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...

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.
