<s>[INST] Prove for all $x$, $x^8+x^6-4x^4+x^2+1\ge0$ Prove for all $x$
$x^8+x^6-4x^4+x^2+1\ge0$
By completing the square you get
$(x^4-2)^2+(x^3)^2+(x)^2-3\ge0$
I'm stuck about the $-3$
 [/INST] Observe that $1$ and $-1$ are roots of your polynomial. You thus find that it is equal to
$$(x-1)^2(x+1)^2(x^4+3x^2+1)$$
and is thus clearly non-negative.
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