For $|a|\le 1$ and $|b|\le 1$ show that
$$\sqrt{1 - b^2}\sqrt{1 - a} \le \frac{\sqrt{6}(1-\frac{3}{4}b)(7a^2b^2 + 96a^2b + 65a^2 + 14ab^2 + 42a + 7b^2 - 96b - 135)}{43a^2b^2 + 129a^2 + 86ab^2 - 86a + 43b^2 - 327} $$
Numerical checks confirm this involved inequality. I tried expansions of the roots and clearing denominators, but didn't succeed. It holds with equality for $a=b=1$ and $a=-b=1$. Hints are welcome. Since numerator and denominator are bivariate quadratic forms, also a geometric interpretation might be helpful.
no idea. The denominator really is nonzero on your square, namely less than zero