If ordered sets $S=\{\frac{1}{n},\frac{2}{n},\frac{3}{n},\cdots,\frac{n-1}{n}\},T=\{\frac{1}{n+1},\frac{2}{n+1},\frac{3}{n+1},\cdots,\frac{n}{n+1}\}$ then $t_1<s_1<t_2<s_2<\cdots$
I cant even see how to apply induction for this, besides showing it is rue for some specific cases up 7 that lead me to believe it should be true in general.
No, this is not home work, it popped into my head, but if there is a text book example somewhere then I take it.
Also I am interested in knowing what type of inequality does this fall under, and want to see some examples.
What you want to show is that for all $n$ and for $k$ from $1$ to $n-1$ there is an inequality
$$\frac{k}{n+1} <\frac{k}{n} <\frac{k+1}{n+1}$$
The first inequality is not hard to see. Since $n < n+1$ we have $1/(n+1) < 1/n$, and multiplying by $k$ on both sides gives
$$\frac{k}{n+1} <\frac{k}{n}$$
To check the second inequality we can try to cross multiply. It amounts to showing that
$$ k(n+1) \leq n(k+1)$$
We can understand this by factoring things out and noting that $k < n$.
$$k(n+1) = kn + k < kn + n = n(k+1)$$