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.
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Can you determine its local minimum values? – Berci Jul 07 '18 at 11:51
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Essentially the same problem as Proving inequality $x^{10}-x^6+x^2-x+1>0$ – Martin R Jul 07 '18 at 11:56
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thanks, after reading the section, i got the answer. thanks Martin R – pheludar Jul 07 '18 at 12:52
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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 $$x^5\underbrace{(x^3-1)}_{>0}+x\underbrace{(x-1)}_{>0}+1>0$$
nonuser
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$$x^8-x^5+x^2-x+1 = \frac{1}{2}\bigg[2x^8-2x^5+2x^2-2x+2\bigg]$$
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 $$\frac{x^8}{2}+\frac{x^2}{2}-x^5+\frac{x^2}{2}-x+\frac{1}{2}+\frac{1}{2}>0\forall x \in \mathbb{R}$$
DXT
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