-2
$\begingroup$

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?

$\endgroup$
5
  • 3
    $\begingroup$ Cauchy-Schwarz yields $\left(\sum x_i\right)\left(\sum x_i^{-1}\right) \geq \left(\sum \sqrt{x_ix_i^{-1}}\right)^2 = N^2$. $\endgroup$ Sep 22, 2010 at 14:54
  • 1
    $\begingroup$ Use the Cauchy-Schwarz inequality. $\endgroup$
    – user6096
    Sep 22, 2010 at 14:55
  • $\begingroup$ darij beat me to it! $\endgroup$
    – user6096
    Sep 22, 2010 at 14:55
  • $\begingroup$ Cauchy-Schwarz is an overkill. This is just the inequality between the arithmetic and harmonic means. $\endgroup$ Sep 22, 2010 at 19:15
  • 5
    $\begingroup$ $a+1/a\ge 2$. Add and divide by 2. $\endgroup$ Sep 23, 2010 at 2:04

1 Answer 1

3
$\begingroup$

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 $\sum x_i$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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