<s>[INST] Prove that $x^{12}-x^9+x^4-x+1>0$ for all $x\in \mathbb{R}$ 
Prove that the expression $x^{12}-x^9+x^4-x+1>0\; \forall x\in \mathbb{R}$

My try:: Using Interval method::
$\bullet \; $If $x\leq 0$, Then $x^{12}-x^9+x^4-x+1>0$
$\bullet \; $If $0<x\leq 1$, Then $x^{12}+x^4.(1-x^5)+(1-x)>0$
$\bullet \; $If $x>1$, Then $x^9.(x^3-1)+x(x^3-1)+1>0$
So the expression $x^{12}-x^9+x^4-x+1>0\; \forall x\in \mathbb{R}$
My question is How can I solve Using $A.M\geq G.M$ method. or How can I complete the 
square so that the expression is $>0$
Thanks 
 [/INST] $$\begin{aligned}
\frac{1}{4}x^{12}+x^6&\ge |x^9|\ge x^9\\
\frac{3}{4}x^{12}+\frac13&\ge x^6\\
x^4+\frac14&\ge x^2\\
x^2+\frac14&\ge |x|\ge x
\end{aligned}$$
Add them all up, you get that 
$$x^{12}-x^9+x^4-x+1\ge \frac16.$$

Another proof:
$$\begin{aligned}
\frac{x^{12}+x^{12}+x^{12}+1}4&\ge \sqrt[4]{x^{36}}\ge x^9\\
\frac{x^4+1+1+1}{4}&\ge \sqrt[4]{x^4}\ge x.
\end{aligned}$$
The inequality follows.
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