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I have a set $x_i$ with length $n$, where $n>1$. The set is composed of real positive numbers, $x_i \in \mathbb{R}_{+}$. The mean of the set is equal to 1.

$$ \bar{x_i}=\frac{1}{n}\sum_{i=1}^{n}x_i=1 $$

I want to prove that the summation of the inverse of this set, minus one, is positive. Basically, proving the following:

$$ \sum_{i=1}^{n}\left(\frac{1}{x_i}-1\right)\geq0 $$

My main problem is that I'm not sure how to simplify the sum of an inverse of a set. Really appreciate any help on this. Not a math expert, so let me know if there is something obvious I am missing.

Clement Yung
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Marco Pastor Mayo
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  • Suppose $n=1$ and $x_1=1$. Note that you refer to the "mean" of the set but then you write out the sum, not the mean. Did you mean to say that the sum was $n$? Or did you mean "sum" when you wrote "mean"? – lulu Feb 15 '23 at 12:26
  • If you really meant the sum, and if $n>1$ then each $x_i<1$, clearly, so each term in your second sum is positive. But it isn't clear what you really meant to ask. – lulu Feb 15 '23 at 12:28
  • I corrected the formula to obtain the mean. I forgot to divide the sum by $n$ so that it is the mean. – Marco Pastor Mayo Feb 15 '23 at 12:31
  • Still need to require that $n>1$ though. – lulu Feb 15 '23 at 12:32
  • Ok, just edited it to specify that $n>1$. Also, I edited the summation so that it is greater or equal to 0, because, as @lulu says, if the set is composed of a series of 1,1,1..., then it the summation would be 0. But, if there is any variance in the set, it would be positive. – Marco Pastor Mayo Feb 15 '23 at 12:35
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    Your desired inequality follows from the HM-AM Inequality, see this And, yes, you were correct to modify it to $≥0$. – lulu Feb 15 '23 at 12:36
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    Cauchy-Schwarz gives $\left(\sum\limits_{k=1}^n1\right)^2\le\sum\limits_{k=1}^nx_k\sum\limits_{k=1}^n\frac1{x_k}$ – robjohn Feb 15 '23 at 12:57
  • This can be cast a convex minimization problem where we want to show that the minimum is nonnegative. – KBS Feb 15 '23 at 13:00

1 Answers1

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Original statement: $$ \sum_{i=1}^{n}\left(\frac{1}{x_i}-1\right)\geq0 $$ Expand the $-1$ as $-n$: $$ \sum_{i=1}^{n}\left(\frac{1}{x_i}\right)-n\geq0 $$ Add $n$: $$ \sum_{i=1}^{n}\left(\frac{1}{x_i}\right)\geq n $$ Divide by $1/n$: $$ \frac{1}{n}\sum_{i=1}^{n}\left(\frac{1}{x_i}\right)\geq1 $$ Invert the equation: $$ \frac{n}{\sum_{i=1}^{n}\left(\frac{1}{x_i}\right)}\leq1 $$

The HM-AM inequalities tell us that $n$ divided by the sum of the set's inverse is smaller or equal to the mean, which is equal to 1:

$$ \frac{n}{\sum_{i=1}^{n}\left(\frac{1}{x_i}\right)}\leq\frac{1}{n}\sum_{i=1}^{n}\left(x_i\right)=1 $$

Therefore, the inequality holds:

$$ \frac{n}{\sum_{i=1}^{n}\left(\frac{1}{x_i}\right)}\leq1 $$

Marco Pastor Mayo
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