# stability analysis help algebra and inequality

Can someone please explain the last three inequality expression to me, as I do not understand how these are obtained, in particular, why is the square root term not included in the inequality for $\xi_+ > -1$? I dont understand the part when they do the analysis for $\xi_{+} \geq - 1$

why is the square root not included in this inequality ?

$$2r\cos \omega \le 2r$$ if $\delta \ge 0$, then $$2r\cos \omega - \delta \le 2r\cos \omega \le 2r$$

In your case, $\delta = \sqrt{1 - 4r^{2}\sin^{2} \omega - 4r \lambda \Delta t-\lambda^{2} \Delta t^{2}}$

(I am assuming, that just like the missing close bracket, the difference in the two denominators is a typo.)

Edit

$$2r\cos \omega \ge -2r$$ if $\delta \ge 0$, then $$2r\cos \omega + \delta \ge 2r\cos \omega \ge -2r$$
In your case, $\delta = \sqrt{1 - 4r^{2}\sin^{2} \omega}$.
• Hi , I edited the question, and put the solution for a similar problem, I dont understand why the square root term is exluded for xi > -1 $\frac{1}{1+2r}[ 2r cos \omega + \sqrt{1-4r^{2} sin^{2} \omega} ] \geq \frac{-2r}{ 1+2r} \geq -1$ can you explain this part? Thankyou – italy May 21 '18 at 20:25
• Thank you, but I still find this hard to understand from here , in your explanation you are starting with $2rcosw \geq -2r$ but this is the last part of the inequality you dont assume this as first, cause how would you know? – italy May 21 '18 at 22:56
• So we start off like this $\frac{1}{1+2r} [ -2r + \delta] \geq -1$ from here the delta ie the square root is excluded and I don’t see why, cosw > -1 , so that term can be replaced what about the square root? What is this equivalent to ? – italy May 21 '18 at 23:00
• $-1 \le \cos \omega \le 1$, regardless of the (real) value of $\omega$. This is a basic fact about cosine that I assumed you were familiar with, Given $r > 0$ (which is not stated in the snippet you've given, but is evident from how it is used), multiplying by $2r$ does not change this inequality: $$-2r \le 2r\cos\omega$$ Adding $\sqrt{1 - 4r^2\sin^2 \omega}$ to $2r\cos\omega$ only increases the value, so it will remain larger than $-2r$. Then dividing both sides by $1+2r$ gives exactly the inequality you are after. – Paul Sinclair May 22 '18 at 3:54