Proving $x^{4}+x^{3}+x^{2}+x+1$ is always positive for real $x$ 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.
 A: The polynomial equation $x^n-1=0$ has n complex roots, one of which is definitely $1$, and, if n is even, then so is $-1$. But if you divide the polynomial with $x-1$ for odd n, that root which is $1$ just disappears. And if you were already taught complex numbers in class, then you already know by now that the complex solutions to $x^n=a$ form a regular n-sided polygon of radius $\sqrt[n]a$, centered in the origin. Now, for $n=3$, we have an equilateral triangle with one tip in $x=1$, and the other two tips obviously not on the real axis. The same for $n=5$, (where we have a regular pentagon), and for all other odd values of n as well. So all other roots, except for $x=1$, which we've eliminated through polynomial division, are non-real. See n-th root and/or cyclotomic polynomial for more details.
A: For the first one, a more elementary approach would be:
$$\begin{align}x^4 + x^3 + x^2 + x + 1 &= x^2\left(x^2 + x + \frac{1}{2}\right) + \frac{1}{2}x^2 + x + 1\\
&\ge\frac{1}{2}\left((x+1)^2 + 1\right)\\
&\ge \frac{1}{2}\\
&> 0 \end{align}$$
A: $$x^4+x^3+x^2+x+1=x^2\left(x^2+x+1+\frac1x+\frac1{x^2}\right)$$
$$=x^2\left[\left(x+\frac1x\right)^2-\left(x+\frac1x\right)+1\right]$$
Now for real $y,$ $$y^2-y+1=\left(y-\frac12\right)^2+\frac34\ge\frac34$$
A: For $n$ odd:


*

*if
$x\neq 1$: $$
1+x+\cdots+x^{n-1} = \frac{x^n-1}{x-1}\neq 0
$$

*
if $x=1$: $$1+x+\cdots+x^{n-1}=n> 0$$


As it is a continuous function the intermediate value theorem concludes that it is $>0$.
A: 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 $x^{2n}+x^{2n-1}+...x+1\geq1>0$
A: It turns out that any polynomial or rational function that is always positive can be written as a sum of squares.
e.g.
$$ x^4 + x^3 + x^2 + x + 1 = \left(\frac{x^2 + x}{\sqrt{2}}\right)^2
+ \left(\frac{x + 1}{\sqrt{2}}\right)^2
+ \left(\frac{x^2}{\sqrt{2}}\right)^2
+ \left(\frac{1}{\sqrt{2}}\right)^2
$$
Alas, I don't know of any systematic way to figure out how to come up with such a representation, although this one is easily extended to the particular family of polynomials you are interested in.
