sempteim245

2022-03-30

Prove an inequality involving a root of a quadratic equation

If$x=\rho$

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

Prove that$\left|\rho \right|-1<\left|b\right|+\left|c\right|$

If

is a solution to:

Prove that

Tristatex9tw

Beginner2022-03-31Added 18 answers

Step 1

Let$\rho}_{1$ and $\rho}_{2$ be the solutions to ${x}^{2}+bx+c=0$ /

Let's assume, without loss of generality, that$|{\rho}_{1}\ge |{\rho}_{2}\mid$

We shall consider two cases:

a) If$\left|{\rho}_{1}\right|<1$ , then

$\left|{\rho}_{2}\right|-1\le \left|{\rho}_{1}\right|-1<0\le \left|b\right|+\left|c\right|$

b) If$\left|{\rho}_{1}\right|\ge 1$ , it follows from Vieta's formulas and the reverse triangular inequality that

$\left|b\right|+\left|c\right|=|{\rho}_{1}+{\rho}_{2}|+\left|{\rho}_{1}{\rho}_{2}\right|\ge |\left|{\rho}_{1}\right|-|{\rho}_{2}\left||+|{\rho}_{1}\right|\left|{\rho}_{2}\right|$

$=\left|{\rho}_{1}\right|-\left|{\rho}_{2}\right|+\left|{\rho}_{1}\right|\left|{\rho}_{2}\right|$

$=(\left|{\rho}_{1}\right|-1)(\left|{\rho}_{2}\right|+1)+1>(\left|{\rho}_{1}\right|-1)(\left|{\rho}_{2}\right|+1)\ge \left|{\rho}_{1}\right|-1\ge \left|{\rho}_{2}\right|-1.$

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

Let

Let's assume, without loss of generality, that

We shall consider two cases:

a) If

b) If

Therefore, in both cases it is true that

microsgopx6z7

Beginner2022-04-01Added 14 answers

Step 1

Consider the matrix

$A=\left[\begin{array}{cc}0& 1\\ -c& -b\end{array}\right]$

then${x}^{2}+bx+c$ is the corresponding characteristic polynomial.

By Gershgorin's theorem, we have$\left|\rho \right|\le 1$ or

$|\rho +b|\le \left|c\right|$

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

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

Consider the matrix

then

By Gershgorin's theorem, we have

By reversed triangle inequality,

hence

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