<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] Dinesh's answer is alright, but there is another approach. Your question can be rephrased as finding the range of $x$ such that $f(x)\geq 0$, where
$$f(x) = \sqrt{8-x^2}-\sqrt{25-x^2}-x.$$
We find the derivative of $f$ then set it to be $0$ to find its extreme values:
$$ f'(x) = -\frac{x}{\sqrt{8-x^2}}+\frac{x}{\sqrt{25-x^2}}-1=0 $$
Which implies that
$$ x\left(\sqrt{8-x^2}-\sqrt{25-x^2}\right)=\sqrt{(8-x^2)(25-x^2)}. $$
Solve this (with some effort) to get that the critical point of $f$, which happens to be a maximum, occurs at about $x=-2.37$ where the $y$ value is $-0.49<0$. Thus $f$ is always less than $0$ and there is no solution.
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