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.

Unfortunately the approach I did there is not useful here. However, here is another approach:
Write the inequality as
$$\sqrt{8-x^2}-x\geq\sqrt{25-x^2}. \tag{1}$$
Now use the Cauchy–Bunyakovsky–Schwarz inequality,
$$(x_1y_1+x_2y_2+\cdots+x_ny_n)^2\leq\left(x_1^2+x_2^2+\cdots+x_n^2\right)\left(y_1^2+y_2^2+\cdots+y_n^2\right),$$
to evaluate the left-side of $(1)$. Using this formula, one gets
$$\left(\sqrt{8-x^2}-x\right)^2\leq\left(1+1\right)\left(8-x^2+x^2\right)=16. $$
Notice we must have $-2\sqrt{2}\leq x \leq 2\sqrt{2}$ since, see Doug M's answer, $>$ and $<$ are not defined concepts over the complex numbers (sorry for that). Then, the right-side of $(1)$ takes
$$\sqrt{25-x^2}\geq \sqrt{25-8}=\sqrt{17}.$$
Futher, notice also that the left-side of $(1)$ is less than or equal to $4$, and that the right-side is greater than or equal to $4$. Therefore, there's no real solution.
