5 added extra information.

Your alleged inequality is the "well-known" Carleman's inequality, for which it is known that $C=e$ is the best constant.

There are several interesting generalizations to this basic inequality; the wikipedia page lists some. Also, one proof of this inequality follows directly from Hardy's inequality.

EDIT. You might also enjoy the survey: Carleman's inequality: history and new generalizations by J. Pečarić (Aequationes Mathematicae, Volume 61, Numbers 1-2, 49-62)

4 cleaned up.

Your alleged inequality is the "well-known" Carleman's inequality, for which it is known that $C=e$ is the best constant.

There are several interesting generalizations to this basic inequality; the wikipedia page lists some. Also, one proof of this inequality follows directly from Hardy's inequality.

some older ruminations

Maybe the following hand-waving helps answer part 2 of your question. First, note that $MG(1,\ldots,n) \le MG(1,\ldots,m)$ for $n\le m$; then observe that

$$\lim_{m\to \infty} \frac{ \tfrac{1}{m}\sum_{i=1}^m i} {\mbox{MG}(1,2,\ldots,m)} = e.$$

3 reworded a bit.

Update

Your alleged inequality is the "well-known" Carleman's inequality, for which it is known that $C=e$ is the best constant.

There are several interesting generalizations available, to this basic inequality; the wikipedia page list lists some. Also, it seems that one proof of this inequality follows directly from Hardy's inequality.

some older ruminations

Maybe the following hand-waving helps answer part 2 of your question. First, note that $MG(1,\ldots,n) \le MG(1,\ldots,m)$ for $n\le m$; then observe that

$$\lim_{m\to \infty} \frac{ \tfrac{1}{m}\sum_{i=1}^m i} {\mbox{MG}(1,2,\ldots,m)} = e.$$

2 fixed url
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