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
 A: First let us check where the above formula is defined: 
$ x^2-3x-4 \geq 0  \; \; \Leftrightarrow  \; \; x\in ] -\infty , -1] \cup [4,+\infty[$.
$x^2-4 \geq  0  \; \; \Leftrightarrow  \; \; x\in ] -\infty , -2] \cup [2,+\infty[ $.
$ \sqrt{21 } - \sqrt{ x^2-4} > 0  \; \; \Leftrightarrow  \; \;  x\in   ]-5,-2[ \cup ]2,5[.$ 
Thus  You set of solution must  be subset of  $]-5, -2] \cup [4,5[ $. Note that whenever $x  \in   ]-5, -2] \cup [4,5[ $ the inequality holds true as  $x-5 < 0 $ and the other side  of the inequality is surely positive. 
A: 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 $x=5$?
A: Well, this is a very messy problem.
We start off with:
$\sqrt[4]{\frac{\sqrt{x^{2}-3x-4}}{\sqrt{21}-\sqrt{x^{2}-4}}}\geqslant x-5$
Now we can raise each side to the power $4$. 
${\frac{\sqrt{x^{2}-3x-4}}{\sqrt{21}-\sqrt{x^{2}-4}}}\geqslant (x-5)^4$
We can then multiply the denominator to the right side:
$\sqrt{x^{2}-3x-4}\geqslant(x-5)^4 * (\sqrt{21}-\sqrt{x^{2}-4})$
From here, again we can square both sides, receiving:
${x^{2}-3x-4}\geqslant(x-5)^8 * (\sqrt{21}-\sqrt{x^{2}-4})²$
Afterwards, you can expand these.
