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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?

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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
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
$a+1/a\ge 2$. Add and divide by 2. – Andrés E. Caicedo Sep 23 '10 at 2:04
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 $\sum x_i$.

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