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I recently came across this problem:

Show that the set $\{x : x^2 < 1-x\}$ is bounded above.

How should I approach any similar problem of the form $\{x : f(x) < g(x)\}$?

Unfortunately, I cannot say what I have already tried; this is a very new concept to me, and I am not sure what things I can even try in the first place. I am not necessarily looking for a solution; I am wondering how I should begin to tackle such a problem.

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2 Answers 2

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You just need to prouve that there exists an $a>0$ such that $f(x)\geq g(x)~~\forall x>a$ or just prouve that $\lim_{x\to+\infty} (f(x)-g(x))>0$. In your case $f(x)-g(x)=x^2+x-1\to+\infty$ when $x\to+\infty$ so you can deduce that the set is bounded from above.

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The first thing I would do, and what Harnack suggested, is solve the given inequality. We are given that $x^2< 1- x$. That is the same as $x^2+ x- 1= 0$. Completing the square, $x^2+ x+ 1/4- 1/4- 1= (x- 1/2)^2- 5/4= 0$. $(x- 1/2)^2= 5/4$ so $x- 1/2= \pm\frac{\sqrt{5}}{2}$ so that $x= \frac{1}{2}\pm\frac{\sqrt{5}}{2}$. $y= x^2+ x- 1$ is a parabola opening upward, crossing the x-axis at $x= \frac{1- \sqrt{5}}{2}$ and $x= \frac{1+ \sqrt{5}}{2}$. The parabola is below the x-axis, so the original inequality is true, between those two numbers.

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