<s>[INST] 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
 [/INST] By using the symmetry in quartic,
\begin{array}
.x^4+3x^3+9x^2+3x+1&=x^2(x^2+3x+\frac{9}{2}+\frac{9}{2}+3x^{-1}+x^{-2})\\
&=x^2\large\left((x+\frac{3}{2})^2+(x^{-1}+\frac{3}{2})^2+\frac{9}{2}\large\right)\geq0
\end{array}
and at $x=0$ it is $1\geq 0$.
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