How do we prove $x^6+x^5+4x^4-12x^3+4x^2+x+1\geq 0$? Question
How do we prove the following for all $x \in \mathbb{R}$ :
$$x^6+x^5+4x^4-12x^3+4x^2+x+1\geq 0 $$
My Progress
We can factorise the left hand side of the desired inequality as follows:
$$x^6+x^5+4x^4-12x^3+4x^2+x+1=(x-1)^2(x^4+3x^3+9x^2+3x+1)$$
However, after this I was unable to make any further progress in deducing the desired inequality.
I appreciate your help

$$
x^4+3x^3+9x^2+3x+1 = x^2\left(x+\frac 32\right)^2 + \frac{27}{4}\left(x+\frac 2 9\right)^2 + \frac 23
$$
is strictly positive for all real $x$.
How did I come up with that? I started by completing the square in
$$
x^4+3x^3+9x^2 = x^2(x^2+3x+9) = x^2\left( (x+\frac 32)^2 + \frac{27}{4}\right)
$$
so that
$$
x^4+3x^3+9x^2 +3x+1= x^2\left( x+\frac 32 \right)^2 +  \frac{27}{4}\left( x^2 +\frac 4 9 x + \frac{4}{27}\right)
$$
and then completed the square in $x^2 +\frac 4 9 x + \frac{4}{27}$.
