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

Square both sides, keeping always into account that we must have $x^2\ge5$:
$$
x^2-5+6\sqrt{x^2-5}+9>x^2-2x+1
$$
which becomes
$$
6\sqrt{x^2-5}>-2x-3
$$
This splits into two parts: if $-2x-3<0$, the inequality is satisfied; otherwise we can square again
$$
36(x^2-5)>4x^2+12x+9
$$
that becomes
$$
32x^2-12x-189>0
$$
The roots of the polynomial are
$$
\frac{6+78}{32}=\frac{21}{8},\qquad \frac{6-78}{32}=-\frac{9}{4}
$$
So we get the two cases:
$$
(1)\quad
\begin{cases}
x\le-\sqrt{5}\text{ or }x\ge\sqrt{5}\\
x>-3/2
\end{cases}
\qquad\qquad
(2)\quad
\begin{cases}
x\le-\sqrt{5}\text{ or }x\ge\sqrt{5}\\
x\le-3/2\\
x<-9/4\text{ or }x>21/8
\end{cases}
$$
and it's easy to deduce that the solutions are
$$
x<-\frac{9}{4}\text{ or }x\ge\sqrt{5}
$$
