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...

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 \ . $
