We show that for $ \alpha \geq \frac{1}{2}$, the following inequality holds: \[ \frac{\alpha}{2}\! \int_{-1}^1\!\! (1-x^2) | g' (x)|^2 dx +\! \int_{-1}^1 \!\!g(x) dx - \log \frac{1}{2}\! \int_{-1}^1\!\! e^{2 g(x)} dx \geq 0,\] for every function $g$ on $(-1, 1)$ satisfying $ \| g \|^2 = \int_{-1}^1 (1-x^2) | g' (x) |^2 dx < \infty $ and $ \int_{-1}^1 e^{ 2 g(x)} x dx =0$. This improves a result of Feldman et al., 1998, and answers a question of Chang and Yang in the axially symmetric case.