TITLE: Proving $x^{4}+x^{3}+x^{2}+x+1$ is always positive for real $x$
QUESTION [15 upvotes]: So I was bored in class  and decided to graph polynomials in geogebra, I noticed that $x^{4}+x^{3}+x^{2}+x+1$ and $x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1$, are all above the x-axis.
Now I am wondering if it is possible to prove that these polynomials (or maybe the first one) are above x-axis without finding the stationary points. (I ask this since I can't solve cubic equations without newtons method).
I am also wondering that if every polynomial that has the format above, which starts with an even number power will be above the x-axis.

REPLY [2 votes]: By pairing monomials differing by one degree it is possible to obtain a solution to the problem.
We show that any polynomial of the following form:
$$x^{2n}+x^{2n-1}+...x+1$$ where $n \in \mathbb {N}$, is strictly positive for all $x \in \mathbb{R}$ 
If $x<-1$, then $x+1<0$, implying that $x^{2i-1}(x+1)>0$, and hence:
$$x^{2n}+x^{2n-1}+...x+1= \sum_{i=1}^n (x^{2i} + x^{2i-1}) +1=\sum_{i=1}^n x^{2i-1}(x+1) +1> 0$$
If $ -1\leq x <0$, then $x+1\geq0$, hence $x^{2i-2}(x+1) \geq 0$. This implies that
$$x^{2n}+x^{2n-1}+...x+1= x^{2n}+ \sum_{i=1}^n (x^{2i-1} + x^{2i-2}) =x^{2n}+\sum_{i=1}^n x^{2i-2}(x+1) \geq x^{2n}+ 0>0$$ 
Finally if $x\geq 0$ it is clear that $x^{2n}+x^{2n-1}+...x+1\geq1>0$