7
$\begingroup$

Let $a_i\gt0$ for all $1\le i\le n$. It is well known that $$ \frac{a_1+a_2+\cdots+a_n}{n}-\frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}}\ge0, $$ with the equality when all $a_i$ are equal. Now let $a_i$ are not equal but satisfy the following condition $|a_{i+1}-a_i|\le \varepsilon$ for some $\varepsilon$. I am trying to find an upper bound depending $\varepsilon$(and maybe some $a_i$) for the above difference, but could not find one so far.

Any hints and suggestions would be appreciated.

Thanks!

Improve this question
$\endgroup$
1
  • 2
    $\begingroup$ The obvious first guess is that you'll have the maximum (say, with a given AM) if $a_i=a_1+(i-1)\epsilon$. The question is whether you can compute the HM for this sequence. $\endgroup$
    – Alex Degtyarev
    Commented Feb 7, 2015 at 21:43

2 Answers 2

Reset to default
10
$\begingroup$

a simple upper bound is $(\sqrt{a_{\rm max}}-\sqrt{a_{\rm min}})^2$, with $a_{\rm max}$ and $a_{\rm min}$ the largest and smallest of the $a_i$'s. So for $a_i=a_1+(i-1)\varepsilon$ this would give as upper bound $(\sqrt{a_1+(n-1)\varepsilon}-\sqrt{a_1})^2$.

see theorem 1 of Some Inequalities for Elementary Mean Values, B. Meyker (1984).

Improve this answer
$\endgroup$
1
  • $\begingroup$ I am sorry @Carlo Beenakker, but it is not obvious for me, could you give some hints how to prove it. $\endgroup$
    – pointer
    Commented Feb 7, 2015 at 21:56
0
$\begingroup$

Some results for differences $A_n-G_n, G_n-H_n$ and so after summing up for $A_n-H_n$ you may find in the book: Classical and new inequalities in analysis by D. S. Mitrinovic; J. E. Pecaric; A. M. Fink on pages 25,39.

Improve this answer
$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .