<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] Ok you already proved it if $x \in (-\infty ,0]\cup [1,+\infty) $ , it suffice to prove it for $x\in (0,1)$ to conclude : you can write it as , 
$(x^{2}-\frac{x}{2})^{2}+\frac{3x^{2}}{4}-x+\frac{1}{2}=(x^{2}-\frac{x}{2})^{2}+\frac{x^{2}}{4}+\frac{1}{2} (x-1)^{2}$
</s>