TITLE: An Entropy Inequality
QUESTION [34 upvotes]: Let $X,Y$ be probability measures on $\{1,2,\dots,n\}$, and set $K=\sum_i\sqrt{X(i)Y(i)}$ so that $Z:=\frac{1}{K}\sqrt{XY}$ is also a probability measure on $\{1,2,\dots,n\}$.  How can we prove the inequality
  $$H(X)+H(Y)\geq 2K^2 H(Z),$$ 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 of $X$ and $Y$, with $n$ up to size $100$, and it always held, which suggests that finding a counter example is unlikely.
Remark: By Cauchy Schwarz, $1\geq K^2,$ so the above inequality would be implied by $H(X)+H(Y)\geq 2H(Z).$  However it is worth noting that this inequality does not hold, so the factor of $K^2$ is important.

REPLY [4 votes]: Here is another alternative proof of 
the inequality
$$
(1+a)\ln(1+b)+(1+b)\ln(1+a)\ge 2(1+c)\ln(1+c) \tag{1}
$$
with 
$$a,b,c>0,\ ab=c^2,\tag{2}$$
proved in the above answer. 
The left-hand side of $(1)$ goes to $\infty$ as $a$ or $b$ does so while keeping condition $(2)$. So, it remains to show that a  conditional local extremum given $(2)$ can only be attained when $a=b$ and hence $a=b=c$. 
Without loss of generality, $a<b$. 
By Lagrange multipliers, for any such local extremum $(a,b)$ one has 
$\ln(1+b)+\frac{1+b}{1+a}=\lambda b$ and $\ln(1+a)+\frac{1+a}{1+b}=\lambda a$ for some real $\lambda$. Multiplying these two equations by $a$ and $b$, respectively, and then subtracting one from the other, for
$$d(a,b):=a\ln(1+b)+a\,\frac{1+b}{1+a}-b\ln(1+a)-b\,\frac{1+a}{1+b}
$$
one has $d(a,b)=0$ if $(a,b)$ is a local extremum. But $d''_{b,a}(a,b)=\frac b{1 + b^2}- \frac a{1 + a^2}$, which vanishes only if $a=b$ or $ab=1$. 
Next, 
$$d'_b(a,b)=\frac{a^2 b+a b (b+3)-1}{(1+a) (1+b)^2}-\ln(1+a). 
$$
So, $d'_b(0+,b)=-1/(1+b)^2<0$, $d'_b(1/b,b)\big|_{b=1/a}=\frac a{1+a}-\ln(1+a)<0$, and $d'_b(b,b)=\frac{2b-1}{1+b}-\ln(1+b)<0$ (the latter two inequalities easily checked by differentiation). 
So, $d'_b(a,b)<0$ whenever $0<a<b$. 
So, $d(a,b)<d(a,a)=0$ whenever $0<a<b$, which contradicts the necessary condition $d(a,b)=0$ for $(a,b)$ to be a  conditional local extremum. This completes the proof.