Let $(n,m)\in(\mathbb{N}\setminus\{0\})^2$ and let $a_{n,m}=\frac{(n+m)^2}{2^{nm}}$. Evaluate $\sup_{(n,m)\in(\mathbb{N}\setminus\{0\})^2}\frac{(n+m)^2}{2^{nm}}$.
I was thinking about an approach with monotonicity for one index of the sequence, as the one that follows: let $m\in\mathbb{N}\setminus\{0,1\}$ be fixed. We have: $$[a_{n+1,m} \le a_{n,m}] \iff \left[\frac{(n+1+m)^2}{2^{(n+1)m}} \le \frac{(n+m)^2}{2^{nm}}\right] \iff \left[\left(1+\frac{1}{n+m}\right)^2 \le 2^m\right]$$ Since $n\in\mathbb{N}\setminus\{0\}$ and $m\in\mathbb{N}\setminus\{0,1\}$, $n+m \ge 3$ and so $\left(1+\frac{1}{n+m}\right)^2 \le 16/9$. Hence, being $2^m \ge 4>16/9$ for each $m \in \mathbb{N}\setminus\{0,1\}$, the latter inequality in the chain of logical equivalences is true.
Thus, by monotonicity we have that $a_{n,m} \le a_{1,m}$ for each $n\in\mathbb{N}\setminus\{0\}$ and for each $m\in\mathbb{N}\setminus \{0,1\}$, and so $a_{n,m} \le \sup_{(n,m)\in(\mathbb{N}\setminus\{0\})^2}\{a_{n,1},a_{1,m}\}$ for each $(n,m)\in(\mathbb{N}\setminus\{0\})^2$. Being $a_{n,m}$ symmetric by exchanging $n$ with $m$, the sequences $a_{1,m}$ and $a_{n,1}$ are the same and so we have $\sup_{(n,m)\in(\mathbb{N}\setminus\{0\})^2}\{a_{n,1},a_{1,m}\}=\sup_{m\in\mathbb{N}\setminus\{0\}} a_{1,m}$.
I now claim that $\sup_{(n,m)\in(\mathbb{N}\setminus\{0\})^2} a_{n,m}=\sup_{m\in\mathbb{N}\setminus\{0\}} a_{1,m}$. From above, we have that $a_{n,m} \le \sup_{m\in\mathbb{N}\setminus\{0\}} a_{1,m}$ and so $\sup_{m\in\mathbb{N}\setminus\{0\}} a_{1,m}$ is an upper bound for $(a_{n,m})_{(n,m)\in(\mathbb{N}\setminus\{0\})^2}$. By the definition of supremum $\sup_{m\in\mathbb{N}\setminus\{0\}} a_{1,m}$, given an arbitrary $\epsilon>0$ there exists $M_{\epsilon} \in \mathbb{N}\setminus\{0\}$ such that $\sup_{m\in\mathbb{N}\setminus\{0\}} a_{1,m} -\epsilon<a_{1,M_{\epsilon}}$. Being $(1,M_{\epsilon})\in(\mathbb{N}\setminus\{0\})^2$, this also holds for $a_{n,m}$ too and so $\sup_{(n,m)\in(\mathbb{N}\setminus\{0\})^2} a_{n,m}=\sup_{m\in\mathbb{N}\setminus\{0\}} a_{1,m}$. Being: $$\sup_{m\in\mathbb{N}\setminus\{0\}}a_{1,m}=\sup_{m\in\mathbb{N}\setminus\{0\}} \frac{(m+1)^2}{2^m}=\frac{9}{4}$$ It follows that $\sup_{(n,m)\in(\mathbb{N}\setminus\{0\})^2}\frac{(n+m)^2}{2^{nm}}=9/4$.
Is this correct? In particular, I am not sure about the use of monotonicity and the equality between the supremum of the double index sequence and the supremum of $a_{1,m}$. Any simpler approach is welcome as an answer.