Question

[Key Concept: $AC$ a diameter, $AB$ tangent, $D$ on $BC$: $tan\ C = \frac{AD}{b} = \frac{a}{1}$ gives $AD = ab < AC = 1$, so $ab < 1$.]

$$tanC = \frac{AD}{b} = \frac{a}{1}$$

$$\Rightarrow AD = ab < AC = 1 \Rightarrow ab < 1$$

$${(AC)^{2} = (AD)^{2} + (CD)^{2} = a^{2}b^{2} + b^{2} = 1 }{\Rightarrow b^{2} = \frac{1}{a^{2} + 1} }{\Rightarrow \frac{b^{2}}{a^{2}} = \frac{1}{a^{4} + a^{2}} > \frac{1}{a^{4} + a^{2} + \frac{1}{4}} = \frac{1}{\left( a^{2} + \frac{1}{2} \right)^{2}}}$$

$$\Rightarrow \frac{b}{a} > \frac{1}{a^{2} + \frac{1}{2}}$$