If exist, find $\sup S,\inf S$ $$S:=\left\{\left(m+n\right)^{\frac{1}{mn}}:\;m,n\in\Bbb N\right\}$$ $$m=1,n=1:\;\left(m+n\right)^{\frac{1}{mn}}=2$$
For a fixed $m\in\Bbb N,n\to\infty\quad$:
Error pointed out by @FimPellizieri and @Peterforeman in the comments corrected: $$(m+n)^{\frac1{mn}}=\left(n\left(1+\frac{m}n\right)\right)^{\frac1{mn}}=\left(n^{\frac1n}\right)^{\frac1m}\cdot\left(\left(1+\frac{m}n\right)^{\frac{n}m}\right)^{\frac1{n^2}}=1$$
Original question (the accepted answer is still correct):
Is this correct?
We have that
$$\frac \partial{\partial m} {\left(m+n\right)}^{1/mn} = {\left(m+n\right)}^{1/mn}\left(\frac1{mn(m+n)}-\frac{\log(m+n)}{m^2n}\right)$$
We see that $\frac \partial{\partial m} {\left(m+n\right)}^{1/mn} < 0$ if and only if
$$\begin{align} &\,\,\,\frac1{mn(m+n)}-\frac{\log(m+n)}{m^2n} < 0 \\\iff&m<(m+n)\log(m+n) \\\iff&e^m<(m+n)^{m+n} \end{align}$$
If $m=n=1$, already this is true, and increasing $n$ or $m$ only makes it more so. Therefore, $\frac \partial{\partial m} {\left(m+n\right)}^{1/mn} < 0$ for all $m,n\geqslant 1$.
It follows that $\sup S$ is attained when $m=n=1$ and $\inf S$ is attained as $m,n\to\infty$, so $\sup S = 2$ and $\inf S = 1$.