<s>[INST] Solution to $\sqrt{x^2-5}+3>|x-1|$ I tried many ways to solve this but I just can't figure it out...
$$\sqrt{x^2-5}+3>|x-1|$$
 [/INST] first the domain of the inequality is $(-\infty, -\sqrt 5] \cup [\sqrt 5, \infty).$
so we will deal with the absolute value sign by breaking the function into two pieces:
on $x \le -\sqrt 5$ we need to solve $\sqrt{x^2 - 5} + 3 > 1 - x$  and on $x ge \sqrt 5$ solve 
$\sqrt{x^2 - 5} + 3 > x - 1$ 
first for $x \le -\sqrt 5:$
$ \sqrt{x^2 - 5} > -x -2 $  implies $x^2 - 5 > x^2 + 4x + 4$ which says that $x < -9/4.$
now for $x \ge \sqrt 5:$
$ \sqrt{x^2 - 5} > x -4 $  implies $x^2 - 5 > x^2 -8x + 16$ which says that $x > 21/8$ but for $x = 21/8$ the right hand side $x - 4 < 0.$ that is $x = 21/8$ solves 
$\sqrt{x^2 - 5} > 4 - x$ instead.
combining the two the solution is $$-\infty < x < -\dfrac{9}{4} \text{ or } \sqrt 5 < x < \infty$$
p.s. every day you learn a new thing.
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