$2ab-2a^2-ac \ge 1/2 \implies 4ab-4a^2-2ac \ge 1 \implies 4ab-4a^2-2ac \ge a^2+b^2+c^2$ $4ab-4a^2-2ac \ge a^2+b^2+c^2 \implies 4a^2-4ab+b^2+a^2+2ac+c^2 \le 0 \implies (2a-b)^2+(a+c)^2 \le 0 $ By Trivial Inequality: $b=2a$ and $c=-a$ Since $a^2+b^2+c^2=1$, $a^2+(2a)^2+(-a)^2=1 \implies 6a^2=1, \implies a^2=\frac{1}{6}$ and $a= \pm \frac{\sqrt{6}}{6}$ Thus we have the following triples: $\left( \frac{\sqrt{6}}{6}, \frac{\sqrt{6}}{3}, -\frac{\sqrt{6}}{6} \right)$ and $\left( -\frac{\sqrt{6}}{6}, -\frac{\sqrt{6}}{3}, \frac{\sqrt{6}}{6} \right)$