<s>[INST] Find the values of $x$ for which $x^{12}-x^9+x^4-x+1>0$ 
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
 [/INST] Another way.
For $x\leq0$ it's obvious.
But for $x>0$  by AM-GM we obtain:
$$x^{12}-x^9+x^4-x+1=x^{12}-x^9-x^4+x+2x^4-2x+1=$$
$$=x(x^8-1)(x^3-1)+2x^4+3\cdot\frac{1}{3}-2x\geq$$
$$\geq4\sqrt[4]{2x^4\cdot\left(\frac{1}{3}\right)^3}-2x=2\left(\sqrt[4]{\frac{32}{27}}-1\right)x>0.$$
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