Prove that a real polynomial $x^n+ a_1x^{n-1}+ \cdots +a_n$ cannot be completely resolved into linear factors if $a_1^2
Prove that a real polynomial $x^n+ a_1x^{n-1}+ \cdots +a_n$ cannot be completely resolved into linear factors if $a_1^2<a_2$.
Here's what I've got. Let $\alpha_1, \dots, \alpha_n$ be the roots of the polynomial. Then $\displaystyle{a_1^2-a_2=(\alpha_1+ \cdots +\alpha_n)^2 - \sum_{k,j \leq n; k \neq j}\alpha_{j}\alpha_{k} = \alpha_1^2+ \cdots +\alpha_n^2 + \sum_{k,j \leq n; k \neq j}\alpha_{j}\alpha_{k}}$, but I have no idea how to prove this nonnegative. Please help.

Let $f(x)=x^n+a_1x^{n-1}+a_2x^{n-2}+\dots+a_n$ has $n$ real roots.
Hence, $f^{(n-2)}$ has two real roots, which says
$$\frac{n!}{2}x^2+(n-1)!a_1x+(n-2)!a_2=0$$ or
$$\frac{n(n-1)}{2}x^2+(n-1)a_1x+a_2=0$$ has two real roots.
Thus, $$(n-1)^2a_1^2-2n(n-1)a_2\geq0,$$ which gives
$$(n-1)a_1^2\geq2na_2>2na_1^2,$$
which is contradiction.
