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

Let us begin with noticing that
\begin{align*}
f(x) & = x^{6} + x^{5} + 4x^{4} - 12x^{3} + 4x^{2} + x + 1\\\\
& = x^{3}\left[\left(x^{3} + \frac{1}{x^{3}}\right) + \left(x^{2} + \frac{1}{x^{2}}\right) + 4\left(x + \frac{1}{x}\right) - 12\right] \tag{$x\neq 0$}
\end{align*}
Hence, if we make the change of variable $u = x + 1/x$, we get that:
\begin{align*}
\left(x^{3} + \frac{1}{x^{3}}\right) + \left(x^{2} + \frac{1}{x^{2}}\right) + 4\left(x + \frac{1}{x}\right) - 12 & = u(u^{2} - 3) + u^{2} - 2 + 4u -12\\\\
& = u^{3} + u^{2} + u - 14\\\\
& = (u^{3} - 2u^{2}) + (3u^{2} - 6u) + (7u - 14)\\\\
& = u^{2}(u - 2) + 3u(u - 2) + 7(u - 2)\\\\
& = (u^{2} + 3u + 7)(u - 2)
\end{align*}
If $x = 0$, the proposed relation clearly holds. If $x > 0$, then $u\geq 2$. Finally, if $x < 0$, then $u\leq -2$.
Gathering all these results, one concludes the validity of the proposed claim.
Hopefully this contributes!
