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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3$\begingroup$ Cauchy-Schwarz yields $\left(\sum x_i\right)\left(\sum x_i^{-1}\right) \geq \left(\sum \sqrt{x_ix_i^{-1}}\right)^2 = N^2$. $\endgroup$– darij grinbergSep 22, 2010 at 14:54
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1$\begingroup$ Use the Cauchy-Schwarz inequality. $\endgroup$– user6096Sep 22, 2010 at 14:55
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$\begingroup$ darij beat me to it! $\endgroup$– user6096Sep 22, 2010 at 14:55
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$\begingroup$ Cauchy-Schwarz is an overkill. This is just the inequality between the arithmetic and harmonic means. $\endgroup$– Sergei IvanovSep 22, 2010 at 19:15
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5$\begingroup$ $a+1/a\ge 2$. Add and divide by 2. $\endgroup$– Andrés E. CaicedoSep 23, 2010 at 2:04
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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 $\sum x_i$.