Find the values of $x$ for which $x^{12}-x^9+x^4-x+1>0$.
I tried to substitute some basic values like $-1,0,1$ and try to find the roots of the function but couldn't.
Then I graphed the function on desmos and this is the graph.
So from this, we can say that $x^{12}-x^9+x^4-x+1>0$ for all values of $x$.
But I want to know how to find the required values of $x$ without graphing
On
Hint: Break it into cases:
If $x \ge 1$, then $x^{12} \ge x^9$ and $x^4 \ge x$.
If $x \le 0$, then $x^{12} \ge 0$ and $x^9 \le 0$ and $x^4 \ge 0$ and $x \le 0$.
If $0 < x < 1$, then $x^{12} > 0$ and $x^4 > x^9$ and $1 > x$.
Can you finish each of these cases?
On
Another way.
For $x\leq0$ it's obvious.
But for $x>0$ by AM-GM we obtain: $$x^{12}-x^9+x^4-x+1=x^{12}-x^9-x^4+x+2x^4-2x+1=$$ $$=x(x^8-1)(x^3-1)+2x^4+3\cdot\frac{1}{3}-2x\geq$$ $$\geq4\sqrt[4]{2x^4\cdot\left(\frac{1}{3}\right)^3}-2x=2\left(\sqrt[4]{\frac{32}{27}}-1\right)x>0.$$
On
The sum of square form $$2(x^{12}-x^9+x^4-x+1)$$ $$=x^6(x^3-1)^2+\left(x^6-\frac{1}{2}\right)^2+2\left(x^2-\frac{1}{4}\right)^2+(x-1)^2+\frac{5}{8}>0.$$
On
My SOS is ugly. $$x^{12}-x^9+x^4-x+1$$ $$={\frac { ( 8x^6-4x^3-1)^2}{64}}+{\frac { \left( 80x^ 2-5x-72 \right) ^{2}}{6400}}+{\frac {2299}{1280} \left( x-{\frac { 712}{2299}} \right) ^{2}}+{\frac {7741}{3678400}}.$$ Remark. From Mr. Mike solution we can get $$x^{12}-x^9+x^4-x+1={\frac {{x}^{13} \left( {x}^{2}+x+1 \right) \left( {x}^{8}+1 \right) + \left( {x}^{2}+x+1 \right) \left( {x}^{6}+{x}^{3}-x+1 \right) }{ \left( {x}^{2}+x+1 \right) \left( x+1 \right) \left( {x}^{2}+1 \right) \left( {x}^{2}-x+1 \right) \left( {x}^{4}-{x}^{2}+1 \right) }}$$ For $x>0,$ it's clearly true!
For $x\geq1$ we obtain: $$x^{12}-x^9+x^4-x+1=x^9(x^3-1)+x(x^3-1)+1>0.$$ For $0<x<1$ we have: $$x^{12}-x^9+x^4-x+1=(1-x)+x^4(1-x^5)+x^{12}>0.$$ For $x\leq0$ it's obvious that $$x^{12}-x^9+x^4-x+1>0.$$