<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] First $x^2 \le 8$. Second, continuing from what you have written, i.e., 
$$\frac{-17}{\sqrt{8-x^2}+\sqrt{25-x^2}}\geq x,$$
we have
$$\frac{-17}{\sqrt{8-0^2}+\sqrt{25-0^2}} \ge \frac{-17}{\sqrt{8-x^2}+\sqrt{25-x^2}} \ge x \implies x \le \frac{-17}{\sqrt{8}+5} \implies x^2 > 8.$$
Contradiction! So no real solutions.
</s>