The first relation also writes as $-\frac{1}{2b} < a - b\sqrt{c} < \frac{1}{2b}$, i.e. $\frac{2ab - 1}{2b^2} < \sqrt{c} < \frac{2ab + 1}{2b^2}$.

The second relation also writes as $-\sqrt{c} < a^2 - b^2c < \sqrt{c}$, or $b^2c + \sqrt{c} - a^2 > 0$ and $b^2c - \sqrt{c} - a^2 < 0$, i.e. $\frac{\sqrt{1 + 4a^2b^2} - 1}{2b^2} < \sqrt{c} < \frac{\sqrt{1 + 4a^2b^2} + 1}{2b^2}$.

Computing the difference between the bounds yields $0 < \delta = \frac{\sqrt{1 + 4a^2b^2} \pm 1}{2b^2} - \frac{2ab \pm 1}{2b^2} = \frac{\sqrt{1 + 4a^2b^2} - 2ab}{2b^2} = \frac{1}{2b^2(\sqrt{1 + 4a^2b^2} + 2ab)} < \frac{1}{2b^2 \cdot 4ab} = \Delta$.

Clearly $\sqrt{c} > \frac{\sqrt{1 + 4a^2b^2} - 1}{2b^2}$ implies $\sqrt{c} > \frac{2ab - 1}{2b^2}$. When $\sqrt{c} > \frac{2ab - 1}{2b^2}$ it follows that $4b^4c > (2ab - 1)^2$, so $4b^4c \ge (2ab - 1)^2 + 1$. On the other hand
$$\begin{aligned} \left( \frac{\sqrt{1 + 4a^2b^2} - 1}{2b^2} \right)^2 < \left( \frac{2ab - 1}{2b^2} + \Delta \right)^2 &= \frac{(2ab - 1)^2 + \frac{2(2ab - 1)}{4ab} + \frac{1}{16a^2b^2}}{4b^4} \\ &= \frac{(2ab - 1)^2 + 1 - \frac{1}{2ab} + \frac{1}{16a^2b^2}}{4b^4} < \frac{(2ab - 1)^2 + 1}{4b^4} \le c, \text{ hence the required} \\ &\quad \sqrt{c} > \frac{\sqrt{1 + 4a^2b^2} - 1}{2b^2}. \end{aligned}$$

Clearly $\sqrt{c} < \frac{2ab + 1}{2b^2}$ implies that $\sqrt{c} < \frac{\sqrt{1 + 4a^2b^2} + 1}{2b^2}$. When $\sqrt{c} < \frac{\sqrt{1 + 4a^2b^2} + 1}{2b^2}$, assume $\sqrt{c} \ge \frac{2ab + 1}{2b^2}$, hence $4b^4c \ge (2ab + 1)^2$, therefore $4b^4c \ge (2ab + 1)^2 + 3$ (because of the divisibility by 4).

On the other hand
$$$$
$$\begin{aligned} \left( \frac{\sqrt{1 + 4a^2b^2} + 1}{2b^2} \right)^2 < \left( \frac{2ab + 1}{2b^2} + \Delta \right)^2 &= \\ \frac{(2ab + 1)^2 + 2(2ab + 1)/4ab + 1/16a^2b^2}{4b^4} &= \frac{(2ab + 1)^2 + 1 + 1/2ab + 1/16a^2b^2}{4b^4} \\ < \frac{(2ab + 1)^2 + 3}{4b^4} \le c, \text{ hence } \sqrt{c} > \frac{\sqrt{1 + 4a^2b^2} + 1}{2b^2}, \text{ contradiction. It means our assumption was wrong, so } \sqrt{c} < \frac{2ab + 1}{2b^2}. \end{aligned}$$