find the solution set of the following inequality with so many radical I get lost
$\sqrt[4]{\frac{\sqrt{x^{2}-3x-4}}{\sqrt{21}-\sqrt{x^{2}-4}}}\geqslant x-5$
Edit
I get online with wolfram
 -5 < x <= -2 || x == -1 || 4 <= x < 5

Clearly $\sqrt[4]{\dfrac{\sqrt{x^{2}-3x-4}}{\sqrt{21}-\sqrt{x^{2}-4}}}\ge0$
So, one immediate solution is $x-5<0\iff x<5$
Otherwise i.e., if  $x\ge5$
$$\dfrac{\sqrt{x^2-3x-4}}{\sqrt{21}-\sqrt{x^2-4}}=\dfrac{\sqrt{(x-4)(x+1)(\sqrt{21}+\sqrt{x^2-4})}}{25-x^2}$$ which will be $<0$ if $x>5$
and what if $x=5$?
