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$\ \frac{4x^2}{\Bigl(1-\sqrt{\ 1\ +2x}\Bigr)^2} < 2x+9$

... IMO-1960

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a nice and simple inequality (+1) – user 1618033 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

$$x\in \left[-\frac{1}{2}, 0\right)\cup \left(0,\frac{45}{8}\right).$$


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Hint: you could look for the important points, such as when the square root is zero , when the denominator is zero, and when the inequality is an equality. The last of these can be found by multiplying out, rearranging and then squaring to remove the radical. That gives you three points and four intervals to consider: by testing, two of the intervals and one of the points satisfy the inequality.

Now you know the answer, you can prove it by carefully manipulating the inequality, making sure that the known solutions remain as solutions and spurious extra solutions are rejected.

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

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