# Show $\{x : f(x) < g(x)\}$ is bounded above

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.

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.
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.