<s>[INST] Solving $\sqrt{8-x^2}-\sqrt{25-x^2}\geq x$ 
I would like to find the solution of
$$\sqrt{8-x^2}-\sqrt{25-x^2}\geq x.$$

My try:
First I used the hint of this answer.
$$ \frac{8-x^2-25+x^2}{\sqrt{8-x^2}+\sqrt{25-x^2}}\geq x \leftrightarrow \frac{-17}{\sqrt{8-x^2}+\sqrt{25-x^2}}\geq x.$$
Then the solution can be found by
$$\left(-17\right)^2\geq \left(x\sqrt{8-x^2}+x\sqrt{25-x^2}\right)^2.$$
But I think this is not the best approach.
 [/INST] Since $>,<$ are not defined concepts over the complex numbers, we can assume that $\sqrt{8-x^2}, \sqrt{25-x^2}$ are real
And $\sqrt{25-x^2} > \sqrt{8-x^2}$
Which implies $x\in [-2\sqrt 2,0]$
but for all $x$ in this interval, $\sqrt{8-x^2} - \sqrt{25-x^2} > x$
there is no solution.
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