How prove $ \int_0^1 f(x)dx-\exp\left(\int_0^1\log(f(x)) dx\right)\le\max_{0\le x,y\le 1}\left(\sqrt{f(x)}-\sqrt{f(y)}\right)^2 $

Question

Consider a continuous function $f:[0,1]\to\mathbb{R}^{+}$. How show that

$\int_0^1 f(x)dx-\exp\left(\int_0^1 \log(f(x)) dx\right)\le \max_{0\le x,y\le 1}\left(\sqrt{f(x)}-\sqrt{f(y)}\right)^2$?

possible duplicate of this question. This link does not have an answer though. — r9m, Dec 28 '14 at 13:40

Jack D'Aurizio · Answer 1 · 2014-12-28T13:49:52.230

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It is the continuous version of a well-known result about the difference between the arithmetic and geometric mean: see, for instance, the article of S. H. Tung, or the article of J. M. Aldaz, proving that:

If $0\leq a_1\leq a_2\leq\ldots\leq a_n$ and $X=(a_1,\ldots,a_n)$, then: $$ AM(X)-GM(X)\leq n \operatorname{Var}(\sqrt{X}).$$

edited Dec 28 '14 at 13:49

answered Dec 28 '14 at 13:38

Jack D'Aurizio

353,855

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Jack san for us who do not have access to jstor contents it will be helpful if you summarized the results of the paper in the answer. Thanks :-) – r9m Dec 28 '14 at 13:39
@r9m: answer updated :) – Jack D'Aurizio Dec 28 '14 at 13:50
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@r9m: You're welcome. – Jack D'Aurizio Dec 28 '14 at 13:52
This is nice! +1. – Olivier Oloa Dec 28 '14 at 13:55
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This is AMM problem (2014),I have a nice methods – math110 Dec 28 '14 at 14:00
@math110 the discrete version was Joszef Wildt math competition 2012 problem W-32 :) – r9m Jan 05 '15 at 03:55

How prove $ \int_0^1 f(x)dx-\exp\left(\int_0^1\log(f(x)) dx\right)\le\max_{0\le x,y\le 1}\left(\sqrt{f(x)}-\sqrt{f(y)}\right)^2 $

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