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.
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.