How to prove $3^\pi>\pi^3$ using algebra or geometry?

Question

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.

Answer 1

Hint: $$ \frac{\log(x)}x $$ is monotonically decreasing for $x\gt e$.

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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 $x\ge e$.

Answer 2

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 $3^{47}\times 7^{45} > 22^{45}$.

Answer 3

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.

Answer 4

Hint: $$\pi >3\log _{ 3 }{ \pi } \\ \log _{ 3 }{ \pi } >1\\ \pi >3\log _{ 3 }{ \pi } >3$$

Answer 5

Here's a proof using the inequalities $3.14\lt\pi\lt3.1416$.

We start with some seemingly random observations:

$$0.0475\times21=0.95+0.0475=0.9975\lt1\implies0.0475\lt{1\over21}\implies1.0475\lt{22\over21}$$

$$1.0475\lt{22\over21}\implies(1.0475)^2\lt{484\over441}=1+{43\over441}\lt1.1$$

(Alternatively you could simply compute $(1.0475)^2=1.0972562\lt1.1$, but I'm trying to stick to arithmetic that can be easily checked by eye.)

$$1.1^2=1.21\lt1.25={5\over4}$$ $$1.1^3=1.331\lt1.333\ldots={4\over3}$$

Putting all this together, we have

$$(1.0475)^{22}\lt(1.1)^{11}=(1.1)^{2+3+3+3}\lt{5\over4}\left(4\over3\right)^3={80\over27}\lt3$$

Now we're getting somewhere:

$${\pi\over3}\lt{3.1416\over3}=1.0472\lt1.0475\implies\left(\pi\over3\right)^{22}\lt3$$

This in turn implies

$$\left(\pi\over3\right)^{150}\lt\left(\pi\over3\right)^{154}=\left(\pi\over3\right)^{22\cdot7}\lt3^7$$ hence $$\left(\pi\over3\right)^{3}\lt3^{7/50}=3^{0.14}$$ and thus

$$\pi^3\lt3^3\cdot3^{0.14}=3^{3.14}\lt3^\pi$$

as desired.

Needless to say, all the trial and error of the sausage making has been swept under the rug; the presentation here pretty much unwinds the thought process that went into it. Also, no justification is given for the inequalities $3.14\lt\pi\lt3.1416$. A proper proof would truly start from scratch.