Shannon entropy of $p(x)(1-p(x))$ is no less than entropy of $p(x)$ 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.
 A: 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 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 $i\leqslant k<j$. 
