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 $-3$

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
$$x^8+x^6-4x^4+x^2+1\ge0$$
