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Let $x_1,\dots, x_n \in \mathbb{R^+}$. Is there any nice way of comparing $$ \left(\sum_{i=1}^n \frac1 x_i\right)^{-1} $$ and $$ \min_{i\in\{1,\dots,n\} }x_i $$

If one has $n$ assets uncorrelated and $x_i$ is their variance, then one can show that the above expression is the variance of the portfolio of assets with least variance. Therefore I would of course like it to be less than the second expression always and if not when they are, but I cant really find out how to compare them at all.

I don't really know which tags to use so feel free to add some.

Edit: just to be clear I would like if $\left(\sum_{i=1}^n \frac1 x_i\right)^{-1}\leq \min_{i\in\{1,\dots,n\} }x_i$ could be proven.

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    $\begingroup$ You can read about the relationship between Harmonic Mean, Geometric Mean, and Arithmetic Mean $\endgroup$ – Geoff Robinson Apr 3 '14 at 17:59
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First, $$ \frac1{x_j} < \sum_{i=1}^n \frac1{x_i} $$ because the LHS is one of the terms in the sum of the RHS, and the other terms are positive. (To get $<$, I'm assuming $n\ge 2$; if $n=1$ we get $=$.) Taking reciprocals yields $$ \Big(\sum_{i=1}^n \frac1{x_i}\Big)^{-1} < x_j $$ Since $j$ was arbitrary, $$ \Big(\sum_{i=1}^n \frac1{x_i}\Big)^{-1} < \min_{j=1}^n x_j $$

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  • $\begingroup$ Awesome! Nice and simple :) $\endgroup$ – Henrik Apr 3 '14 at 18:12
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at least you can do that :

$$ \sum_{i=1}^n \frac1 x_i \leq \left( \max_{i\in\{1,\dots,n\} } \frac 1 x_i \right) \sum_{i=1}^n 1 $$

$$ \sum_{i=1}^n \frac1 x_i \leq \frac n {\min_{i\in\{1,\dots,n\} }x_i} $$

so

$$ \left(\sum_{i=1}^n \frac1 x_i\right)^{-1} \geq \left( \frac n {\min_{i\in\{1,\dots,n\} }x_i} \right)^{-1}$$

$$ \left(\sum_{i=1}^n \frac1 x_i\right)^{-1} \geq \frac {\min_{i\in\{1,\dots,n\} }x_i} {n} $$

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