<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] You can note that the inequality holds for $x=-1$. For $x\ne-1$ it is equivalent to
$$
\frac{x^5+1}{x+1}>\frac{1}{2}
$$
that can be rewritten as
$$
\frac{2x^5-x+1}{x+1}>0
$$
Let's analyze $f(x)=2x^5-x+1$, with $f'(x)=10x^4-1$, which vanishes for $x=\pm10^{-1/4}$. So $f$ has a relative maximum at $-10^{-1/4}$ and a relative minimum at $10^{-1/4}$. Let $a=10^{-1/4}$, for simplicity; then $-1<-a$, so we just need to analyze
$$
f(a)=2a^5-a+1=\frac{1}{5}a-a+1=1-\frac{4}{5}a>0
$$
(prove it).
Since $f(-1)=0$, we conclude that $f(x)<0$ for $x<-1$ and $f(x)>0$ for $x>-1$, so
$$
\frac{f(x)}{x+1}>0
$$
for every $x\ne-1$.
</s>