$|Z_1| = | \frac{v(1+\alpha)+ \sqrt{v^2(1+\alpha)^2-4\alpha}}{2}|$

I know that using triangle inequality method $|Z_1|$ is:

$|Z_1|= |\frac{v(1+\alpha)}{2}| + |\frac{\sqrt{v^2(1+\alpha)^2-4\alpha}}{2}|$

Case I: $-1 \leq \alpha \leq 0$ and $0 < v <1$

Prove that $|Z_1| \leq 1$

Triangle inequality |x+y|=|x|+|y|

I've been stuck on this problem now for a couple of days and I'm having a difficult time proving this case. I was wondering if anybody can assistance me on this problem. I want to thank you ahead of time for your cooperation.

  • $\begingroup$ What is the question? $\endgroup$ – DanielWainfleet Feb 19 '16 at 17:25
  • $\begingroup$ Corrections have been made. $\endgroup$ – D.d.C Feb 19 '16 at 18:25
  • 1
    $\begingroup$ What is "triangle inequality method" ? Besides, I don 't see any inequality . $\endgroup$ – Jean Marie Feb 19 '16 at 18:28

For $v\in (0,1)$ and $a\in [-1,0]$ we have $\;-4 a\geq 0\;$, so $\; v^2(1+a)^2- 4 a\geq 0.$

For brevity let $v(1+a)=p.$ We have $p\geq 0.$

Therefore $|Z_1|=|\;\frac {1}{2}(p+\sqrt {p^2-4 a} \;)\;|=\frac {1}{2}(p+\sqrt {p^2-4 a}\;) =Z_1.$

$$\text {We have }\quad |Z_1|\leq 1 \iff Z_1\leq 1\iff \frac {1}{2}\sqrt {p^2-4 a}\leq 1-\frac {p}{2} \iff $$ $$\iff \sqrt {p^2-4 a}\leq 2-p\iff p^2-4 a\leq (2-p)^2\iff $$ $$\iff p^2-4 a \leq 4-4 p +p^2 \iff -a\leq 1-p\iff $$ $$\iff -a\leq 1-v(1+a) \iff v(1+a)\leq 1+a $$ which holds because $v>0$ and $1+a\geq 0 $.


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