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eπ > πe.
The underlying idea here is that ab > ba whenever b > a, given that a and b are both greater than 1.
If you look at the graphs of y=xa and y=ax, they will cross, of course, at x=a.
For higher values of x (again, given that a>1), the graph of ax rises faster than xa.
Thus, when x>a, ax > xa.
1 < e < π, so eπ > πe.
Another way to prove this directly follows:
π > e, so ln(π) > 1
e(1-1) = 1
e(x-1) > x (for x>1)
e[ln(π)-1] > ln(π)
(eln(π))/e > ln(π)
eln(π) > e ln(π)
π > e ln(π)
π > ln(πe)
eπ > eln(πe)
eπ > πe |
