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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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What about $f=0$ and $g=HM$? You need some conditions to avoid trivialities. –  AD. Nov 4 '10 at 10:39
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@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. –  Qiaochu Yuan Nov 4 '10 at 11:11
    
@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? –  AD. Nov 4 '10 at 12:30
    
@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. –  Qiaochu Yuan Nov 4 '10 at 14:15
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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. –  Qiaochu Yuan Nov 4 '10 at 15:53

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