How must I proceed to solve inequalities of the following form:


This is the final step for another problem that I'm attempting to solve and thus far, this is where I've gotten:

By taking the L.C.M and multiplying both sides by $r$ (given to take strictly positive (or zero) values), I got-


Taking the left inequality separately, I found that $$(r-1)(r+2) \geq 0$$

which gives me

$$r \in (-\infty, -2] \cup [1, \infty)$$

(Clearly, the only valid region is $[1,\infty)$, since $r$ can only take strictly positive/zero values but let me leave that there until the very end, making a mental note of it.)

The right half of the inequality gives me $r²-2r-2≤0$ and I have no insights as to how to solve this.

Please help! Thanks in advance :) Regards.


Let's consider two cases, $r > 0$ and $r < 0$.

Case 1

Suppose $r > 0$. Multiplying both sides of the inequality by $r$, $$-r \leq r^2-1 \leq r.$$

Rearranging the left-side inequality, $$r^2+r-1 \geq 0.$$

Using the quadratic equation to find the roots of $r^2+r-1$, $$r = \frac{-1 \pm \sqrt{1^2-4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{5}}{2} .$$

Thus, our inequality becomes, $$\left(r-\frac{-1 + \sqrt{5}}{2}\right)\left(r-\frac{-1 - \sqrt{5}}{2}\right) \geq 0.$$

From here, we get $r \in (-\infty, \frac{-1 - \sqrt{5}}{2}]$ or $r \in [\frac{-1 + \sqrt{5}}{2}, \infty).$ However, we assumed $r > 0$, thus we only have $r \in [\frac{-1 + \sqrt{5}}{2}, \infty).$

Now let's rearrange the right hand side of the inequality to get $$r^2-r-1 \leq 0.$$

The roots of $r^2-r-1$ are, $$\frac{-(-1) \pm \sqrt{(-1)^2-4(1)(-1)}}{2(1)} = \frac{1 \pm \sqrt{5}}{2}.$$

Thus, the right hand side inequality becomes, $$\left(r-\frac{1 + \sqrt{5}}{2}\right)\left(r-\frac{1 - \sqrt{5}}{2}\right) \leq 0.$$

From here, we get $r \in [\frac{1 - \sqrt{5}}{2}, \frac{1 + \sqrt{5}}{2}]$. But remember $r > 0$, so this means $r \in (0, \frac{1 + \sqrt{5}}{2}]$.

Case 2

Let's multiply both sides of the original inequality by $r$, but remember to flip the direction of the inequalities, $$-r \geq r^2 - 1 \geq r.$$

Rearranging the left and right side inequalities, $$r^2+r-1 \leq 0 \text{ and } r^2-r-1 \geq 0.$$

These inequalties are the same as before except with the direction of the inequalties switched around so I'm not going to go through the same level as detail as in case 1.

For the left-hand inequality, we end up with $r \in [\frac{-1 - \sqrt{5}}{2}, \frac{-1 + \sqrt{5}}{2}]$; however, since $r < 0$, this becomes $r \in [\frac{-1 - \sqrt{5}}{2}, 0)$. For the right-hand inequality, we end up with $r \in (-\infty, \frac{1 - \sqrt{5}}{2}]$ or $r \in [\frac{1 + \sqrt{5}}{2}, \infty).$ However; since $r < 0$, this becomes just $r \in (-\infty, \frac{1 - \sqrt{5}}{2}].$

Putting everything together

Looking at the result of everything, we have, $$\left(r \in [\frac{-1 + \sqrt{5}}{2}, \infty) \text{ AND } r \in [0, \frac{1 + \sqrt{5}}{2}] \right) \text{ OR } \left( r \in [\frac{-1 - \sqrt{5}}{2}, 0) \text{ AND } r \in (-\infty, \frac{1 - \sqrt{5}}{2}] \right).$$

More succintly, $$r \in \left([\frac{-1 + \sqrt{5}}{2}, \infty) \cap [0, \frac{1 + \sqrt{5}}{2}] \right) \cup \left([\frac{-1 - \sqrt{5}}{2}, 0) \cap (-\infty, \frac{1 - \sqrt{5}}{2}] \right)$$ $$= [\frac{-1 + \sqrt{5}}{2},\frac{1 + \sqrt{5}}{2}] \cup [\frac{-1 - \sqrt{5}}{2}, \frac{1 - \sqrt{5}}{2}]$$

  • $\begingroup$ Oh my God, thanks so so much! Now I can solve any inequality of this kind! $\endgroup$ – user361896 Aug 23 '16 at 11:04
  • $\begingroup$ No problem! :) This method is a bit long-winded (it took awhile to write up) but it gets the job done! $\endgroup$ – benguin Aug 23 '16 at 11:08
  • $\begingroup$ Yes, but I wonder why this solution doesn't match with that given by Josy Etxezarreta Martinez... $\endgroup$ – user361896 Aug 23 '16 at 11:09
  • $\begingroup$ I believe his answer is only addressing the part of the question where you have $r^2-2r-2 \leq 0$. His solution to THAT inequality is correct. However, that inequality doesn't follow from the original inequality...I believe you might have accidentally added instead of subtracting somewhere or vice versa when rearranging your inequalities. $\endgroup$ – benguin Aug 23 '16 at 11:13
  • $\begingroup$ Oh, OK, I'll check :) $\endgroup$ – user361896 Aug 23 '16 at 11:15

If you solve the equation with an equality, you get $1+\sqrt{3}$ and $1-\sqrt{3}$, so, and as 1 gives the output $1-2-2=-3$, then the region that you are looking for in the inequality is $[1-\sqrt{3},1+\sqrt{3}]$.

  • $\begingroup$ What are the steps to find out this solution region? $\endgroup$ – user361896 Aug 23 '16 at 10:52
  • $\begingroup$ First you solve the equation $r^2-2r-2 = 0$ to see when the function is crossing zero. Then as the inequality given is a parabola, you have to find when the parabola is negative and when it is positive, have doen that using the fact that as r=1 is between the solutions of the equation, if it is negative, then the region inside the solved values will be the region satisfying the inequality, and if it is bigger than zero, then the region outside the values would be the one you are looking for. $\endgroup$ – Josu Etxezarreta Martinez Aug 23 '16 at 10:57
  • $\begingroup$ What about other possible regions of solution? $\endgroup$ – user361896 Aug 23 '16 at 10:57
  • $\begingroup$ There are three regions: $(-\infty,1-\sqrt{3})$, where the function is positive, $[1-\sqrt{3},1+\sqrt{3}]$, where the function is negative or zero, and $(1+\sqrt{3},\infty)$ where the function is positive again. That's why I chose that region, because it is the only negative one. $\endgroup$ – Josu Etxezarreta Martinez Aug 23 '16 at 11:00
  • $\begingroup$ Oh, OK, thanks :) I'll keep this in mind. $\endgroup$ – user361896 Aug 23 '16 at 11:01

The simplest way to solve is to note first the domain of this inequality is $r\ne 0$, and to replace it with the equivalent: $$\Bigl(r-\frac1r\Bigr)^2<1\iff (r^2-1)^2<r^2$$ Now set $R=r^2$. We have to solve for $\;(R-1)^2<R,\enspace R>0$. So this comes down to finding the positive solutions of a quadratic inequation.

This quadratic inequation rewrites as $\;R^2-3R+1<0$. The roots of the quadratic polynomial are $\;\dfrac{3\pm\sqrt5}2=\biggl(\dfrac{1\pm\sqrt5}2\biggr)^{\!\!2}$. Both are positive, hence the required solutions are given by $$\dfrac{3-\sqrt5}2<R<\dfrac{3+\sqrt5}2\iff \begin{cases}\dfrac{1-\sqrt 5}2<r<\dfrac{1+\sqrt 5}2\\ \qquad\text{or}\\-\dfrac{1+\sqrt 5}2<r<\dfrac{-1+\sqrt 5}2\end{cases}$$


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