>Let $X,Y$ be probability measures on $\{1,2,\dots,n\}$, and set $Z=\frac{\sqrt{XY}}{\mathbb{E}(\sqrt{XY})}$.  How can we prove the inequality
$$H(X)+H(Y)\geq 2\mathbb{E}(\sqrt{XY})^2H(Z),$$ where $H(X)=-\sum_{i=1}^n X(i)\log X(i)$ is the [Entropy function][1].

The problem originates from this [math stack exchange post][2], and [cardinal's rewording][3] of it in the comments.  Despite having being asked over two years ago, with numerous bounties posted, the problem was never solved, and for that reason I am posting it here.

I checked the inequality numerically on matlab for millions of choices, with $n$ up to size $100$, and it always held, which suggests that finding a counter example is unlikely.

**Remarks:** By Cauchy Schwarz, $1\geq \mathbb{E}(\sqrt{XY})^2,$ so the above inequality would be implied by $H(X)+H(Y)\geq 2H(Z).$  However, this inequality is false, so the factor of $\mathbb{E}(\sqrt{XY})^2$ is important.


  [1]: http://en.wikipedia.org/wiki/Entropy_%28information_theory%29
  [2]: http://math.stackexchange.com/questions/41340/im-not-sure-about-this-inequality-how-to-prove-or-disprove-it
  [3]: http://math.stackexchange.com/questions/41340/im-not-sure-about-this-inequality-how-to-prove-or-disprove-it#comment92485_41340