How to prove $3^\pi>\pi^3$ using algebra or geometry? It's a question of a some time ago test,
I've found a way to solve the problem using calculus, but always I've thought that exist a solution with algebra and geometry.
Thank you for your time.
 A: Hint:
$$
\frac{\log(x)}x
$$
is monotonically decreasing for $x\gt e$.

Pre-calculus Approach
We will use the fact that $e^h\ge1+h$ for $h\ge0$.
If $x\ge1$ and $h\ge0$, then
$$
\begin{align}
\frac{x+h}{e^{x+h}}-\frac{x}{e^x}
&=\frac{x+h-xe^h}{e^{x+h}}\\
&=\frac{\overbrace{\vphantom{\left(e^h\right)}\ \ \ \ x\ \ \ \ }^{\ge1}\overbrace{\left(1+h-e^h\right)}^{\le0}+\overbrace{\vphantom{\left(e^h\right)}(1-x)}^{\le0}\overbrace{\vphantom{\left(e^h\right)}\ \ \ \ h\ \ \ \ }^{\ge0}}{e^{x+h}}\\\\
&\le0
\end{align}
$$
Thus, $\frac x{e^x}$ is monotonically decreasing for $x\ge1$. Since $\log(x)$ is monotonically increasing, substituting $x\mapsto\log(x)$ says that $\frac{\log(x)}x$ is monotonically decreasing for $x\ge e$.
A: Warning: not pretty.
Since $3^{47}\times 7^{45} \approx 2.85\times 10^{60}$ and $22^{45}\approx 2.56\times 10^{60}$, it follows that $3^{47}\times 7^{45} > 22^{45}$, and hence
$$ 3^{\frac{47}{15}} > \left(\frac{22}{7}\right)^{\frac{45}{15}} = \left(\frac{22}{7}\right)^3. $$
Since $\frac{47}{15}<\pi<\frac{22}{7}$, the result follows.
EDIT: A slightly less bashy way to get the result $3^{47}\times 7^{45} > 22^{45}$:
Note the inequalities
\begin{matrix}
17010 > 16384 &\implies& 3^5\times7\times10 &>& 2^{14} \\
4000 > 3993 &\implies& 2^2\times10^3 &>& 3\times11^3 \\
2401 > 2400 &\implies& 7^4 &>& 2^3\times3\times10^2 \\
243 > 242 &\implies& 3^5 &>& 2\times11^2
\end{matrix}
Raising the first inequality to the 1st power, the second to the 7th, the third to the 11th, the fourth to the 12th, and multiplying, we have
$$ 2^{14}\times 3^{65}\times 7^{45}\times 10^{22} > 2^{59}\times 3^{18}\times 10^{22}\times 11^{45}$$
which simplifies to $3^{47}\times 7^{45} > 22^{45}$.
A: Now you want 
$$3^{\pi}>\pi^3$$
Taking logarithm, it suffices to show
$$\frac{\ln3}{3}>\frac{\ln\pi}{\pi}$$
It again suffices to show that
$$f(x)=\ln x/x$$
is strictly decreasing at least when $x\ge3$. 
From a geometric point of view, $\ln x/x$ is equal to  the slope $k(x)$ of the line that connects the origin and the point $(x,\ln x)$ located on  the curve  $y=\ln(x)$. Now from the graph it is obvious that $k(x)$ assumes maximum at some $x_0$ where the line is tangent to the curve, and $k(x)$ is strictly decreasing when $x>x_0$ (strictly speaking it is due to concavity of the logarithm function). Thus all you have to do is to find the $x_0$ (how? Well, try to give it a shot!) and you'll find out that $x_0=e<3$, so...
EDIT @robjohn pointed out that finding the exact value of $x_0$ requires calculus, so I'm "de-calculizing" this part:
Well, you don't have to find the exact value, just making sure $x_0\le3$ is sufficient. Thus, if you have a calculator at hand, by verifying the following 
$$\ln2.8/2.8>\ln2.9/2.9$$
You can verdict that $x_0\le 2.9$. If not, $f(x)$ should be strictly increasing when $x<2.9$ and hence contradiction.  (However this trick is a little bit empirical I confess).  
NOTE Although it seems to be a de-calculized answer now, it is, strictly speaking, still implicitly based on calculus - without calculus, we won't even have the notion of the concavity of a graph, or the geometrical intuition that $k(x)$ hits max at the tangent position. So if you are really in search of a literally non-calculus answer, @robjohn sure has the best one. 
A: Hint:
$$\pi >3\log _{ 3 }{ \pi  } \\ \log _{ 3 }{ \pi  } >1\\ \pi >3\log _{ 3 }{ \pi  } >3$$
