<s>[INST] Use Triangle Inequality to Solve Inequality $\left|x+\frac{1}{2}\right| > \frac{\sqrt{5}}{2}$ 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.
 [/INST] 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. 
</s>