How do I solve the inequality $x^{12}−x^9+x^4−x+1>0$ using intervals?

Question

My book has the question $x^{12}−x^{9}+x^{4}−x+1>0$. The solution given gives three cases, when $x \le 0$, when $0 < x \le 1$ and when $x > 1$. How did they get these intervals? What is the method used?

Answer 1

Clearly for $x=0$, We have $$x^{12}-x^{9}+x^4-x+1>0$$

Now for $x\neq 0$, Using Arithmetic Geometric Inequality

$$\frac{x^{12}+x^{12}+x^{12}+1}{4}\geq \sqrt[4]{x^{12}\cdot x^{12}\cdot x^{12}\cdot 1}=|x^9|>x^9$$

$$\frac{x^4+1+1+1}{4}\geq \sqrt[4]{x^4\cdot 1 \cdot 1\cdot 1}=|x|>x$$

So $\displaystyle \frac{3x^{12}}{4}-x^9+\frac{x^4}{4}-x+\frac{1}{4}+\frac{3}{4}>0$(Equality hold when $x=1$)

So $$ x^{12}-x^9+x^4-x+1>\frac{3x^{12}}{4}-x^9+\frac{x^4}{4}-x+1>0\forall x\in\mathbb{R}.$$

Answer 2

Hint: When $x=0$ we get $1>0$ and when $x<0$ then $-x^9,-x>0$ so $$x^{12}-x^9+x^4-x+1>0$$ For $$0<x\le 1$$ we get $$x=1$$ then we get $$1>0$$ and we substitute $$x=\frac{1}{y}$$ so we obtain $$1+y^3(y^5-1)+y^{11}(y-1)>0$$ and for $x>1$ we have $$x^9(x^3-1)+x(x^3-1)+1>0$$

Answer 3

If $x=0$ or $x=1$ we get $1>0$ so both are OK

If $x>1$ then write like this $$x^{9}(x^3-1)+x(x^3−1)+1>0$$

If $x<0$ then write like this $$x^{12}+x^4+1-(x^9+x^3)>0$$

If $0<x<1$ then write it like this $$x^{12}+x^4(1-x^5)+(1-x)>0$$