I was recently introduced to this topic and I'm trying to prove Tue following statement.
For most of numbers, m^n => n^m
So I derived this into something that could be proved by induction... The other problem is I realized I have to work with m and n, and in the past I only worked with a single n for most of the proofs.
Can someone guide me?
By previously known facts m > 1 so in other words I have to prove that:
For all m, there exists a k such that for all n, if m >1 ^ n>= k) then m^n => n^m
Here's what you want to prove (I think):
For all $m$, there exists a $k =k(m)$ such that if $n>= k$ then $m^n \ge n^m$.
This is equivalent to $m^{1/m} \ge n^{1/n}$. But $x^{1/x}$ is decreasing for $x > e$. (There are lots of proofs of that on this site.)
Therefore, if $3 \le m < n$, $n^{1/n} < m^{1/m}$.
Therefore, if $m \ge 3$, chosing $k(m) = m+1$ works.
If $m=2$, $n > 4$ works, since $n^{1/n} < 4^{1/4} = 2^{1/2}$.