Prove that the algebraic expression is greater than zero Prove that $x^8 - x^5 + x^2 -x + 1 >0$ for $x \in \mathbb{R}$.
It's from the (junior) high-school competition and the idea is that not everyone there knows calculus, so that I'm looking for more "basic" justification.
My idea was to use AM-GM: $x^8 + x^2 \geq 2\sqrt{x^{10}} = 2x^5$. Thus,
$$x^8 - x^5 + x^2 -x + 1 \geq  x^5-x + 1 $$
So that we can now focus on proving that $x^5 - x + 1 > 0 $ but I don't really see how to do so without calculus...
 A: $f(x)=x^8-x^5+x^2-x+1$
If $x>1$, then
$$f(x)=x^5(x^3-1)+x(x-1)+1>0$$
If $x<1$, then
$$f(x)=x^8+x^2(1-x^3)+1-x >0$$
ans for $x=1$,
$$f(x)=1>0$$
Hence for all real values $f(x)>0$.
A: An alternative way which, whilst not as elegant as the others, doesn't really require factorisation, which can sometimes be tricky to spot.
Rearrange to get: $x^8 + x^2 + 1 > x^5 + x$. Now consider $x$ in different cases:
If $x \leq 0$, the the RHS is clearly less than or equal to zero (as $x^5+x = x(x^4+1)$, which is either $0$ or a product of positive and negative number), whilst the LHS is always greater than zero (negative numbers to the power of even degree). So the inequality holds for $x \leq 0$.
If $x\geq1$, it's straightforward to argue the inequality holds again.
Final case is $0<x<1$. Now the LHS will always be greater than $1$ (because of the $+1$), and the RHS will always be less than $1$. So again the inequality holds.
This shows it for all $x\in\mathbb{R}$.
A: You can rewrite this $ x^8 - x^5 + x^2 -x + 1 >0 $
as:
$x^8+(x^4-x)^2+(x-1)^2+1.$
You can figure out the rest ;)
A: The AM-GM says
$$
\frac12x^8+\frac12x^2\ge\left(x^8\cdot x^2\right)^{1/2}=|x|^5\tag1
$$
and
$$
\frac12x^2+\frac12\ge\left(x^2\cdot1\right)^{1/2}=|x|\tag2
$$
Therefore,
$$
\begin{align}
x^8-x^5+x^2-x+1
&=\frac12x^8+\overbrace{\left(\frac12x^8+\frac12x^2-x^5\right)}^{\text{(1)}\implies\ge0}+\overbrace{\left(\frac12x^2+\frac12-x\right)}^{\text{(2)}\implies\ge0}+\frac12\tag{3a}\\
&\ge\frac12x^8+\frac12\tag{3b}\\[6pt]
&\gt0\tag{3c}
\end{align}
$$
A: 0)For $x\le 0:$  obvious, since all terms are non-negative.
1)$x\ge 1$: $x^8\ge x^5,$ and $x^2 \ge x;$
2)$0<x<1:$ $x^2 >-x^5;$ and
$x^8-x+1= x^8 +(1-x)>0,$ and we are done(Why?)
A: Let's complete the alternate sign polynomial with  $\begin{cases}a=x^{8}-x^{5}+x^{2}-x+1\\b=-x^{7}+x^{6}+x^{4}-x^{3}\end{cases}$
When $x\le 0$ then $(x^5+x)\le 0$ and since $(x^8+x^2+1)\ge 1$ we conclude easily that $a>0$.
So let's focus on the case $x>0$.
Notice that $a+b=\dfrac{x^9+1}{x+1}>0$
But since $b=(1-x)(x^6-x^3)=-x^3(1-x)^2(x^2+x+1)\le 0$
We can conclude that $a=(a+b)+(-b)>0$ in this interval.
A: I considered parsing the polynomial as  $ \ (x^8 - x^5) \ + \ (x^2 -x + 1) \ $ .  Restricting techniques to ones that would likely appear in junior high school and high school curricula, we can show that $ \ x^2 - x + 1 \ $ is never equal to zero since its discriminant is $ \ -3 \ $ . So its value can never change sign; since it equals $ \ 1 \ $ for  $ \ x \ = \ 0 \ $ , it is always positive.  We can factor
$ \ (x^8 - x^5)  \ = \ x^5  ·  (x^3 - 1)  \ = \ x^5  ·  (x - 1)  ·  (x^2  +  x  + 1)  \ . \ \ \  $ [using "difference of two cubes"]
By an argument similar to the above, $ \ x^2 + x + 1 \ $ is also always positive.  The product of the first two factors is positive for  $ \ x \ > \ 1 $ and for  $ \ x \ < \ 0 \ $ , hence the complete polynomial is positive in these intervals.  It is equal to $ \ 1 \ $ at  $ \ x \ = \ 0 \ $ and  $ \ x \ = \ 1 \ $ , so it remains to establish the proposed inequality for  $ \ 0 \ < \ x \ < \ 1 \ $ .
Using "completion of the square", we write our polynomial as  $ \ (x \ - \ \frac{1}{2}  )^2 \ + \ \frac{3}{4} \ - \ (x^5 \ - \ x^8) \ $ . From what we know about power-functions, we can say that $ \ x^5 \ - \ x^8 \ > \ 0 $ in the interval  $ \ 0 \ < \ x \ < \ 1 \ $ , but I don't think there's a convincing argument using the available means to show that  $ \ 0 \ < \ x^5 \ - \ x^8 \ < \ \frac{3}{4} \ $ .  [We know that calculus will give us a maximum value for $ \ x^5 \ - \ x^8 \  $ of $ \ ( \frac{5}{8} )^{5/3}  ·  \frac{3}{8} \ \approx \ 0.171 $ at $ \ x_{min} \ = \ (\frac{5}{8} )^{1/3} \ \approx \ 0.855 \ $ . ]
Instead we may exploit symmetry and separate the polynomial into "even" and "odd" components as $ \ (x^8  +  x^2  + 1) \  - \ (x^5 + x ) \ . $ [This is a variation on what user143137 does.]  For  the first polynomial, $ \ x^8  +  x^2  + 1 \ \geq \ 1 \ $ for all real numbers (the graphs of power functions should be familiar at this level).  The second polynomial, $ \ x^5 + x  \ $ , is negative for $ \ x \ < \ 0 \ $ , so we only need to be concerned about $ \ 0 \ < \ x \ < 1 \ $ .  In that interval, we have  $ 1 \ < \ x^8  +  x^2  + 1 \ < \ 3 \ $ and $ \ 0 \ < \ x^5 + x  \ < 2 \ $ .  Each of these polynomials increases "monotonically" (although we might not use that word at this level), so we can conclude that $ \ (x^8  +  x^2  + 1) \  - \ (x^5 + x ) \ > \ 0 \ . $
