For this isosceles triangle, the expression $(a/2)^{2} + v^{2} = b^{2}$ holds by the Pythagorean Theorem. If we square both sides of the given inequality, then: $(a/2)^{2} + av + v^{2} \ge 2b^{2}$; or $av + b^{2} \ge 2b^{2}$; or $av \ge b^{2}$ (i) and multiplying both sides of (i) by $1/2$ yields $\frac{av}{2} \ge \frac{b^{2}}{2}$ (ii), the LHS being the triangle's area. This implies that the triangle is at the very minimum a right-isosceles triangle (i.e. the angles are $\pi/4, \pi/4$, and $\pi/2$). If $b = 8\sqrt{2}$, then the area is just $(1/2)[8\sqrt{2}]^{2} \cdot sin(\pi/2)$ = $\fbox{64}$.