We have $a^2 > 2b^2$ and so $a^2 \geq 2b^2 + 1$, i.e. $b^2 \leq \frac{a^2-1}{2}$. The desired inequality is equivalent to $2a^2 - 1 > 2\sqrt{2}ab$, i.e. $b < \frac{(2a^2 - 1)^2}{8a^2}$. So it now suffices to show $\frac{a^2-1}{2} < \frac{(2a^2 - 1)^2}{8a^2}$ - but this is equivalent to $4a^4 - 4a^2 < 4a^4 - 4a^2 + 1$, which is true.

If $\frac{a}{b}-\frac{1}{2ab}\leq \sqrt{2}$ one can prove $a^2<2b^2+1$ (using quadratic formula and doing bounding of the discriminant) meaning forcing $a^2=2b^2$ which is not possible. Thus, the problem is still true on positive rationals.