An entropy inequality 
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.

Remark: This is to give an alternative proof of the inequality in fedja's nice answer:
$$
(1+a)\log(1+b)+(1+b)\log(1+a)\ge 2(1+c)\log(1+c)
$$
where $a, b, c>0$ with $ab=c^2$.
Proof.
WLOG, assume that $a \le b$.
Let $x = \frac{a}{c} \in (0, 1]$.
It suffices to prove that
$$f(x) := (1 + cx)\ln(1 + c/x) + (1 + c/x)\ln(1 + cx) - 2(1 + c)\ln(1 + c) \ge 0.$$
We have
$$f'(x) = c\ln(1 + c/x) - \frac{c(1+cx)}{x(x + c)}
- \frac{c\ln(1 + cx)}{x^2} + \frac{c(x+c)}{x(1+cx)}$$
and
$$f''(x) = \frac{2c}{x^2(x+c)} - \frac{c^2(1+cx)}{x^2(x+c)^2} + \frac{2c\ln(1+cx)}{x^3} - \frac{2c^2}{x^2(1+cx)} - \frac{c^2(x+c)}{x(1+cx)^2}.$$
Let $g(x) := x^3 f''(x)$. We have
\begin{align*}
    g'(x) &= \frac{c^2(1-x^2)}{(x+c)^3(1+cx)^3}\\
    &\qquad \times [cx^4 + (c^2 + 3)x^3 + (2c^5 - 6c^3 + 10c)x^2 + (c^2 + 3)x + c]\\
    & \ge 0.
\end{align*}
Also, $g(0) = 0$. Thus, $g(x) \ge 0$ on $(0, 1]$. Thus, $f''(x) \ge 0$ on $(0, 1]$. Also, $f'(1) = 0$. Thus, $f'(x) \le 0$ on $(0, 1]$.
Also, $f(1) = 0$. Thus, $f(x) \ge 0$ on $(0, 1]$.
We are done.
