Find $n \in \mathbb{Z}$ that maximizes $ (n+10)/(n^3+10)$ $(n+10)/(n^3+10)=(n+10)/(n^3+1000-990)$
I have thought about something like this. But am not sure about how to proceed with this. So I would like some help.

You have that for $n=1$ that $11/11=1$ so you want to find $$\frac{n+10}{n^3+10}> 1$$
Otherwise the maximum value is $1$.
If $n^3+10>0$ we can multiply both sides of inequality to get
$$n+10> n^3+10\\n> n^3\\0> n(n-1)(n+1)$$
This happens when $n< -1$ or $0< n< 1$ The only integer satisfying $n^3+10>0$ and $n< -1$ is $-2$ and there are no integers satisfying $0<n<1$. For $n=-2$ we get $8/2=4$ so it's greater.
Now if $n^3+10<0$ then when we multiply we must also flip signs so
$$n+10<n^3+10\\n(n-1)(n+1)>0 $$
This happens when $-1<n<0$ or $n>1$ none of those satisfy $n^3+10<0$ hence for $n=-2$ is the maximum.
