Solve $\sqrt{n^3+1}+\sqrt{n^3}>10^3$ I was given this task:
"Find $n_0$ so that for all $n > n_0$:
$$\frac{1}{\sqrt{n^3+1}+\sqrt{n^3}}<\ 10^{-3}$$
This should be equal to:
$$\sqrt{n^3+1}+\sqrt{n^3}>10^3$$
But this is where I am already stuck. Wolfram Alpha gave me this solution for n:
$$n\ >\ \frac{9}{100}\left(\frac{37037}{2}\right)^{\frac{2}{3}}$$
And this is equal to $n_0$ = 62
But how did wolfram Alpha get that solution? 
 A: $$\sqrt{n^3+1}+\sqrt{n^3}\gt 10^3$$
is equivalent to
$$\sqrt{n^3+1}\gt 10^3-\sqrt{n^3}$$
For $n$ such that $10^3-\sqrt{n^3}\ge 0$, i.e. $n\le 10^2$, the both sides are non-negative, so squaring the both sides gives
$$n^3+1\gt (10^3-\sqrt{n^3})^2,$$
i.e.
$$\sqrt{n^3}\gt \frac{10^6-1}{2\times 10^3}$$
So, we get
$$n\gt \left(\frac{10^6-1}{2\times 10^3}\right)^{2/3}=\frac{9}{100}\left(\frac{37037}{2}\right)^{2/3}$$
A: So we first notice that $n>0$ for the relation to be defined. Also $f(x)=x^3+1$ and $g(x)=x^3$ are positive and strictly increasing in this range. This means that as long as we can find some $k$ that satisfy the relation
$$\sqrt{k^3+1}+\sqrt{k^3}=10^3$$
then any $n>k$ will satisfy your inequality.
After this, you should be able to solve for $k$ easily.
Hint: Move one of the term to the right so that it is easier to square through and solve for k easily.
A: For big enough $n^3$ we have that $\sqrt{n^3+1}\approx \sqrt{n^3}$ so it's enough to consider $2\sqrt{n^3}>10^3 \implies n>500^{2/3}$ the difference is pretty small with the actual answer $$\frac9{100}(\frac{37037}2)^{2/3}-500^{2/3}\approx-0.00004$$
Since $500^{2/3}$ is close to but smaller then $63$ you get $n_0=62$
