a) Assume that the equation $f(x) = x^n + a_1 x^{n-1} + \dots +a _n = 0$ has $n$ real roots $x_1,\dots x_n$ (that means $f(x_i) = 0$ and $x_i \in \mathbb{R}$ $\forall i$). Prove that:
$$\prod\limits_{i = 1}^{n}(1+ x_i^2) = (1 – a_2 + a_4 – \dots)^2 + (a_1- a_3 + a_5 + \dots)^2.$$
b) Prove that $f(x) = 0$ cannot have $n$ real roots if $(1 – a_2 + a_4 – \dots)^2 + (a_1- a_3 + a_5 + \dots)^2 < 2^n a_n$.
I got stuck and I don't have any idea that can be used to solve this. Can anybody help me?
Best Answer
Hint. Rewrite left part using complex numbers: $$ \prod_{j=1}^n\left(1+x_j^2\right)= \prod_{j=1}^n\left[\left(1+\textbf i \:x_j\right)\left(1-\textbf i \:x_j\right)\right] $$ Then try to use Vieta's formulas. General idea is to find the way to use $x_j$ instead of $x_j^2$.
UPD (using @Exodd hint):
Let's define the following function: $$ g(z):=\prod_{j=1}^n\left(1-\frac{x_j}{z}\right)= \prod_{j=1}^n\frac{z-x_j}{z}=\frac{1}{z^n}\prod_{j=1}^n\left(z-x_j\right)= \frac{f(z)}{z^n} $$ Then $$ \prod_{j=1}^n\left(1+x_j^2\right)= \prod_{j=1}^n\left[\left(1+\textbf i \:x_j\right)\left(1-\textbf i \:x_j\right)\right]= \prod_{j=1}^n\left(1+\textbf i\:x_j\right)\cdot\prod_{j=1}^n\left(1-\textbf i\:x_j\right)= $$ $$ =\prod_{j=1}^n\left(1-\frac{x_j}{\textbf i}\right)\cdot\prod_{j=1}^n\left(1+\frac{x_j}{\textbf i}\right)= g(\textbf i)g(-\textbf i) $$ Now we notice that $g(\bar z)=\overline{g(z)}$, so $$ g(\textbf i)g(-\textbf i) = |g(\textbf i)|^2= \left|\frac{f(\textbf i)}{\textbf i^n}\right|^2 = \left|f(\textbf i)\right|^2 $$ That's it. Not to deal with some technical stuff (general $n$ case, a lot of indices...) let's show for small $n=4$: $$ \left|f(\textbf i)\right|^2= \left|\textbf i^4+a_1\textbf i^3+a_2\textbf i^2+a_3\textbf i+a_4\right|^2= $$ $$ =\left|1-a_1\:\textbf i-a_2+a_3\:\textbf i+a_4\right|^2= \left|(1-a_2+a_4)-\textbf i\:(a_1-a_3)\right|^2= $$ $$ =(1-a_2+a_4)^2+(a_1-a_3)^2 $$