Calculate the number of real roots of $x^8-x^5+x^2-x+1 = 0$ Calculate the number of real roots of $x^8-x^5+x^2-x+1 = 0$
My try: $$\left(x^4-\frac{x}{2}\right)^2+\frac{3}{4}x^2-x+1 = \left(x^4-\frac{x}{2}\right)^2+\frac{3}{4}\left(x^2-\frac{4}{3}x+\frac{4}{3}\right)$$
$$\implies \left(x^4-\frac{x}{2}\right)^2+\frac{3}{4}\left(x-\frac{2}{3}\right)^2+1-\frac{4}{9}>0\quad \forall x\in \mathbb{R}$$
My question is any other method like Using Inequality to solve Given Question
If Yes then please explain here.
 A: AM-GM Inequality gives us,
$$\large\frac{1}{2} \left(x^{8}+x^{2}\right)\geq x^{5}$$
$$\large\frac{1}{2} \left(x^{2}+1\right)\geq x$$
$$\large\therefore x^{8}+x^{2}+1>\frac{1}{2} \left(x^{8}+x^{2}\right)+\frac{1}{2} \left(x^{2}+1\right)\geq x^{5}+x$$
$$\large\implies x^{8}-x^{5}+x^{2}-x+1> 0$$
Hence there cannot exist any real solution for $\large x$.
A: I was curious to see whether even the homely methods we teach early on can resolve this.  Descartes' Rule of Signs tells us that there are 4, 2, or 0 positive real roots and no negative real roots.  So we already know there are at least two complex conjugate pairs of zeroes.
The Rational Zeroes Theorem only suggests the candidates $\pm 1$, which plainly don't work.  Any of the four remaining prospective real zeroes must be irrational.
As a polynomial of even degree, $f(x)$ has an absolute extremum; the positive leading coefficient tells us this is an absolute minimum.  The leading term so dominates the behavior of the polynomial that it is only worth looking for that and any real zeroes fairly close to $x = 0$ .  We find  $f( -1 ) = 5 , f ( 0 ) = 1 $, and $f ( 1 ) = 1 $ [ $f(\pm 2)$ are relatively colossal positive values], so the Intermediate Value Theorem isn't much help (though we suspect the function isn't "wiggling around" much in the interval $( -1 , 1 )$ ) .
We can check that, though now we need derivatives:  $f'(x) = 8x^7 - 5x^4 + 2x - 1$ and $f''(x) = 56x^6 - 20x^3 + 2$ .  We can test $f''(x)$ by solving it like a quadratic function with the substitution $t = x^3$ to find that it has no real zeroes.  Hence, $f''(x)$ is always positive, making $f(x)$ concave upward everywhere; thus, there is only one "turning point".  
We also have $f'(-1) = -16 , f'(0) = -1 ,$ and $f'(1) = +3$, so it seems pretty safe to say the one local minimum in $( 0 , 1 )$ is positive.  So the absolute minimum of $f(x)$ appears to be positive, meaning there are no real zeroes.
(Thank you for the link to MathOverflow.  I have been studying theory of equations on my own and have been interested in the analysis that can be done without resorting to graphing.)
