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 $ Consider a continuous function  $f:[0,1]\to\mathbb{R}^{+}$. How show that
$\int_0^1 f(x)dx-\exp\left(\int_0^1 \log(f(x)) dx\right)\le \max_{0\le x,y\le 1}\left(\sqrt{f(x)}-\sqrt{f(y)}\right)^2$?

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: $$ AM(X)-GM(X)\leq n \operatorname{Var}(\sqrt{X}).$$

