Can one compare $\left(\sum_{i=1}^n \frac1 x_i\right)^{-1}$ with $\min x_i, x_i>0 \forall x_i=1,..,n$

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.

• You can read about the relationship between Harmonic Mean, Geometric Mean, and Arithmetic Mean – Geoff Robinson Apr 3 '14 at 17:59

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, $$\Big(\sum_{i=1}^n \frac1{x_i}\Big)^{-1} < \min_{j=1}^n x_j$$

• Awesome! Nice and simple :) – htd Apr 3 '14 at 18:12

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

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

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