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Let $\mathbf{x} =$ {$x_{i}$} be a set of $n$ positive reals. In every good book on inequalities, one finds the classical result \begin{eqnarray} AM(\mathbf{x}) \geq GM(\mathbf{x}) \geq HM(\mathbf{x}), \end{eqnarray} where $AM(\mathbf{x}) = \frac{1}{n} \sum_{i = 1}^{n} x_{i}$ is the arithmetic mean, $GM(\mathbf{x}) = \sqrt[n]{x_{1} \cdots x_{n}}$ is the geometric mean and $HM(\mathbf{x}) = n (\sum_{i = 1}^{n} \frac{1}{x_{i}})^{-1}$ is the harmonic mean of $\mathbf{x}$, respectively.

Question: I'm curious about sharp bounds of the form: \begin{eqnarray} HM(\mathbf{x}) \geq f(\mathbf{x}) AM(\mathbf{x}) + g(\mathbf{x}), \end{eqnarray} where $f$ and $g$ are some functions which do not imply the classical result above or render the inequality trivial. Do such results exist in the literature (or mathematical folklore)? (References are welcome.)

Thanks!

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  • $\begingroup$ What about $f=0$ and $g=HM$? You need some conditions to avoid trivialities. $\endgroup$ – AD. Nov 4 '10 at 10:39
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    $\begingroup$ @AD.: I think it is pretty clear what the intent of the problem is. @user02138: Your best bet would be to look in books of Olympiad inequality problems; sometimes they will pose problems like this. $\endgroup$ – Qiaochu Yuan Nov 4 '10 at 11:11
  • $\begingroup$ @Qiaochu Yuan: To me it is not, another choice could be $f =HM/(2AM)$ and $g=HM/2$, could you please explain what you mean? $\endgroup$ – AD. Nov 4 '10 at 12:30
  • $\begingroup$ @AD.: I interpret the problem as meaning "is there a bound of this form which does not already immediately follow from the classical result?" For example, let QHM be the quadratic harmonic mean; then HM(x) \ge QHM(x) but HM(x) \le AM(x), so any result of the form HM(x) \ge p AM(x) + (1-p) QHM(x) where 0 < p < 1 would be nontrivial and require a new proof. $\endgroup$ – Qiaochu Yuan Nov 4 '10 at 14:15
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    $\begingroup$ @AD.: Yes, but I think it's facetious to pretend that ignoring the implied conditions would constitute a satisfying answer to the OP's question. $\endgroup$ – Qiaochu Yuan Nov 4 '10 at 15:53
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This bound isn't exactly in the format that you want, but it's likely useful nonetheless. Suppose $\max{\vec{x}} / \min{\vec{x}}=r$.

$$\dfrac{(r-1)^2}{n(r+1)}\le AM - HM \le (\sqrt{r}-1)^2$$

See here for the paper that goes through the proof. I used the same technique to prove that

$$\dfrac{1}{HM}-\dfrac{1}{AM} \le \dfrac{(r-1)^2}{nr(r+1)}$$

because (in terms of their notation) I found $x=A_{n-1}$ maximized the difference of the reciprocals.

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  • $\begingroup$ Your first upper bound seems to be violated by $x = \{2,3,5\}$. I get $\{ 0.214286, 0.430108, 0.337722 \}$. Am I missing something? $\endgroup$ – user02138 Sep 13 '19 at 14:00
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The suggested inequalities in the previous answer are incorrect if $\min \vec{x} > 1$. (Counter-example: $\vec{x} = \{ 2, 3, 5 \}$.) The corrected inequalities are \begin{align} \frac{m(r-1)}{n(r+1)} \leqslant AM - HM \leqslant m(\sqrt{r} - 1)^2, \end{align} where $m = \min \vec{x}$.

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