Very challenging: max{floor,ceil}=? I spotted a pattern while trying to generalize a problem. (EDIT: said problem has been removed from this post to avoid confusion. EDIT(2): Here is the problem again: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=367432&sid=356f2a2b519ab007bb648bf00e84934d#p367432 ) <--- That isnt the problem I want to solve, it is just where I got the pattern below from. I know how to solve the problem in the link already; I am only here trying to solve the problem below:
Here is the pattern I found, which I cannot prove. If you prove this, you win the bounty, and if there are multiple proofs, the most detailed (including extra information, etc) wins:
If $$\{k:k\in \mathbb{Z}^+,k\ne 4, k\ge 3\}$$,
$$\max\left\{\left(\frac{k}{\lfloor \frac{k}{\text{e}}\rfloor }\right)^{\lfloor \frac{k}{{\text{e}}}\rfloor},\left(\frac{k}{\lceil \frac{k}{\text{e}}\rceil }\right)^{\lceil \frac{k}{{\text{e}}}\rceil}\right\}=\begin{cases}\left(\frac{k}{\lfloor \frac{k}{\text{e}}\rfloor }\right)^{\lfloor \frac{k}{{\text{e}}}\rfloor}\quad\text{if}\ \{\frac{k}{\text{e}}\}<0.5\\ \\\left(\frac{k}{\lceil \frac{k}{\text{e}}\rceil }\right)^{\lceil \frac{k}{{\text{e}}}\rceil}\quad\text{if}\ \{\frac{k}{\text{e}}\}\ge 0.5\end{cases}$$.
I honestly don't know where to begin.
EDIT: Here $\{x\}$ denotes the fractional part of $x$, e.g. $\{5.4\}=0.4$.
 A: Not an answer, just an observation - hence CW.
The plot of 
$\text{max}
\begin{cases}
\dfrac{e}{k}\log\left(\dfrac{k}{\lfloor k/e\rfloor}^{\lfloor k/e\rfloor}\right)\\
\\
\dfrac{e}{k}\log\left(\dfrac{k}{\lceil k/e\rceil}^{\lceil k/e\rceil}\right)\\
\end{cases}$
against background $\mod \dfrac{e}{2}$

shows that what you conjecture is not strictly true, but it certainly approaches that limit (if I have understood the problem correctly).
Update
For $e\pm\frac{1}{2}$, fractional part $\pm\frac{1}{4}$:

Manipulate[s = E + m; pr2 = 1.005; pr1 = 0.94; r1 = s; c = m/E;Show[Plot[{If[FractionalPart[k/s] < 0.5 - m/2, 1.1, 0], 1}, {k, r1, (r1 + 5 s)}, PlotStyle -> {{Opacity[0]}, {Opacity[0]}},Filling -> {1 -> {2}}, FillingStyle -> {Opacity[0.2]},Frame -> True, Axes -> False, GridLines -> {{}, {1}},GridLinesStyle -> {Gray, Dashing[0.01]},PlotRange -> {{r1, (r1 + 5 s)}, {pr1, pr2}}], Plot[{If[Log[(k/Floor[k/s])^Floor[k/s]]/(k/s) - c >Log[(k/Ceiling[k/s])^Ceiling[k/s]]/(k/s) - c,Log[(k/Floor[k/s])^Floor[k/s]]/(k/s) - c, I],If[Log[(k/Ceiling[k/s])^Ceiling[k/s]]/(k/s) - c >Log[(k/Floor[k/s])^Floor[k/s]]/(k/s) - c,Log[(k/Ceiling[k/s])^Ceiling[k/s]]/(k/s) - c, I]}, {k,r1, (r1 + 5 s)}, PlotStyle -> {{Red, Thick}, {Blue, Thick}}, Frame -> True,Axes -> False, PlotRange -> {{r1, (r1 + 5 s)}, {pr1, pr2}}]], {{m, 0}, -0.5, 0.5}]

A: Not an answer, but an approach: Fix a $k$ and let $K := \lceil \frac{k}{e} \rceil$. Suppose $\left\{ \frac{k}{e} \right\} \ge \frac12$. We want to show that
\begin{align*}
\left(\frac{k}{\lfloor \frac ke \rfloor}\right)^{\lfloor \frac ke \rfloor} \le \left(\frac{k}{\lceil \frac ke \rceil}\right)^{\lceil \frac ke \rceil}
\end{align*}
Basic transformations (and the fact that $K = \lceil \frac ke \rceil = \lfloor \frac ke \rfloor + 1$) yield the equivalent inequality
\begin{align*}
K^K \le k \cdot (K - 1)^{K - 1}
\end{align*}
Now our assumption $\left\{ \frac{k}{e} \right\} \ge \frac12$ means $\frac ke \ge \lfloor \frac ke \rfloor + \frac 12 = K - \frac 12$. Hence
\begin{align*}
k \cdot (K - 1)^{K - 1} \ge e \cdot \left(K - \frac12\right) \cdot (K - 1)^{K - 1}
\end{align*}
Plugging this in above and transforming, we see that it is sufficient to show that
\begin{align*}
\left(\frac{K}{K-1}\right)^K \le e \cdot \frac{K - \frac12}{K - 1}
\end{align*}
Taking the reciprocal and multiplying by $e$, we get the equivalent inequality
\begin{align*}
\left(1 - \frac1K\right)^K \cdot e \ge \frac{K - 2}{K - 1}
\end{align*}
Both sides converge to $1$ as $K \to \infty$ (here the role of the constant $e$ becomes apparent).
If we want to prove the other case, i. e., $\left\{ \frac ke \right\} < \frac12$, we analogously arrive at
\begin{align*}
\left(1 - \frac1K\right)^K \cdot e < \frac{K - 2}{K - 1}
\end{align*}
I could not proceed from here since I don't know enough about the convergence of $(1+\frac xn)^n$. I am not even sure whether this helps at all, but I wanted to post it since it explains where the $e$ comes from.
