<s>[INST] If $f(x) = x^4 - x^2 + 1$, find the values of $x$ such that $f(f(f(x))) \le x^8$ 
If $$f(x) = x^4 - x^2 + 1$$ find the values of $x$ such that $f(f(f(x))) \le x^8$

I noticed that $f(\pm 1) = 1 \implies \underbrace{f(f(f(... f(\pm 1)..)))}_{\text{n times}} = 1$. thus, $f(f(f(x))) = x^8$ at $x = \pm 1$. Fortunately, these were the only two solutions to this problem. However, this is hacky at best and incomplete at worst.
Can somebody provide a rigorous proof for this problem?
 [/INST] $f(x)$=$x^2(x^2+1/x^2-1)$ ,
Now notice that
$(x^2+1/x^2-1)\ge1$ for all real x . (Apply $AM \ge GM$ for $x^2$ and $1\over x^2$)
That is $f(x)\ge x^2$ , which implies that $f(f(f(x))) \ge x^8$
But in the question it is given that $f(f(f(x))) \le x^8$
Therefore , $f(f(f(x))) = x^8$ , which will happen when $(x^2+1/x^2-1)=1$ , hence $x = \pm 1$ are the only solutions.
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