Local Mathematical Competitions

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

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

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

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

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