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.

The problem originates from this math stack exchange post, and cardinal's rewording 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.