1. ## Inequality

I have the following inequality

$\displaystyle -1 \leq \frac{p2R\cos(\phi)+R^2-p-2R\cos(\phi)}{1-pR^2} \leq 3$

where $\displaystyle p\in(-\infty,\infty)$, $\displaystyle R\in(0,\infty)$ and $\displaystyle \phi\in(0,\pi)$.

Given a $\displaystyle \phi$ and a $\displaystyle R$ my goal is to find which values of $\displaystyle p$ satisfies the inequality.
I have tried to follow the procedure described in the document here: Solving inequalities
but have not found a way to create a table such as seen in example 5.
Does anyone have an idea how to proceed with this inequality?

2. ## Re: Inequality

For given $\displaystyle \phi$ and R, this double inequality has the form $\displaystyle -1\le\frac{ap+b}{cp+d}\le 3$ for some constants a, b, c and d. You need to take the intersection of the sets of solutions of these two inequalities. For example, $\displaystyle \frac{ap+b}{cp+d}\le 3$ <=> $\displaystyle \frac{ap+b}{cp+d}-3\le 0$ <=> $\displaystyle \frac{(a-3c)p+(b-3d)}{cp+d}\le0$ <=> [$\displaystyle (a-3c)p+(b-3d)\ge0$ and $\displaystyle cp+d< 0$] or [$\displaystyle (a-3c)p+(b-3d)\le0$ and $\displaystyle cp+d> 0$].

3. ## Re: Inequality

Thanks for feedback.

However, it gets rather messy rather quickly.

First case

(a): $\displaystyle 1-pR^2\geq0$
and
(b): $\displaystyle p(2R\cos(\phi)-1+3R^2) \leq 2R\cos(\phi)+3-R^2$

(a) is easy, we get $\displaystyle p\leq\frac{1}{R^2}$

(b) is not so easy, we don't know the sign of $\displaystyle 2R\cos(\phi)-1+3R^2$.
If $\displaystyle R\leq1$ then the RHS $\displaystyle 2R\cos(\phi)+3-R^2 \geq 0$, so for $\displaystyle R\leq1$ we must study the sign of the LHS $\displaystyle 2R\cos(\phi)-1+3R^2$. On the other hand, $\displaystyle R\geq1$ the LHS is always positive and we must study the sign of the RHS.

$\displaystyle R\leq1$: (LHS)
R must reach a minimum value before LHS can be positive. That value is $\displaystyle \frac{1}{3}$.

For $\displaystyle R\leq\frac{1}{3}$ we have
$\displaystyle p\geq \frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2}$ which is negative.
So for $\displaystyle R\leq\frac{1}{3}$ we can't met equality[/TEX].

For $\displaystyle \frac{1}{3}<R\leq1$

if $\displaystyle R<\frac{\sqrt{\cos^2(\phi)+3}-\cos(\phi)}{3}$ then $\displaystyle 2R\cos(\phi)-1+3R^2 < 0$
if $\displaystyle R>\frac{\sqrt{\cos^2(\phi)+3}-\cos(\phi)}{3}$ then $\displaystyle 2R\cos(\phi)-1+3R^2 > 0$
Then
For $\displaystyle \frac{1}{3}<R\leq\frac{\sqrt{\cos^2(\phi)+3}-\cos(\phi)}{3}$
$\displaystyle p\geq \frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2}$ which is negative
So for $\displaystyle \frac{1}{3}<R\leq\frac{\sqrt{\cos^2(\phi)+3}-\cos(\phi)}{3}$ we require $\displaystyle \frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2}<p\leq\frac{1}{R^2}$.
For $\displaystyle \frac{\sqrt{\cos^2(\phi)+3}-\cos(\phi)}{3}<R<1$
$\displaystyle p\leq \frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2}$ which is positive

For $\displaystyle \frac{\sqrt{\cos^2(\phi)+3}-\cos(\phi)}{3}<R\leq 1$ we require
$\displaystyle p \leq min(\frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2},\frac{1}{R^2})$

$\displaystyle R>1$: (RHS)
When $\displaystyle R>3$ the RHS is always negative.

For $\displaystyle 1<R\leq\3$ we have

if $\displaystyle R<\cos(\phi)+\sqrt{\cos^2(\phi)+3}$ then $\displaystyle 2R\cos(\phi)+3-R^2 > 0$
if $\displaystyle R>\cos(\phi)+\sqrt{\cos^2(\phi)+3}$ then $\displaystyle 2R\cos(\phi)+3-R^2 < 0$
Then
For $\displaystyle 1<R\leq\cos(\phi)+\sqrt{\cos^2(\phi)+3}$
$\displaystyle p\leq \frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2}$ which is positive
For $\displaystyle \cos(\phi)+\sqrt{\cos^2(\phi)+3}<R<3$
$\displaystyle p\leq \frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2}$ which is negative

For $\displaystyle R>3$ we have

$\displaystyle p\leq \frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2}$ which is negative

We therefore conclude

For $\displaystyle R\leq\frac{\sqrt{\cos^2(\phi)+3}-\cos(\phi)}{3}$ we require $\displaystyle \frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2}<p\leq\frac{1}{R^2}$
For $\displaystyle \frac{\sqrt{\cos^2(\phi)+3}-\cos(\phi)}{3}<R\leq \cos(\phi)+\sqrt{\cos^2(\phi)+3}$ we require $\displaystyle p \leq min(\frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2},\frac{1}{R^2})$
For $\displaystyle R\geq\cos(\phi)+\sqrt{\cos^2(\phi)+3}$ we require $\displaystyle p \leq \frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2}$.

Second case

(a): $\displaystyle 1-pR^2\leq 0$
and
(b): $\displaystyle p(2R\cos(\phi)-1+3R^2) \geq 2R\cos(\phi)+3-R^2$

(a) is easy, we get $\displaystyle p\geq\frac{1}{R^2}$

(b)
$\displaystyle R\leq1$: (LHS)
R must reach a minimum value before LHS can be positive. That value is $\displaystyle \frac{1}{3}$.

For $\displaystyle R\leq\frac{1}{3}$ we have
$\displaystyle p\leq \frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2}$ which is negative.
So for $\displaystyle R\leq\frac{1}{3}$ we require $\displaystyle p\leq \frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2}$.

For $\displaystyle \frac{1}{3}<R\leq1$

if $\displaystyle R<\frac{\sqrt{\cos^2(\phi)+3}-\cos(\phi)}{3}$ then $\displaystyle 2R\cos(\phi)-1+3R^2 < 0$
if $\displaystyle R>\frac{\sqrt{\cos^2(\phi)+3}-\cos(\phi)}{3}$ then $\displaystyle 2R\cos(\phi)-1+3R^2 > 0$
Then
For $\displaystyle \frac{1}{3}<R\leq\frac{\sqrt{\cos^2(\phi)+3}-\cos(\phi)}{3}$
$\displaystyle p\leq \frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2}$ which is negative
For $\displaystyle \frac{\sqrt{\cos^2(\phi)+3}-\cos(\phi)}{3}<R<1$
$\displaystyle p\geq \frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2}$ which is positive

For $\displaystyle \frac{1}{3}<R\leq \frac{\sqrt{\cos^2(\phi)+3}-\cos(\phi)}{3}$ we can't meet equality.
For $\displaystyle \frac{\sqrt{\cos^2(\phi)+3}-\cos(\phi)}{3}<R<1$ we require $\displaystyle p \geq max(\frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2},\frac{1}{R^2})$

$\displaystyle R>1$: (RHS)
When $\displaystyle R>3$ the RHS is always negative.

For $\displaystyle 1<R\leq\3$ we have

if $\displaystyle R<\cos(\phi)+\sqrt{\cos^2(\phi)+3}$ then $\displaystyle 2R\cos(\phi)+3-R^2 > 0$
if $\displaystyle R>\cos(\phi)+\sqrt{\cos^2(\phi)+3}$ then $\displaystyle 2R\cos(\phi)+3-R^2 < 0$
Then
For $\displaystyle 1<R\leq\cos(\phi)+\sqrt{\cos^2(\phi)+3}$
$\displaystyle p\geq \frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2}$ which is positive
For $\displaystyle \cos(\phi)+\sqrt{\cos^2(\phi)+3}<R<3$
$\displaystyle p\geq \frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2}$ which is negative

For $\displaystyle R>3$ we have

$\displaystyle p\geq \frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2}$ which is negative

We therefore conclude

For $\displaystyle R\leq\frac{\sqrt{\cos^2(\phi)+3}-\cos(\phi)}{3}$ we can't meet inequality
For $\displaystyle \frac{\sqrt{\cos^2(\phi)+3}-\cos(\phi)}{3}<R\leq \cos(\phi)+\sqrt{\cos^2(\phi)+3}$ we require $\displaystyle p \geq max(\frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2},\frac{1}{R^2})$
For $\displaystyle R\geq\cos(\phi)+\sqrt{\cos^2(\phi)+3}$ we require $\displaystyle p \geq \frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2}$.

For the other inequality
$\displaystyle \frac{ap+b}{cp+d}\geq-1$
$\displaystyle \frac{ap+b}{cp+d}+1\geq0$
$\displaystyle \frac{p(a+c)+b+d}{cp+d}\geq 0$