<s>[INST] Showing $x^4-x^3+x^2-x+1>\frac{1}{2}$ for all $x \in \mathbb R$ 
Show that
  $$x^4-x^3+x^2-x+1>\frac{1}{2}. \quad \forall x \in \mathbb{R}$$

Let $x \in \mathbb{R}$,
\begin{align*}
&\mathrel{\phantom{=}}x^4-x^3+x^2-x+1-\frac{1}{2}=x^4-x^3+x^2-x+\dfrac{1}{2}\\
&=x^2(x^2-x)+(x^2-x)+\dfrac{1}{2}=(x^2-x)(x^2+1)+\dfrac{1}{2}\\
&=x(x-1)(x^2+1)+\dfrac{1}{2}.
\end{align*}
Is there any way to solve this question?
 [/INST] For $0 \leqslant x \leqslant 1$,$$
x(x - 1)(x^2 + 1) \geqslant \left(-\frac{1}{4}\right)(x^2 + 1) > \left(-\frac{1}{4}\right) \times 2 = -\frac{1}{2}.
$$
For $x < 0$ or $x > 1$,$$
x(x - 1)(x^2 + 1) > 0 > -\frac{1}{2}.
$$
</s>