prove that the polynomial $x^8 -x^5 +x^2 -x +1$ is positive for all real values of x

Question

is there any factorization possible of the above expression or can it be shown that is a a sum of two or three squares? i tried various factorizations but none of them were conclusive.

Answer 1

$$2(x^8-x^5+x^2-x+1)=x^8+(x^4-x)^2+(x-1)^2+1.$$

Answer 2

For $x<1$ we have $$x^8+\underbrace{(1-x)}_{>0}+x^2\underbrace{(1-x^3)}_{>0}>0$$

and for $x>1$ we have $$x^5\underbrace{(x^3-1)}_{>0}+x\underbrace{(x-1)}_{>0}+1>0$$

Answer 3

$$x^8-x^5+x^2-x+1 = \frac{1}{2}\bigg[2x^8-2x^5+2x^2-2x+2\bigg]$$

So $$ = \frac{1}{2}\bigg[x^8+(x^8-2x^5+x^2)+(x^2-2x+1)+1\bigg]$$

So $$ = \frac{1}{2}\bigg[x^8+(x^4-x)^2+(x-1)^2+1\bigg]>0\forall x \in \mathbb{R}.$$

Added:: For $x=0,x^8-x^5+x^2-x+1>0$

Using Arithmetic Geometric Inequality $(x\neq 0)$

$$\frac{x^8}{2}+\frac{x^2}{2}\geq |x^5|\geq x^5$$

$$\frac{x^2}{2}+\frac{1}{2}\geq |x|\geq x$$

So $$\frac{x^8}{2}+\frac{x^2}{2}-x^5+\frac{x^2}{2}-x+\frac{1}{2}+\frac{1}{2}>0\forall x \in \mathbb{R}$$