Use the triangle inequality and the reverse triangle inequality to find an upper bound for the set of all numbers of the form $\left\lvert\frac{x^2-3}{x-2}\right\rvert$ as $x$ ranges over the interval defined by $\lvert x-1\rvert < \frac 23$.

I know you need to use the reverse inequality as well but I'm not too sure what to do

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    $\begingroup$ Finding an upper bound (and proving that it is) is not difficult. Getting one triangle inequality into the game, let alone two, seems not so easy. $\endgroup$ Sep 17 '15 at 5:04
  • $\begingroup$ @KevinSheng Isn't the upper limit for $\,x\,$ equal to $\,5/3\,$? $\, \big\lvert\, x-1\,\big\rvert < \dfrac{2}{3} \implies -\dfrac{2}{3} < x-1 < \dfrac{2}{3} \implies \boxed{\;\dfrac{1}{3} < x < \dfrac{5}{3}\;} \,$ $\endgroup$
    – Vlad
    Sep 17 '15 at 5:09
  • $\begingroup$ @Vlad, yes you are correct; I made a typo. It has been removed, thanks! $\endgroup$ Sep 17 '15 at 5:11

We will use the triangle inequality on the numerator, and the reverse triangle inequality on the denominator.

Note that $|x^2 - 3| \le x^2 + 3$, we know that $|x-1| < \frac{2}{3} $, so that $\frac{1}{3} < x < \frac{5}{3}$ . Therefore it is now an easy task to find an upper bound for $ x^2 + 3 $

Lets use the reverse triangle inequality now.

Note that the reverse triangle inequality says that $||a| - |b|| \le |a-b|$ for all real a and b.

We will apply this to the denominator as I said, so that $||x|- |2|| \le |x-2| $,

this is actually a very vacuous application, because $x$ is always positive, for $|x-1| < \frac{2}{3} $ so that $|x| = x$ and $|2| = 2$ and the inequality we have produced is $|x- 2| \le |x-2| $

Regardless, we will now find a lower bound for $|x-2|$ .

We note that $\frac{1}{3} < x < \frac{5}{3}$ , so that $\frac{-7}{3} < x-2 < \frac{-1}{3}$ .

This is sufficient to show that $|x-2| > \frac{1}{3} $

Now since $|x-2| > \frac{1}{3} , |x-2| > 0 , \frac{1}{3} > 0$ we know that

$\frac{1}{|x-2|} < 3$

So we have bounded both $\frac{1}{|x-2|}$ and $|x^2 + 3|$ above using the triangle inequality and sorta using the reverse triangle inequality.

Could you perhaps elaborate on why you insist on using the reverse triangle inequality in this question?


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