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.
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$\begingroup$$x^4+2x^3+3x^2+4x+5=x^2(x+1)^2+2x^2+4x+5=x^2(x+1)^2+2（x+1)^2+3>0$
$\endgroup$ $\begingroup$Just another way: Using AM-GM, $$x^6+5=x^6+1+1+1+1+1\geqslant 6|x|\geqslant 6x$$
with equality iff $x=1$.
$\endgroup$ 0 $\begingroup$Because \begin{aligned}x^6-6x+5=3(x-1)^2+(x^2+2)(x^2-1)^2 \geqslant 0.\end{aligned}
$\endgroup$ $\begingroup$For $x=1$ our inequality is wrong.
For $x<0$ our inequality is obvious.
For $x\geq0$ by AM-GM we have: $x^6+5\geq6x$.
Id est, the answer is $\mathbb R\setminus\{1\}$.
Also, for $x\geq0$ we have: $$x^4+2x^3+3x^2+4x+5=(x^2+x+1)^2+2x+4>0$$ and for $x<0$ the inequality is obviously true.
$\endgroup$ $\begingroup$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 $x \neq 1$.
$\endgroup$ 1 $\begingroup$For $x< 0$ the inequality is obviously true.
For $x \geq 0$. We're only need to prove: $f(x)=x^4 +2x^3 +3x^2 +4x+5 > 0$
Which clearly true!
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