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5 | added extra information. | ||
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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) |
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4 | cleaned up. | ||
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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 $$ \lim_{m\to \infty} \frac{ \tfrac{1}{m}\sum_{i=1}^m i} {\mbox{MG}(1,2,\ldots,m)} = e.$$ |
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3 | reworded a bit. | ||
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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 $$ \lim_{m\to \infty} \frac{ \tfrac{1}{m}\sum_{i=1}^m i} {\mbox{MG}(1,2,\ldots,m)} = e.$$ |
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2 | fixed url | ||
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