$|Z_1| = | \frac{v(1+\alpha) + \sqrt{v^2(1+\alpha)^2-4\alpha}}{2}|$
Triangle inequality |x+y|=|x|+|y|
Where x= $\frac{v(1+\alpha)}{2}$ and $y= \frac{\sqrt{v^2(1+\alpha)^2-4\alpha}}{2}$
I've been stuck on this problem now for a couple of days and I'm having a difficult time proving this case.
I'm trying to prove that $|Z_1| \leq 1$ while using triangle inequality.
Where in Case I: $−1\leq \alpha \leq 0$ and $0<v<1$
I was wondering if anybody can assistance me on this problem. I want to thank you ahead of time for your cooperation.