TITLE: Real roots of a polynomial of real co-efficients , with the co-efficients of $x^2 , x$ and the constant term all $1$
QUESTION [7 upvotes]: Can all the roots of the polynomial equation (with real co-efficients) $a_nx^n+...+a_3x^3+x^2+x+1=0$ be real ? I tried using Vieta's formulae $\prod \alpha=\dfrac {(-1)^n}{a_n}$ , $(\prod \alpha )(\sum \dfrac 1 \alpha)=(-1)^{n-1}\dfrac 1 a_n$ , so $\sum \dfrac1\alpha = -1$ but this is not going anywhere , please help .

REPLY [3 votes]: Here is another approach. Rolle's Theorem (a special case of the Mean Value Theorem) states that for a function $f$ continuous on $[a,b]$ and differentiable on $(a,b)$, $f(a) = f(b)$ implies the existence of a $c \in (a,b)$ such that $f(c) = 0$. 
We can then say that if a polynomial has $k$ real roots, its derivative must have $k-1$ real roots where each root of the derivative lies between the roots of the original polynomial. Now every root $r$ of a polynomial $P(x)$ of degree $n$ corresponds to a root $r' = \frac{1}r$ of $x^nP\left( \frac{1}x \right)$. Thus, $$P(x) = \sum_{k = 3}^n a_kx^k+x^2+x+1$$ has all real roots iff $$G(x) = x^n+x^{n-1}+x^{n-2} + \sum_{k = 3}^na_kx^{n-k}$$ has all real roots. Taking the $n-2$th derivative of $G$ gives us that $$\frac{n!}{2}x^2+(n-1)!x+(n-2)!1$$ has $2$ real roots or $n(n-1)x^2+(n-1)x+1$ has real roots. However, $$G_{\delta} = (n-1)^2-4n(n-1) < 0$$ for all $n > 1$. Thus the result follows.