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

Your quartic polynomial is called self-reciprocal or palindromic, in mathematics.
Let,
$$
\begin{align}P(x):=&x^4+3x^3+9x^2+3x+1\end{align}
$$
The case $x=0$ is trivial. Therefore, we can divide all terms of the polynomial $P(x)$ by $x^2\thinspace : (x\neq 0\thinspace)$
$$
\begin{align}\frac {P(x)}{x^2}=&\left(x^2+\frac {1}{x^2}\right)+3\left(x+\frac 1x\right)+9\end{align}
$$
Then, using the standard substitution $x+\frac 1x=u$, you have:
$$
\begin{align}&\frac {P(x)}{x^2}=\underbrace{ u^2+3u+7}_{\Delta_u<0}\thinspace >0\\
\implies &P(x)>0\thinspace , \forall x\neq 0\end{align}
$$
This means: $\thinspace P(x)>0,\thinspace \forall x\in\mathbb {R}\thinspace.$
