To prove this lemma I use the relationship between roots and coefficients of a quadratic equation but did not get the result.
Please help me prove this lemma.
If $ – \theta_{2}x^2 – \theta_{1}x + 1 = 0$ and
$\left| \frac{ \theta_{1} \mp \sqrt{(\theta_{1}^2+4\theta_{2})}}{-2\theta_{2}} \right|< 1$ then:
$1. \theta_{2}+\theta_{1}<1$
$2. \theta_{2}-\theta_{1}<1$
$3. -1<\theta_{2}<1$
Best Answer
$\frac{\theta_1\pm \sqrt {\theta_1^2+4\theta_2}}{2}$ are the two roots of $z^2-\theta_1 z -\theta_2=0$, which are the reciprocals of the roots of $1-\theta_1 x-\theta_2x^2=0$. Thus the premise is that the roots of $z^2-\theta_1 z -\theta_2$ are smaller in absolute value than ${\theta_2}$. Since $\theta_2$ is also the (negative) product of the roots, each root is $>1$ in absolute value.
Note that $\frac{\theta_1+ \sqrt {\theta_1^2+4\theta_2}}{-2\theta_2}\cdot \frac{\theta_1- \sqrt {\theta_1^2+4\theta_2}}{-2\theta_2} = -\frac1{\theta_2}$. Since the numbers on the left are $<1$ in absolute value, we conclude that $|\theta_2|>1$, hence claim 3 is clearly wrong.
The numbers $\frac{\theta_1\pm \sqrt {\theta_1^2+4\theta_2}}{2}$ are also the roots of $f(z)=z^2-\theta_1 z -\theta_2$. Claim 1 is that $f(1)>0$, claim 2 is that $f(-1)>0$. Note that $f$ is a parabola oben upwards and assumes its minimum on the real line at $z=\frac{\theta_1}2$, with value $-\frac{\theta_1^2+4\theta_2}{4}$. If this number is positive, then even more so must we have $f(\pm1)>0$. And if $-\frac{\theta_1^2+4\theta_2}{4}$ is nonpositive, then the roots of $f$ are real and the absolute value of their product equals $|\theta_2|>1$, hence at least one of the roots is $>1$ in absolute value; this seems to work agains the claim that $f(\pm1)>0$. Indeed, you should easily be able to construct counterexamples based on the above calculations.