Prove that $x^{2020} + x^{1011} – 2 x^{1010} + x^2 – x + 1$ does not have a real root.
$$x^{1010}(x^{1010} + x + 2) + x^2 + x + 1$$
Since $x^2+x+1>0$ for all $x,$ it suffices to prove that $x^{1010}+x+2$ is also $>0.$
If $x\le-1,$ $x^{1010}+x+2\ge|x|+x+2>0.$
If $x>-1,$ $x^{1010}+x+2\ge x+2>0.$
I tried the same way but I failed.
$$x^{1010}(x^{1010} + x – 2) + x^2 – x + 1$$
Since $x^2-x+1$ is nonnegative. So it suffices to prove that $x^{1010} + x – 2>0$. But this is wrong by Wolfram Alpha.
Best Answer
want $$ x^{4n} + x^{1+2n} - 2 x^{2n} + x^2 - x + 1 $$
Well, $$ \left(x^{2n} + \frac{x}{2} -1\right)^2 = x^{4n} + x^{1+2n} - 2 x^{2n} + \frac{x^2}{4} - x + 1 $$
so that $$ x^{4n} + x^{1+2n} - 2 x^{2n} + x^2 - x + 1 = \left(x^{2n} + \frac{x}{2} -1\right)^2 + \frac{3x^2}{4} $$