Try: $-H(Z) = \mathbb{E} [\log Z] $ we have an equality:
$$ H(Z) = -\mathbb{E} [\log \tfrac{1}{K}] - \alpha \mathbb{E} [\log X]-(1-\alpha) \mathbb{E} [\log Y] = \log K + \alpha H(X) +(1-\alpha) H(Y)$$
rearrange a bit to look like the original problem:
$$ \alpha H(X) +(1-\alpha) H(Y) = H(Z) - \log K \; ?\geq \; K^2H(Z)$$
We seem to have to prove this strange inequality.
$$ H(Z) \geq \frac{\log K}{1 - K^2}$$
I think we can argue $K \leq 1$ as in the other question using Hölder and so $H(Z) \geq 0$ and $\log K < 0$ is negative.
Please help me check signs!