So this is a really simple question, but I can't seem to work it out. I started with the equation

$$ x^2 + x + 1 > 2 $$

and, by completing the square and taking the square root, was able to simplify it to

$$ \left|x+\frac{1}{2}\right| > \frac{\sqrt{5}}{2} \tag{1} $$

Next, I want to use the triangle inequality to solve for $x$. I noted that

$$ |a + b| \le |a| + |b| \iff |a| + |b| \ge |a + b| $$

Now, letting $a = x$ and $b = \frac{1}{2}$, I got

$$ |x| + \frac{1}{2} \ge \left|x + \frac{1}{2}\right| $$

Thus, combining this with (1), I reached

$$ |x| + \frac{1}{2} > \frac{\sqrt{5}}{2}, \tag{2} $$ which is easily solvable:

$$ x < \frac{1-\sqrt{5}}{2} \text{ or } x> \frac{\sqrt{5} - 1}{2} $$

By plugging each step into Wolfram Alpha, I've determined that while (1) is correct, (2) produces a different solution, which means the error must be somewhere in my use of the triangle inequality. The solution that both Wolfram Alpha and my textbook give is

$$ x < \frac{-1-\sqrt{5}}{2} \text{ or } x> \frac{\sqrt{5} - 1}{2}, $$ the difference being the negative in front of the $1$ in the first term.

  • $\begingroup$ If you have $x\ge -\frac{1}{2}$, then the inequality becomes $x+\frac{1}{2}>\frac{\sqrt{5}}{2}$, otherwise it becomes $-x-\frac{1}{2}>\frac{\sqrt{5}}{2}$ $\endgroup$ – Peter Mar 29 '18 at 17:30
  • $\begingroup$ Without the triangle inequality, $|a| > b$ iff $a > b$ or $a < -b$. Hence, $x+1/2 > \sqrt{5}/2$ or $x +1/2 < -\sqrt{5}/2$... $\endgroup$ – gt6989b Mar 29 '18 at 17:31
  • $\begingroup$ @gt6989b Hmm, I see this, and it makes sense (and gives the right answer), but how do I reconcile it with my work? I'd like to see where my error is so I don't make a similar mistake again. $\endgroup$ – Calico Mar 29 '18 at 17:34

You shouldn't use the triangle inequality here. If $|x+\frac{1}{2}|>\frac{\sqrt{5}}{2}$, then you are really considering two inequalities: $$ x+\frac{1}{2}>\frac{\sqrt{5}}{2} \text{ when } x+1/2\geq 0 $$ $$ -\left(x+\frac{1}{2}\right)>\frac{\sqrt{5}}{2} \text{ when } x+1/2< 0 $$ These two inequalities then read $x>\frac{\sqrt{5}-1}{2}$ and $x<\frac{-\sqrt{5}-1}{2}$, as the other two inequalities become redundant.

  • $\begingroup$ Yes, this makes sense, but what went wrong in my work? My work produced an incorrect solution, so that means there must be an error in my logic. I would like to avoid making such an error again. $\endgroup$ – Calico Mar 29 '18 at 17:38
  • $\begingroup$ @Calico You are using the triangle inequality, which introduces a new term which we never wanted to consider. Your derived conditions are true if we want $|x|+1/2>\sqrt{5}/2$, but nowhere in the problem did we actually care about that inequality. $\endgroup$ – user546996 Mar 29 '18 at 17:41
  • $\begingroup$ My confusion is that I thought the triangle inequality implied $|x| + 1/2 > \sqrt{5}/2$ since it says that $|x| + 1/2 > |x + 1/2| > \sqrt{5}/2$, and thus, since $a>b>c$ implies $a>c$, $|x| + 1/2 > \sqrt{5}/2$? $\endgroup$ – Calico Mar 29 '18 at 17:50
  • $\begingroup$ @Calico The triangle inequality will imply that $|x|+1/2| \geq |x+1/2|$, but to get the statement that $|x|+1/2| \geq |x+1/2|>\sqrt{5}/2$ you still have to determine which $x$ make that second inequality true. $\endgroup$ – user546996 Mar 29 '18 at 17:53
  • $\begingroup$ @Calico You seem to be forgetting that you do not know that $|x+1/2|>\sqrt{5}/2$, and that you have to prove it to be true. $\endgroup$ – user546996 Mar 29 '18 at 17:54

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