$$\frac{kb(b-1)}{2}\ge \frac{a(2{{x}^{2}}-2xb+{{b}^{2}}-b)}{2}\Rightarrow \frac{k}{a}\ge \frac{2{{x}^{2}}-2xb+{{b}^{2}}-b}{b(b-1)}$$
suy ra $$2{{x}^{2}}-2bx+{{b}^{2}}-b=\frac{{{(2x-b)}^{2}}+{{(b-1)}^{2}}-1}{2}\ge \frac{{{(b-1)}^{2}}}{2}-\frac{1}{2}$$