# 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

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a nice and simple inequality (+1) –  Chris's sis Oct 23 '12 at 7:12
I've added contest-math tag, since you wrote that it is from IMO. I don't think that elementary-number-theory tag is a good fit for this question. –  Martin Sleziak Oct 23 '12 at 7:14

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.

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For what it is worth, the important points are $\frac{-1}{2}, 0, \frac{45}{8}$.