Prove an inequality involving a root of a quadratic equation

Question

If $ x=ρ$

is a solution to: $ x^2 + bx + c = 0 $

Prove that $|ρ|-1<|b|+ |c|$

I tried $$ρ^2 + bρ + c=0$$ with no results.

Answer 1

Let $\rho_1$ and $\rho_2$ be the solutions to $x^2 + bx + c = 0$. Let's assume, without loss of generality, that $|\rho_1|\geq|\rho_2|$. We shall consider two cases:

(a) If $|\rho_1|<1$, then

$|\rho_2|-1 \leq |\rho_1|-1 < 0 \leq|b|+|c|$.

(b) If $|\rho_1|\geq 1$, it follows from Vieta's formulas and the reverse triangular inequality that

$|b|+|c| = |\rho_1+\rho_2|+|\rho_1\rho_2| \geq ||\rho_1|-|\rho_2||+|\rho_1||\rho_2| = |\rho_1|-|\rho_2|+|\rho_1||\rho_2| \\ = (|\rho_1|-1)(|\rho_2|+1)+1 > (|\rho_1|-1)(|\rho_2|+1) \geq |\rho_1|-1 \geq |\rho_2|-1.$

Therefore, in both cases it is true that $|\rho|-1<|b|+|c|$.

Answer 2

Consider the matrix $A=\begin{bmatrix}0 & 1 \\ -c & -b \end{bmatrix}$, then $x^2+bx+c$ is the corresponding characteristic polynomial.

By Gershgorin's theorem, we have $|\rho| \le 1$ or $|\rho+b|\le |c|$,

By reversed triangle inequality, $|\rho| \le 1$ or $|\rho| \le |c| + |b|$,

hence $|\rho| \le \max(1, |b|+|c|) \le 1 + |b|+|c|$