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
In answer to your question asked in the comment:
$$2r\cos \omega \ge -2r$$ if $\delta \ge 0$, then $$2r\cos \omega + \delta \ge 2r\cos \omega \ge -2r$$
In your case, $\delta = \sqrt{1 - 4r^{2}\sin^{2} \omega}$.