If $p(x)$ is a probabilistic density function with finite support, one could construct another probabilistic density function proportional to $p(x)[1-p(x)]$ with a corresponding partition function (to make the new function integrate to 1). It seems that the new constructed p.d.f has a bigger or equal entropy than $p(x)$? How to prove that? Suggestions are welcomed for any references or direct solution.

If $p(x)$ is a discrete probabilistic density function, one could construct another discrete probabilistic density function proportional to $p(x)[1-p(x)]$ with a corresponding partition function to make the new function sum to $1$. (Rule out the degenerate case where $p(x_0) = 1$ for a unique $x_0$.)

It seems that the new constructed p.d.f has a bigger or equal entropy than $p(x)$. How to prove that? Suggestions are welcomed for any references or direct solution.

if p(x) is a probabilistic density function on finite support, one could construct another probablistic density function proportional to p(x)[1-p(x)] with a corresponding partition function (to make the new funciton integrate to 1). It seems that the new constructed p.d.f has a bigger or equal entropy than p(x)? How to prove that? Suggestions are welcomed for any references or direct solution. Thanks.

If $p(x)$ is a probabilistic density function with finite support, one could construct another probabilistic density function proportional to $p(x)[1-p(x)]$ with a corresponding partition function (to make the new function integrate to 1). It seems that the new constructed p.d.f has a bigger or equal entropy than $p(x)$? How to prove that? Suggestions are welcomed for any references or direct solution.