TITLE: How to arrange $e^3,3^e,e^{\pi},\pi^e,3^{\pi},\pi^3$ in the increasing order?
QUESTION [9 upvotes]: For these six numbers, $e^3,3^e,e^{\pi},\pi^e,3^{\pi},\pi^3$, how to arrange them in the increasing order?
This problem is taken from the today test: National Higher Education Entrance Examination.
I think we can consider
$$f(x)=\dfrac{\ln{x}}{x}$$
or perhaps other methods, so I am looking forward to seeing other methods to solve this problem. Thank you.

REPLY [2 votes]: Here's a fairly low-tech argument.
A quick summary is that three inequalities follow from $u^v > v^u$
for $v > u \geqslant e,$ and the two trickier ones that remain
follow from $e > 2\frac7{10}$ and:
$$
\log3 > 1\frac1{12} > 4(2 - \sqrt3) > \frac\pi3 > \sqrt6 - \sqrt2 >
3^{\frac1{39}}.
$$
If $C$ is a circle of unit radius, then a regular dodecagon
inscribed in $C$ has perimeter $6(\sqrt6 - \sqrt2),$ and a regular
dodecagon circumscribed about $C$ has area $12(2 - \sqrt3),$
therefore:
\begin{equation}
\label{823770:eq:1}\tag{1}
3(\sqrt6 - \sqrt2) < \pi < 12(2 - \sqrt3).
\end{equation}
Note the implication $\pi > 3.$ Of course, this could have been
proved more simply by inscribing a regular hexagon, but we will need
the better lower bound.
Because $3\cdot48^2 = 6912 > 6889 = 83^2,$ we have:
$$
\sqrt3 > \frac{83}{48} = 2 - \frac{13}{48},
$$
and therefore:
$$
\pi < 3\frac14.
$$
If $|x| < 1,$ we have:
\begin{equation}
\label{823770:eq:2}\tag{2}
\log\frac{1 + x}{1 - x} = 2\left(x + \frac{x^3}3 + \frac{x^5}5 +
\frac{x^7}7 + \frac{x^9}9 + \cdots\right).
\end{equation}
Taking $x = \frac12,$ and using only the first two terms of
\eqref{823770:eq:2}, we get:
$$
\log3 > 1\frac1{12}.
$$
(Note the implication $e < 3.$)
Combining this with the upper bound for $\pi$ gives:
$$
3\log3 > 3\frac14 > \pi,
$$
whence:
$$
\boxed{e^\pi < 27 < \pi^3.}
$$
The lower bound for $\log3$ gives us an upper bound for $e$:
$$
e < 3^{\frac{12}{13}}.
$$
In addition to that, we will need this lower bound:
$$
e > 1 + 1 + \frac12 + \frac16 + \frac1{24} = 2\frac{17}{24}
> 2\frac7{10}.
$$
By the lower bound in \eqref{823770:eq:1}:
$$
\pi^3 > 27(12\sqrt6 - 20\sqrt2).
$$
Because $6\cdot24^2 = 3456 < 3481 = 59^2,$ and
$2\cdot40^2 = 3200 < 3247 = 57^2,$ we have:
$$
12\sqrt6 < \frac{59}2 \text{ and } 20\sqrt2 < \frac{57}2,
$$
and therefore:
$$
12\sqrt6 - 20\sqrt2 = \frac{64}{12\sqrt6 + 20\sqrt2}
> \frac{64}{58} = 1\frac3{29} > 1.1.
$$
But:
\begin{gather*}
1.1^3 = 1.331 > 1.33, \\
1.33^2 = 1.7689 > \frac74, \\
\left(\frac74\right)^2 = \frac{49}{16} > 3,
\end{gather*}
therefore $1.1^{12} > 3,$ which gives us the estimate we need:
$$
\left(\frac\pi3\right)^3 > 3^{\frac1{12}} > 3^{\frac1{13}}.
$$
We now have:
$$
e^{\frac{10}3} < 3^{\frac{40}{13}} = 27\cdot3^{\frac1{13}} < \pi^3,
$$
whence:
$$
\boxed{e^3 < \pi^{\frac{27}{10}} < \pi^e.}
$$
Finally, if $1 \leqslant x < y,$ then:
\begin{gather*}
e^{y - x} > 1 + (y - x) \geqslant 1 + \frac{y - x}x = \frac{y}x, \\
\therefore\ \frac{x}{e^x} > \frac{y}{e^y} \quad (1 \leqslant x < y).
\end{gather*}
Writing $x = \log u,$ $y = \log v$ ($e \leqslant u < v$), we get:
$$
\frac{\log u}u > \frac{\log v}v \quad (e \leqslant u < v).
$$
Equivalently:
$$
u^v > v^u \quad (e \leqslant u < v).
$$
Taking in turn:
$$
(u, v) = (e, 3), \, (e, \pi), \, (3, \pi),
$$
we get:
$$
e^3 > 3^e, \ e^\pi > \pi^e, \ 3^\pi > \pi^3.
$$
In conjunction with the earlier inequalities $e^\pi < \pi^3$ and
$e^3 < \pi^e,$ this gives:
$$
\boxed{3^e < e^3 < \pi^e < e^\pi < \pi^3 < 3^\pi.}
$$