Let $\mu,\nu$ be two distributions on the same discrete space. Is it true that


$$\mathrm{H}\left(\frac{\mu+\nu}{2}\right) \ge \mathbb{E}\_{xy}-\log\left(\frac{\sqrt{\mu(x)\nu(y)}}{2} + \frac{\langle\mu, \nu\rangle}{2}\right)$$


where $x\sim\mu$ and $y\sim\nu$ independently. Here $\mathrm{H}(\rho) = -\sum\_{z}\rho(z)\log\rho(z)$ is the Shannon's entropy function and $\langle\mu,\nu\rangle =\sum\_z{\mu(z)\nu(z)}$.


Note that if $\langle\mu,\nu\rangle = 0$, then both LHS and RHS are $\frac{\mathrm{H}(\mu)+\mathrm{H}(\nu)}{2}+1$, so the inequality holds true.


On the other extreme, if $\mu = \nu$, then by convexity of $z\mapsto-\log z$, we have
$$\frac{1}{2}\mathrm{H}(\mu) + \frac{1}{2}\mathrm{H}\_2(\mu) \ge \mathrm{RHS}$$
where $H\_2$ is the second order Renyi entropy which is never bigger than the Shannon entropy. Since LHS is $\mathrm{H}(\mu)$ in this case, this shows the inequality is again true.


I can also verify the inequality when $\mu,\nu$ are flat distributions.


I wonder if this holds true for all $\mu, \nu$.



Entropy of average of two distributions
Unfortunately, this is not true in general. Let $\mu=(x,1-x)$ for $x\in[0,1]$ and let $\nu=(1,0)$.


Now LHS is certainly at most 1 (assume the logarithms are to the base 2), however, the maximum of RHS over $x$ is bigger than one [as can be see here](https://www.wolframalpha.com/input/?i=maximize%20-x*log2(sqrt(x)%2F2%20%2B%20x%2F2)%20%20-%20(1-x)*log2(sqrt(1-x)%2F2%20%2B%20x%20%2F2)%20for%20x%3D0%20to%201).


