<s>[INST] For what values of the variable x does the following inequality hold: $\ \frac{4x^2}{\Bigl(1-\sqrt{\ 1\ +2x}\Bigr)^2} <
2x+9$
... IMO-1960
 [/INST] Maybe you wanna write everything as
$$\ \frac{((2x+1)-2)(2x+1)+1}{\Bigl(1-\sqrt{2x+1}\Bigr)^2} <
2x+1+8$$
Then, let's denote $2x+1=y$ that yields
$$\ \frac{(y-2)y+1}{\Bigl(1-\sqrt{y}\Bigr)^2} <
y+8$$
$$\ \frac{(1-y)^2}{\Bigl(1-\sqrt{y}\Bigr)^2} <
y+8$$
$$\ \left(\frac{(1-\sqrt{y})(1+\sqrt{y})}{1-\sqrt{y}}\right)^2 <
y+8$$
$$\ (1+\sqrt{y})^2 <
y+8$$
$$y<\frac{49}{4}$$
Or 
$$2x+1<\frac{49}{4}$$
$$x<\frac{45}{8} \tag1$$
At the same time we know that 
$$x\ge -\frac{1}{2} \tag2$$
and pay attention at $x=0$.
From $(1)$ and $(2)$ we conclude that 
$$x\in \left[-\frac{1}{2}, 0\right)\cup \left(0,\frac{45}{8}\right).$$
Chris.
</s>