<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] Let $$P(x)=x^8+x^6-4x^4+x^2+1.$$
then
$$P'(x)=8x^7+6x^5-16x^3+2x.$$
but
$$P(\pm 1)=P'(\pm 1)=0$$
$$\implies P(x)=(x^2-1)^2(x^4+3x^2+1)$$
$$\implies \forall x\in\mathbb R \;\;P(x)\geq 0.$$
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