Find $\sup$ and $\inf$ of $\left\{\log_2\frac{10mn}{(2m+5n)^2}:m,n\in\mathbb N\right\}$

Question

Given $$S:=\left\{\log_2\frac{10mn}{(2m+5n)^2}:m,n\in\mathbb N\right\}$$ If exist, find $\sup S,\inf S$

My attempt:

$$\frac{10mn}{(2m+5n)^2}=\frac{1}{\frac{2m}{5n}+2+\frac{5n}{2m}}=\left[t=\frac{2m}{5n}\implies\frac{1}{t}=\frac{5n}{2m}\right]=\frac{1}{t+2+\frac{1}{t}}$$ $$t+2+\frac{1}{t}=0\Big/\cdot t\iff t^2+2t+1=(t+1)^2=0\implies t=-1\ne\frac{2m}{2n}\Leftarrow m,n\in\mathbb N$$ $$f(t):=(t+1)^2\implies\min f(t)<\inf\{f(t):t>0\}=1$$$$\nexists\max f(t)$$ $$\lim_{t\to 0^+}\log_2\frac{1}{t+2+\frac{1}{t}}=0$$ $$\lim_{t\to+\infty}\log_2\frac{1}{t+2+\frac{1}{t}}=-\infty$$ $$\implies \sup S=0\;\land\;\nexists\inf S$$

Is this correct and are there any other efficient methods?

Answer 1

$\frac{10mn}{(2m+5n)^2}\leq 1$, So $ \log_2 \frac{10mn}{(2m+5n)^2}\leq \log_2 (1)=0 $. Therefore $\sup S\leq 0$

But $\sup (S)\neq 0$. In fact $\sup (S)=\log_2 \sup_{m,n\in \mathbb{N}} \frac{10mn}{(2m+5n)^2} .$

If $n=km\ (k\in \mathbb{N})$ then $\frac{10mn}{(2m+5n)^2}=\frac{10km^2}{(5k+2)^2m^2}= \frac{10k}{(5k+2)^2} $. If $k\to \infty $ then $\log_2 \frac{10mn}{(2m+5n)^2} \to -\infty$. So $\inf (S)=-\infty $

Answer 2

Using your substitution $t:=\frac{2m}{5n}$ we automatically have $t>0$ as you also noted. The expression then reads $$\log_2\frac{10mn}{(2m+5n)^2} = \log_2 \frac1{t+2+1/t}.$$ Since we know how $\log_2 x$ and $1/x$ behave, it is sufficient to look at the denominator. Suppose, for a while that $t>0$ is a real variable (not just a fraction). Then we can use basic theorems of calculus to find the extremes of the function $d(t) := t +2 + 1/t$. Using $d'(t) = 1 - t^{-2}$ and $d''(t) = 2 t^{-3} > 0$ we get that the only extreme of $d(t)$ is at $t=1$, equals $d(1) = 4$ and it's minimum. We can easily realize $t=1$ as eg. $\frac{2\cdot 5}{5 \cdot 2}$. Because $1/x$ is decreasing, we have that $$\max \left\{ \frac{10mn}{(2m+5n)^2}: m,n \in \mathbb N \right\} = \frac14$$ and since $1/4 = 2^{-2}$ and $\log_2 x$ is an increasing function, we finally get that the maximum, ie. automatically also supremum of the examined set is $-2$.

The infimum is $-\infty$ as shown by Jo Jomax.