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?

So you have done some good work. Now the function is always positive for $x \not \in (0,1)$. Now lets analyse in this interval.
In $(0,1)$, $x^2+1$ lies in $(1,2)$ and $x(x-1)$ has a minimum of only $-1/4$ at $x=1/2$. Thats enough to conclude that $x(x-1)(x^2+1) > -1/2$.
