2010 at 2:04

Denis Serre · Accepted Answer · 2010-09-22 14:57:31Z UTC

This is Cauchy-Schwarz inequality. Set $a_i=x_i^{1/2}$ and $b_i:=x_i^{-1/2}$. Then $$N=(a,b)\le\|a\|\cdot\|b\|,$$ with equality if and only if $a$ and $b$ are colinear vectors. With your assumption, the right-hand side is precisely $\sum x_i$.

Little conjecture about sums of reciprocals

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

1 Answer

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.