I'm trying to solve the following problem:
Find $\sup$ and $\inf$ of $A=\{\frac{nk}{1+2n+3k} : n,k \in \Bbb{N}\}$ and maximal and minimal element of this set.
As for $\sup(A)$ and $\max(A)$ I tried to prove they don't exist - here is how I did it. If you take $k=n$ you get $\frac{n^2}{1+5n}$ Now when we take any $C>0$, we can show that for sufficiently large n
$\frac{n^2}{1+5n}>C$, because $n^2>C+5nC$ and we can find n such that $n^2-5nC-C>0$ because on the left we have a quadratic function and for some n bigger than one of its zeroes we always have a number bigger than 0 so the inequality holds.
When it comes to $\inf(A)$ and $\min(A)$ I decided that $\frac{1}{6}$ might be a good candidate for them. As we take $k=1$ and $n=1$ we get that value so when I prove that it is a lower bound of this set, I've proven that it's $\inf(A)$ as well as $\min(A)$, right?
So we're trying to show that $\frac{1}{6} <= \frac{nk}{1+2n+3k}$ which is the same as showing that $1+2n+3k<=6nk$. I tried to show that using induction. Let's fix $k$. For $n=1$ we get $6<=6k$ and as $k$ is natural (in my definition $0\notin\Bbb{N}$), the base case holds.
So now let's use the inductive hypothesis to prove that $1+2(n+1)+3k<=6(n+1)k$.
After some changes $2+2n+3k<=6nk+6k$, so $2+2n-3k<=6nk$. And as $2+2n-3k<1+2n+3k$ ($k$ is at least $1$ so I think we may say this statemtnt is true) using inductive hypothesis we get $2+2n-3k<1+2n+3k<=6nk$, which ends the proof.
Ok, this is my first try at trying to solve and prove such a complicated statement so I ask you the following questions:
Is eveything right with my proof?
Could this problem be solved faster?
Your argument for $\sup A$ is valid - although it's worth noting that this hinges on the fact that for a polynomial $p(x)$, $\lim_{x\to\infty} p(x) = \infty$ if and only if the leading coefficient is positive (which in this case it is).
Further, your reasoning that if $1/6$ is an element of the set and if it is a lower bound for the set, then it is the greatest lower bound of the set is also correct. However, there is a simple solution here: to prove that $1 + 2n + 3k \leqslant 6nk$ for $n,k\geqslant1$, observe that $1\leqslant nk$, $n\leqslant nk$, and $k\leqslant nk$. Therefore $1+2n+3k\leqslant nk + 2nk + 3nk = 6nk$.