IMO 1960 - Problem 2

Problem: For which real numbers $x$ does the following inequality hold: $\frac{4x^2}{(1-\sqrt{1+2x})^2}<2x+9?$
Solution: The LHS term is well-defined for $x\ge \frac{-1}{2}$ and $x\ne 0$. Furthermore, $\frac{4x^2}{(1-\sqrt{1+2x})^2}=(1+\sqrt{1+2x})^2$. Since $f(x)=(1+\sqrt{1+2x})^2-2x-9=2\sqrt{1+2x}-7$ is increasing and since $f(\frac{45}{8})=0$, it follows that the inequality holds precisely for $\frac{-1}{2}\le x<\frac{45}{8}$ and $x\ne 0$.