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?
 A: Hint:  Alternately, by AM-GM, note $x^4+\frac14x^2\geqslant x^3$ and $\frac12x^2+\frac12\geqslant x$, so $x^4-x^3+x^2-x+1 \geqslant \frac14x^2+\frac12>\frac12$ as equality is not possible.
A: 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 , 
$(x^{2}-\frac{x}{2})^{2}+\frac{3x^{2}}{4}-x+\frac{1}{2}=(x^{2}-\frac{x}{2})^{2}+\frac{x^{2}}{4}+\frac{1}{2} (x-1)^{2}$
A: 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 $x(x-1)(x^2+1) > -1/2$.
A: For $0 \leqslant x \leqslant 1$,$$
x(x - 1)(x^2 + 1) \geqslant \left(-\frac{1}{4}\right)(x^2 + 1) > \left(-\frac{1}{4}\right) \times 2 = -\frac{1}{2}.
$$
For $x < 0$ or $x > 1$,$$
x(x - 1)(x^2 + 1) > 0 > -\frac{1}{2}.
$$
A: 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 :-)
A: 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 $x\ne-1$.
A: We have to prove $$x^4-x^3+x^2-x+0.5>0\forall x\in\mathbb{R}$$
So let $$f(x)=\frac{1}{2}\bigg[2x^4-2x^3+2x^2-2x+1\bigg]$$
$$f(x)=\frac{1}{2}\bigg[x^4+(x^2-x)^2+(x-1)^2\bigg]>0\forall x\in\mathbb{R}$$
