$ A: \{\frac{m^2n^3 + 3m^2n^2+2mn^3+6mn^2}{(m^2 + 2m)^2+(n^3 + 3n^2)^2}, n, m \in \mathbb{N} \}$ Got limitation need superemum. I need to find supremum of:
$$ A: \{\frac{m^2n^3 + 3m^2n^2+2mn^3+6mn^2}{(m^2 + 2m)^2+(n^3 + 3n^2)^2}, n, m \in \mathbb{N} \}$$
I found out that:
$$ \frac{m^2n^3 + 3m^2n^2+2mn^3+6mn^2}{(m^2 + 2m)^2+(n^3 + 3n^2)^2} = \frac{m^2(n^3 + 3n^2) + 2m(n^3+3n^2)}{(m^2 + 2m)^2+(n^3 + 3n^2)^2}= \frac{(m^2+2m)(n^3 + 3n^2)}{(m^2 + 2m)^2+(n^3 + 3n^2)^2}$$
$$\frac{(m^2+2m)(n^3 + 3n^2)}{(m^2 + 2m)^2+(n^3 + 3n^2)^2}= \frac{\sqrt{(m^2+2m)^2(n^3 + 3n^2)^2}}{(m^2 + 2m)^2+(n^3 + 3n^2)^2}$$
And from inequality of means (AM - GM) I know that:
$$\frac{\sqrt{(m^2+2m)^2(n^3 + 3n^2)^2}}{(m^2 + 2m)^2+(n^3 + 3n^2)^2} \leq \frac{1}{2}$$
Therefore I found a limitation of my set A. Now I need a sequence of elements belonging in $A$ that would be convergent to $\frac{1}{2}$. That is the part I can't do myself.
From graph I know that if I take $m = n$ then for $n \in [1 ; \infty)$ I get decreasing function and for that function $\frac{1}{2}$ is reached by $n < 1$. A maximal value of that function for $n \in  [1 ; \infty)$ is equal to $\frac{12}{25}$ and is reached by $n = 1$. But that is not a solution.
There is no ANY fact that would legitimise value $\frac{12}{25}$ as the supremum of that set since there always might by something bigger if $n \neq m$.
What I need is a prove that in the supremum of set $A$: $m = n$ (I don't know why that would be the case) or that there is a bigger value reached for $n$ and $m$ that are not equal.
It might be important to add that I can not use integrals nor derivatives.

We pick a sub-sequence of $A$ in the following fashion: $n=1, 2, \cdots, $, and $m$ depends on $n$ via $m=\lfloor \sqrt{a_n+1} \rfloor$.
Define $a_n = n^3+3n^2, b_m=m^2+2m=(m+1)^2-1=\lfloor\sqrt{a_n+1}+1 \rfloor^2-1$. Then $\frac{a_n b_m}{a_n^2+b_m^2} \in A$ and we will prove that $\lim_{n\to \infty} \frac{a_n b_m}{a_n^2+b_m^2} \to \frac 12.$
Note that $x-1 < \lfloor x \rfloor \le x, \forall x$, then
$$a_n=(\sqrt{a_n+1})^2-1 < \lfloor \sqrt{a_n+1}+1\rfloor^2-1 \le (\sqrt{a_n+1}+1)^2-1 \\
\implies 1 < \frac{b_m}{a_n} \le \frac{(\sqrt{a_n+1}+1)^2-1}{a_n} = 1 + \frac{1}{a_n}+\frac{2\sqrt{a_n+1}}{a_n}$$
Therefore $\lim_{n\to\infty} \frac{b_n}{a_n} = 1 \implies \lim_{n\to\infty} \frac{a_n b_n}{a_n^2+b_n^2} = \frac 12. \blacksquare$
