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It is well known that, if HM denotes the harmonic mean and AM the arithmetic mean, we have $$ AM(x) \ge HM(x) $$

Now I am dealing with the expression $$ \frac{1}{HM(x)} - \frac{1}{AM(x)} $$ A trivial lower bound for this expression is $0$, but is there also a nice upper bound?

Cheers!

EDIT: Or, if there's no general upper bound, might there be one if all numbers involved are positive?

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Please note that the HM-AM inequality holds under the implict assumption that all numbers involved are nonnegative. –  cirpis Jan 6 at 15:46

2 Answers 2

up vote 2 down vote accepted

No, consider just two numbers.

$$\frac{x+y}{2xy} - \frac{2}{x+y} $$

Fix $x \gt 0$ and as $y \to 0+$, this is unbounded.

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Could it be that you've got a typo in your answer? Because if I plug $\pi/2$ into the first formula above, I don't see how this would be unbounded. –  Lagerbaer Jan 26 '11 at 22:54
    
@Lager: Yes there was typo. It is $a \to \pi/2$. –  Aryabhata Jan 26 '11 at 22:56
    
@Lager: I also had the formula for HM wrong. I have corrected. –  Aryabhata Jan 26 '11 at 23:03

I do not have enough reputation to comment, so I give an answer to this old question instead. It is quite easy to understand the problem even for general $n$. Assume, we have $n$ positive numbers $a_j$. Then the easiest upper bound is

$$ \frac{1}{\min(a_j)} \, ,$$

while

$$ \frac{1}{\min(a_j)} - \frac{1}{\max(a_j)} $$

is slightly better. We see that there is no problem, if the $a_j$ uniformly stay away from $0$, which is fulfilled, if you only have one specified set of positive numbers. If we want to improve upon those two bounds, we need more knowledge about the $a_j$ than just their min and max.

To prove the asserted inequalities, use the inequality

$$ \max \ \geq \ AM \ \geq \ HM \ \geq \ \min \, .$$

To show, that we can not do significantly better, call $AM_1$ the usual arithmetic mean and $HM_1$ the usual harmonic mean. By contrast, $AM_{n-1}$ and $HM_{n-1}$ are formed by piking $n-1$ distinct of the $n$ numbers, example for $n = 4$:

$$AM_{1} \ = \ \frac{a_1 + a_2 + a_3 + a_4}{4} \, ,$$

$$AM_{3} \ = \ \frac{a_1a_2a_3 + a_1a_2a_4 + a_1a_3a_4 + a_2a_3a_4}{4} \, ,$$

correspondingly for the harmonic mean. Then you find

$$ \frac{1}{HM_1} \ = \ \frac{AM_{n-1}}{a_1 \cdot \ldots \cdot a_n} $$

and

$$ \frac{1}{AM_1} \ = \ \frac{HM_{n-1}}{a_1 \cdot \ldots \cdot a_n} \, ,$$

so

$$ \frac{1}{HM_1} - \frac{1}{AM_1} \ = \ \frac{AM_{n-1} - HM_{n-1}}{a_1 \cdot \ldots \cdot a_n} \, .$$

To bound this expression, you can refer to the inequality between geometric and harmonic mean, because this works just as well for our $n$ positive numbers $\prod_{\neq i} a_j$, but this gives no upper bound, and a nontrivial bound between AM and HM seems difficult to establish. But you can use the inequality

$$ \max \ \geq \ AM \ \geq \ HM \ \geq \ \min$$

to see

$$ \frac{\max \prod_{\neq i} a_j - \min \prod_{\neq i} a_j}{a_1 \cdot \ldots \cdot a_n} \ \geq \ \frac{AM_{n-1} - HM_{n-1}}{a_1 \cdot \ldots \cdot a_n} \, .$$

Furthermore a moment's thought reveals

$$ \frac{\max \prod_{\neq i} a_j}{a_1 \cdot \ldots \cdot a_n} \ = \ \frac{1}{\min a_j} \, .$$

and

$$ \frac{\min \prod_{\neq i} a_j}{a_1 \cdot \ldots \cdot a_n} \ = \ \frac{1}{\max a_j} \, ,$$

so we get

$$ \frac{1}{\min a_j} - \frac{1}{\max a_j} \ \geq \ \frac{AM_{n-1} - HM_{n-1}}{a_1 \cdot \ldots \cdot a_n} \, .$$

Furthermore it is clear, that we can not improve on this bound, if we know nothing further about the $a_j$, for example compare $a_1 \, = \, \min a_j \ \leq \ a_2 \, = \, a_3 \, = \, \cdots \, = \, a_n \ = \ \max a_j$ and $a_1 \, = \, \max a_j \geq \ a_2 \, = \, a_3 \, = \, \cdots \, = \, a_n \ = \ \min a_j$.

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