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Suppose $x_1,\dots,x_n$ are all positive real numbers. The Arithmetic Mean is $\frac{\sum_{i=1}^n x_i}{n}$. The Geometric Mean is $\sqrt[n]{\prod_{i=1}^n x_i}$. Is there a constant $C$ depending on $n$, $\min x_i$, $\max x_i$ such that $$\frac{\sum_{i=1}^n x_i}{n} \leq C \sqrt[n]{\prod_{i=1}^n x_i}\ \ ?$$ There is a related question which gives a conclusion but it requires $x_1\,\dots,x_n$ to be non decreasing while $x_1,x_2/2,x_3/3,\dots,x_n/n$ to be non increasing. How prove Reversing the Arithmetic mean – Geometric mean inequality？

Is there a generic inequality without any conditions on $x_1,\dots,x_n$ except $x_i>0$?

Zifeng Zhang
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    A look at the usual upper half unit circle picture for the $n=2$ case shows there are cases where AM is $1$ and GM is as small as one wants. – coffeemath Nov 07 '21 at 20:42
  • Is there a possibility to have the reversed inequality above if $C$ is depending on $n$, $\max{x_i}$, and $\min{x_i}$? – Zifeng Zhang Nov 07 '21 at 20:58
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    Consider $x_1=x$ and all other $x_i=1$, then what you ask for is finding some $C$ s.t $(x+n-1)^n-n^nC^nx \leqslant 0$ for all positive $x$, which clearly isn't possible, the LHS is a polynomial with no upper bound. – Macavity Nov 07 '21 at 22:01

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