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?

We have to prove $$x^4-x^3+x^2-x+0.5>0\forall x\in\mathbb{R}$$
So let $$f(x)=\frac{1}{2}\bigg[2x^4-2x^3+2x^2-2x+1\bigg]$$
$$f(x)=\frac{1}{2}\bigg[x^4+(x^2-x)^2+(x-1)^2\bigg]>0\forall x\in\mathbb{R}$$
