Which number its greater $\pi^3$ or $3^\pi$?

Question

$ \pi^3$ or $3^\pi$ using algebra please, I arrive the solution whith $a^x > 1 + x$ but I am interested in more solutions.

Answer 1

Your inequality $\pi^3<3^\pi$ is equivalent to $\frac{\ln(\pi)}{\pi}<\frac{\ln(3)}{3}$.

Now use the function $f(x)=\frac{\ln(x)}{x}$ for $e<x$.

Answer 2

Set $f(x)=\frac{\ln x}{x}$. Then, we have $$f'(x)=\frac{1-\ln x}{x^2}.$$ Since $f'(x)=0\iff x=e$, we know that $f(x)$ is decreasing for $x\gt e$. So, we get $$e\lt 3\lt \pi\Rightarrow \frac{\ln 3}{3}\gt \frac{\ln \pi}{\pi}\Rightarrow \pi^3\lt 3^\pi.$$

Answer 3

The function $x \mapsto x^{1/x}$, $x > 0$, increases from $0$ to $e$, achieves its maximum at $x = e$, and decreases after that. So $3^{1/3} > \pi^{1/\pi}$. This also helps to do the first max/min quiz problem on I assign my students in a calculus class: Find all pairs on positive integers $m, n$, $m \neq n$, $m^n = n^m$.

Answer 4

Note: I've added a new and better proof. To read it, jump to the "Added later" below.

If you are permitted to use the inequalities $$\ln3\gt1$$ and $$\ln(1+x)\lt x$$ for $x\gt0$, then

$$\log_3\pi={\ln\pi\over\ln3}={\ln3+\ln\left({\pi\over3}\right)\over\ln3}=1+{\ln\left(1+{\pi-3\over3}\right)\over\ln3}\lt1+\left({\pi-3\over3}\right)={\pi\over3}$$

upon which one has

$$\log_3\pi\lt{\pi\over3}\implies3\log_3\pi\lt\pi\implies\log_3\pi^3\lt\pi\implies\pi^3\lt3^\pi$$

(To be sure, the key step in this derivation requires knowing that $3\lt\pi$.)

Added later: It sort of bugged me to have to make use of an inequality for the natural logarithm in a problem that only uses $\pi$, so I finally figured out a way to get around it. Here, with all the scaffolding removed, is what I came up with.

First, comparing $\pi\approx3.14159$ with $22/7\approx3.14286$ and $69/22\approx3.13636$, we have

$${69\over22}\lt\pi\lt{22\over7}$$

Suppose we can show that

$$\left({22\over21} \right)^{22}\lt 3$$

Then it's easy to see that

$$\pi^3\lt\left({22\over7}\right)^3 =3^3\left({22\over21}\right)^3 =3^3\left({22\over21}\right)^{22(3/22)} \lt3^33^{3/22} =3^{69/22} \lt3^\pi$$

Now the key inequality $(22/21)^{22}\lt3$ could be proved simply by doing a horrendous calculation. But let's be a little more clever. Note first that

$${22\cdot2\over21^2}={44\over441}\lt{44\over440}={1\over10}\implies \left({22\over21}\right)^2\lt{11\over10}$$

so it suffices to show $(11/10)^{11}\lt3$. This could be done directly with a somewhat less horrendous calculation, but it's easier to note that

$$\left({11\over10}\right)^3=1.331\lt{4\over3}$$ and $$\left({11\over10}\right)^4={11\over10}\left({11\over10}\right)^3\lt{11\over10}\cdot{4\over3}={44\over30}\lt{45\over30}={3\over2}$$

and thus

$$\left({11\over10}\right)^{11}=\left({11\over10}\right)^3\left({11\over10}\right)^4\left({11\over10}\right)^4\lt{4\over3}\cdot{3\over2}\cdot{3\over2}=3$$