How to arrange $e^3,3^e,e^{\pi},\pi^e,3^{\pi},\pi^3$ in the increasing order? 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.
 A: $\bf{My\; Solution\; for \; e^{\pi}\; and \; \pi^e}$
Using $\displaystyle e^x = 1+\frac{x}{1!}+\frac{x^2}{2!}+...............+\infty > 1+x\; \forall x>0$
So $e^x > 1+x$ for $x>0$
Now Put $\displaystyle x = \left(\frac{\pi}{e}-1\right)>0$
So $\displaystyle e^{\frac{\pi}{e}-1}>1+\frac{\pi}{e}-1\Rightarrow \frac{e^{\frac{\pi}{e}}}{e}>\frac{\pi}{e}\Rightarrow e^{\pi}>\pi^e$
A: We start by assuming that somehow we know that
$$ \color{green}{e < 3 < \pi}$$.
This immediately gives the relations
\begin{align*}
\color{blue}{e^3 }&\color{blue}{< e^\pi < 3^\pi}\\
\color{blue}{3^e }&\color{blue}{< \pi ^ e < \pi^3}
\end{align*}
Next, we can consider $$f(x)  = \frac {\ln x}x$$ as you stated.
Differentiating with respect to $x$ we have
$$ f'(x) = -\frac{\ln x}{x^2} +\frac 1 {x^2} = \frac{1-\ln x}{x^2} < 0\quad \forall x > e.$$
Therefore, for values of $x$ larger than $e$, this function is decreasing, which gives
\begin{align*}
  \frac{\ln 3}{3} &> \frac{\ln \pi}{\pi} &\implies\\
  \pi \ln 3 &> 3 \ln \pi &\implies\\
     \color{blue}{3^\pi} &\color{blue}{> \pi^3},
   \end{align*}
and similarly $\color{blue}{e^3 > 3^e}$ and $\color{blue}{e^\pi > \pi^e}$
The two inequalities we have left are $e^\pi$ vs.  $\pi^3$ and $\pi^e$ vs. $e^3$. They're taking longer than I thought...
Edit, some years later. (I was browsing through MSE and came across this question again.)
Suppose we can prove that $\color{green}{6 > e + \pi}$. Then we can show the following:
\begin{align*}
6 &> e + \pi & \Leftrightarrow \\
3 - e &> \pi - 3 & \Leftrightarrow \\
3\cdot\left[1 - \frac{e}{3}\right] &> \pi - 3 & \Leftrightarrow
\end{align*}
For all $x > 0$, we have $\log x > 1 - \frac{1}{x}$. From the preceding, we therefore see that $3 \log(3/e) > 3\cdot\left[1 - \frac{e}{3}\right] > \pi - 3$. We thus have
\begin{align*}
3 \log(3/e) &> \pi - 3  &\Leftrightarrow\\
3 \log(3) &> \pi  &\Leftrightarrow\\
3 ^{3} &> e^{\pi}.
\end{align*}
Since $\pi^3 > 3^3$, we conclude $\pi^3 > e^{\pi}$. To verify the initial inequality, we can pick our favourite upper bounds for $e$ and $\pi$. Archimedes' $\pi < {22}/{7}$ will do for $\pi$, while for $e$ we can use the fact that
$$
e = 3 - \sum_{k=2}^{\infty} \frac{1}{k!(k-1)k} < 3 - \frac{1}{4}.
$$
For the last inequality, we start by showing that $\color{green}{\pi > \frac{e^2}{2e - 3}} \implies \color{blue}{\pi^e > e^3}$ as long as $\color{green}{2e > 3}$:
\begin{align*}
  \pi &> \frac{e^2}{2e - 3} &\Leftrightarrow \\
  \left(2e - 3\right) \pi &> e^2 &\Leftrightarrow \\
-\frac{e^2}{\pi} + e &> 3 - e &\Leftrightarrow\\
  e\left(1 - \frac{e}{\pi}\right) &> 3 - e.
\end{align*}
We now use the same trick as before, that for all $x$, $\log x > 1 - \frac{1}{x}$
\begin{align*}
    e\left(1 - \frac{e}{\pi}\right) &> 3 - e &\Rightarrow \\
 e\log {\pi}/{e} &> 3-e &\Leftrightarrow\\
 e\log \pi &> 3 &\Leftrightarrow\\
 \pi^e &> e^3.
\end{align*}
To finish, we just need to show the initial inequality. The function $f : x \mapsto \frac{x^2}{2x -3}$ is decreasing for $\frac{3}{2}<x<3$, so we'll need a lower bound for $e$. From the series $e = \sum_{k=0}^{\infty} \frac{1}{k!}$ we see $e > {8}/{3}$. For $\pi$, we need a lower bound, and we can again channel Archimedes who proved $\pi > \frac{223}{71}$. We then have
\begin{align*}
  \pi > \frac{223}{71} &\operatorname{?}  \frac{\frac{64}{9}}{\frac{16}{3} - 3} > \frac{e^2}{2e-3}\\
  \pi > \frac{223}{71} &\operatorname{?}  \frac{64}{21} > \frac{e^2}{2e-3},
\end{align*}
and the conclusion follows. The final ordering of the six numbers is therefore $\color{blue}{3^e< e^3 < \pi^e < e^\pi < \pi^3 < 3^\pi}$.
As an aside, we've used a number of the properties of $e$ and $\log$ during the derivations, so an incorrect estimate for $e$ could easily throw the order off. However, we haven't used any of the specific properties of $3$ or $\pi$. Therefore, for any numbers $x$, $y$, such that
\begin{align*}
  e &< x < y,\\
  \frac{e^2}{2e - y} &< x < 2 y - e,\\
  y &< 2e,
\end{align*}
the ordering $x^e< e^x < y^e < e^y < y^x < x^y$ will hold.
