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.

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 $\large x$.
