# Inequality involving $x^n$

This is probably not very exciting, but can't get my head around this for a while. Here's the inequality: $$x_1^3-\epsilon x_1^2 -(2+\epsilon(1-x_1^n))x_1+(1+\epsilon)>0$$ where $\epsilon>0, n \in \mathbb{Z^{+}}$. Please just give me hints, don't solve it for me.

EDIT: $x_1>0$

EDIT2: it seems that the solution to inequality $1-x_1-x_1^2>0$ is the trick. For (see EDIT 1) $0<x_1<\frac{\sqrt{5}-1}{2} \approx 0.618$ the main inequality above is true for all $n, \epsilon$. If $x_1$ is larger than $\approx 0.618$, then it is true for some $\epsilon>0$ iff $n$ is not very large. The condition $\epsilon>0$ is crucial.

• Presumably this is not the full question, since in particular there can be problems if $x_1$ is negative. – André Nicolas Oct 8 '12 at 1:14
• I deleted this condition, but it seems to be important. Add it back. – Alex Oct 8 '12 at 1:23

If it is wished that the inequality be true for all positive $x_1$, it will need modification. For example, let $n=1$. Then we are looking at $$x^3-(2+\epsilon)x+1+\epsilon.$$ This is not necessarily positive. Imagine $\epsilon$ very close to $0$. The function reaches a minimum at $x=\sqrt{(2+\epsilon)/3}$, and the minimum value is negative, though not by much. I chose a small $\epsilon$, because presumably that is what is intended. But if we pick $\epsilon$ large, like $10$, we are looking at $z^3-12x+11$, which is $-5$ at $x=2$.
There will be similar difficulties with $n=2$. And for any $n\ge 3$, and small $\epsilon$, we can reproduce the same problem.
• @Alex: This was a reply to a comment by alex.jordan, who then deleted the comment! So it doesn't make much sense by itself. I assumed in my answer that you wanted to prove that the inequality holds for all $x_1$. alex.jordan pointed out that you could be asking for the solutions to the inequality. That would be of course very difficult except approximately, since we are dealing with possibly a high degree polynomial. – André Nicolas Oct 8 '12 at 3:21
• OK, sorry, I didn't express myself properly: what I mean, is to find for what values of $x_1$ this inequality holds. Does this make better sense now? – Alex Oct 8 '12 at 3:46
• @Alex: Yes, it is clearer. For $n\ge 5$, one cannot expect a pleasant closed form solution. Already there will be some difficulty with $n=3$. – André Nicolas Oct 8 '12 at 3:51
• Ok, I see that for $x=0$ or $x=1$ this inequality holds for all $\epsilon, n$. Is it possible to extend it other subsets of $x$? An approximation is OK. – Alex Oct 8 '12 at 4:22