How to solve $x^6 - 6x + 5 > 0$? Is there any elementary way to solve (i.e. not involving derivatives etc.) $x^6 - 6x + 5 > 0$ ? I can divide it two times by $(x - 1)$ to get $x^6 - 6x + 5 = (x - 1)^2 (x^4 + 2 x^3 + 3 x^2 + 4 x + 5)$, but it is not very helpful, because there is still this $(x^4 + 2 x^3 + 3 x^2 + 4 x + 5)$ part and I don't know how to prove that this is greater than $0$ without using calculus. What I want to say is that $x^6 - 6x + 5$ looks very specific and I am wondering if there exist any simple, yet smart way to solve it.

It's not possible to solve an inequality, but it can be reduced. The best way to do this is by factoring the left side of this inequality. By finding $x^6-6x+5 = (x-1)^2(x^4+2x^3+3^2+4x+5)$, you're most of the way there. You can first solve $x^6-6x+5 = 0$, in saying that $x = 1$ is a solution, which is correct. Now we have to find the root of $f(x) = x^4 + 2x^3 + 3x^2 + 4x + 5$. Unfortunately, there is not a particularly great way of doing this other than the Binary-Ansatz method if we are not using calculus. What this means is we just want to guess and check pretty much. Let us look at the graph of $x^4 + 2x^3 + 3x^2 + 4x + 5$ and find a range where the function crosses $f(x) = 0$. Looking at the graph we find that this point does not exist, and there is as such no $x$ such that $f(x) = 0$. We can then see that $1$ is the only root for this function. Moreover, we can say that as $f(x) > 0$ and $(x-1)^2 > 0$, the only time when this is not greater than $0$ is when $x = 1$ as $f(x) * (x-1)^2$ at $x = 1$ is $f(1) * 0 = 0$. So we can say that this reduces to $x \neq 1$.
