<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] Let $P(x) = x^4 + 3x^3 + 9x^2 + 3x + 1$
$P(x≥0) \;≥\; 1 \;>\; 0\;$
$P(y=-x) \;=\; y^4 - 3y^3 + 9y^2 - 3y + 1 = (y-1)^4 + (y^3+3y^2+y)$
$P(x<0 \;⇒\;y>0) \;>\; 0 \quad\quad $ // RHS first term ≥ 0, second term > 0
$\forall x\in\mathbb {R} \;⇒\; P(x)>0$
</s>