<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] $x^8+x^6−4x^4+x^2+1$
$=(x^8−2x^4+1)+(x^4-2x^2+1)x^2$
$=(x^4-1)^2+(x^2-1)^2x^2$
$≥0 $
</s>