<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] Factorise and complete the square of the first bit to understand function:
$$\begin{align*}
x^4-x^3+x^2-x+0.5&=x^4+x^2-(x^3+x)+\frac12\\
&=x(x^2+1)(x-1)+\frac12\\
&=(x^2+1)(x^2-1)+\frac12\\
&=(x^2+1)\left[(x-\frac12)^2-\frac14\right]+\frac12
\end{align*}$$
Then $(x^2+1)\left[(x-\frac12)^2-\frac14\right]\geq-\frac14$, and $\frac12-\frac14>0$ as required.
I will leave you to work out why $(x^2+1)\left[(x-\frac12)^2-\frac14\right]\geq-\frac14$ holds - it just requires a little thought :-)
</s>