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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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    @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.2010-11-04

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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.