Here's the question of the day: What's bigger, eπ, or πe?
Of course, proofs for this can be easily found on the internet, but the fun part is to make it as short as possible. So here goes:
Let f(x) = ex-ex.
By examining the function's first and second derivative, it is clear that it achieves its only minimum at x = 1, for which f(x) = 0.
Thus, ex > e x for all x not equal to one.
Substituting x = ln(π) in the inequality gives π > e ln(π) .
Noting that e ln(π) = ln(πe) and exponentiating both sides of the previous inequality gives the result.
So now you have your sliver of π e. Hopefully next time I'll remember to bring the Java!
