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 $-3$
 A: 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.
A: What you have looks like 
$$x^8+x^6-4x^4+x^2+1$$
Which on rearranging becomes $$(x-1)^2(x+1)^2(x^4+3x^2+1)$$
Since the terms in the product are greater than or equal to  zero so the product itself is  greater than or equal to zero. i.e
$$x^8+x^6-4x^4+x^2+1\ge0$$
A: $x^8+x^6−4x^4+x^2+1$
$=(x^8−2x^4+1)+(x^4-2x^2+1)x^2$
$=(x^4-1)^2+(x^2-1)^2x^2$
$≥0 $
A: Let $$P(x)=x^8+x^6-4x^4+x^2+1.$$
then
$$P'(x)=8x^7+6x^5-16x^3+2x.$$
but
$$P(\pm 1)=P'(\pm 1)=0$$
$$\implies P(x)=(x^2-1)^2(x^4+3x^2+1)$$
$$\implies \forall x\in\mathbb R \;\;P(x)\geq 0.$$
A: By Descartes'  rule of signs, $x^8+x^6-4x^4+x^2+1$ has either 0 or 2 positive roots (counted with multiplicity). You can check that $x=1$ is a root and it is in fact a double root, which you can confirm either by factoring or by showing $x=1$ is also a root of the derivative. By symmetry, the only negative root is $-1$ (a double root).
Positivity thus follows from evaluating expressions at a few points. At 0 it is 1, at $\pm 10$ it is 100960101 and it is indeed positive in the three intervals $(-\infty,-1)$, $(-1,1)$ and $(1,\infty)$.
