Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
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 '10 at 14:54
1  
Use the Cauchy-Schwarz inequality. –  Byron Schmuland Sep 22 '10 at 14:55
    
darij beat me to it! –  Byron Schmuland Sep 22 '10 at 14:55
    
Cauchy-Schwarz is an overkill. This is just the inequality between the arithmetic and harmonic means. –  Sergei Ivanov Sep 22 '10 at 19:15
5  
$a+1/a\ge 2$. Add and divide by 2. –  Andres Caicedo Sep 23 '10 at 2:04

1 Answer 1

up vote 3 down vote accepted

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 $\sum x_i$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.