Solution to $\sqrt{x^2-5}+3>|x-1|$ I tried many ways to solve this but I just can't figure it out...
$$\sqrt{x^2-5}+3>|x-1|$$

You can eliminate the square roots or absolute values provided you know the signs of the expressions under the square roots or between the absolute value signs.
What you can use : if $a,b\ge 0$, $a>b\iff a^2 >b^2$ ; if $a,b\le 0$, $a>b\iff a^2 <b^2$ ; if $a> 0 >b$, well… $a>b$.
Here it's fairly simple: first the square root is defined if and only if $x\ge \sqrt 5$ or $x\le-\sqrt 5$. 
In the first case, $x >1$ so the inequation amounts to $\sqrt{x^2-5} > x- 4$.
Two subcases: 


*

*either $x<4$, and the LHS ($\ge 0$) is greater than the RHS ($<0$).

*or $x\ge 4$, and $\sqrt{x^2-5} > x- 4\iff x^2-5 > (x-4) ^2 \iff 8x >21$,, which is indeed true if $x\ge4$.


In the second case, the inequation amounts to  $\sqrt{x^2-5} > -(x+2)$. Since $x<-\sqrt 5$, we know $-(x+2)>0$, 
so that  $\sqrt{x^2-5} > -(x+2)\iff 4x <-9 $, so that $ x < -\dfrac94\bigl(<-\sqrt 5\bigr)$.
As a conclusion we have the following solutions:
$$ x< -\dfrac94\quad \text{or}\quad x\ge \sqrt 5.$$
