# Range of values which satisfy this inequality

Consider the following inequality:

$$|f(a)| = \left|\frac{1}{2}(a \pm \sqrt{a^{2}-2})\right| \leq 1$$

I got this inequality while doing stability analysis of a fixed point of a certain discrete dynamical system. We can easily see that $f(a)$ is complex valued in $(-\sqrt{2}, \sqrt{2})$. But this is not the case outside this range. Because of this, I think that I cannot write:

$$-1 \leq f(a) \leq 1$$

Then how can I find the range of values of $a$ which satisfy this inequality?

Treat $a\ge \sqrt{2}$ and $a\le -\sqrt{2}$ separately. The analyses are much the same.
For $a\ge \sqrt{2}$, since $0\le a-\sqrt{a^2-2}\le a+\sqrt{a^2-2}$, we want $a+\sqrt{a^2-2}\le 2$. Equivalently, we want $a^2-2\le (2-a)^2$, since clearly $a\lt 2$.
Manipulation gives $a\le 3/2$. We obtain the inequality $\sqrt{2}\le a\le 3/2$.
• The inequality $a+\sqrt{a^2-2}\le 2$ is equivalent to $\sqrt{a^2-2}\le 2-a$. Since both sides are non-negative, the inequality is equivalent to $\sqrt{a^2-2}\le (2-a)^2$. As to $a\le -\sqrt{2}$, let $b=-a$. We want the absolute value of $(-b-\sqrt{b^2-2})/2$ to be $\le 1$, so we want $b+\sqrt{b^2-2}\le 2$, and we are back to the positive case. The appropriate values of $b$ are $\sqrt{2}$ to $3/2$, so the appropriate values of $a$ are $-3/2$ to $-\sqrt{2}$. – André Nicolas Feb 29 '16 at 17:59
• If $a+\sqrt{a^2-2}\le 2$, then in particular we have $a\lt 2$. That is why $2-a$ is positive. – André Nicolas Feb 29 '16 at 18:09