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.



Shannon entropy of $p(x)(1-p(x))$ is no less than entropy of $p(x)$
Denote $f(p)=p(1-p)$, $H(p)=-p\log p$, let $p\_1,\dots,p\_n$ denote all positive probabilities of our distribution, then $\sum p\_i=1$, finally denote $s=\sum\_i f(p\_i)$. Then we need to prove the inequality
$$
\sum\_i H(f(p\_i)/s)\geqslant \sum\_i H(p\_i).
$$
Since $H$ is concave, it suffices to prove that
the multiset $\{p\_1,\dots,p\_n\}$ [majorizes](https://en.wikipedia.org/wiki/Karamata%27s_inequality) the multiset $\{f(p\_1)/s,f(p\_2)/s,\dots,f(p\_n)/s\}$. We may assume that $p\_1\geqslant p\_2\geqslant \ldots\geqslant p\_n$, it implies $f(p\_1)\geqslant f(p\_2)\geqslant \ldots\geqslant f(p\_n)$, since $f(p\_i)-f(p\_j)=(p\_i-p\_j)(1-p\_i-p\_j)$, and $1-p\_i-p\_j\geqslant 0$ for $i\ne j$. Therefore we need to check for every $k$ that $$\frac{p\_1+\dots+p\_k}{p\_1+\dots+p\_n}=p\_1+\dots+p\_k\geqslant \frac{f(p\_1)+\dots+f(p\_k)}s,$$
that is equivalent (by multiplying through by the denominators on the far left and far right) to 
$$(p\_1+\dots+p\_k)(p\_{k+1}^2+\dots+p\_n^2)\leqslant (p\_1^2+\dots+p\_k^2)(p\_{k+1}+\dots+p\_n),$$
which is a sum of obvious inequalities $p\_ip\_j^2\leqslant p\_i^2 p\_j$ over all pairs $i\leqslant k<j$. 


