How to rigorously prove $n = o(k\log k)$ iff $k = \omega  \left(\frac{n}{\log n} \right)$ How can one prove that
$$ n = o(k\log k)$$  
if and only if 
$$k = \omega \left( \frac{n}{\log n} \right) .$$ 
where $k$ and $n$ are functions of the same variable.
Here $o$ represents the small-o notation and $\omega$ represents the small-omega notation. I do not know which way to attack this.
 A: The statement is not true.  Suppose e.g. $n = n(k) = 1 + 1/k^2$.  As $k \to \infty$ we certainly have $n = o(k \log k)$, but $n/\log n \sim k^2 > k$ so it's not true that $k = \omega(n/\log n)$.  You need to make assumptions such as that both $k$ and $n$ (as functions of the common variable) go to $+\infty$.
A: Make the additional assumptions that both $n$ and $k \to +\infty$ (as the common variable goes to some limit). Let $g(x) = x/\log x$.  Note that this is increasing for $x \ge e$
and $g(t \log t) = \frac{t \log t}{\log t + \log \log t} < t $ for $t > e$, and $g(t \log t) = t + o(t)$ as $t \to \infty$.   
1)  Suppose $n = o(k \log k)$.
Given $K > 0$, take $\delta = 1/(2 K)$.
For $k$ and $n$ sufficiently large we have $n < \delta k \log k$ so
$$ k > g(k \log k) > g(n/\delta) = \frac{n}{\delta \log(n/\delta)}
> \frac{K n}{\log n}$$ 
Therefore $k = \omega(n/\log n)$. 
2) Conversely, suppose $k = \omega(n/\log n)$.  Given $\epsilon > 0$, 
take $B = 2/\epsilon$.  For $k$ and $n$ sufficiently large we have
$k > 2 B g(n)$.  Now $g(k/B \log (k/B)) = k/B + o(k/B) > g(n)$ , so $n < k/B \log(k/B) < \epsilon k \log k$, for sufficiently large $k$ and $n$.  Thus $n = o(k \log k)$
