$\pi ^e$ vs $e^\pi$

This is a popular problem with several known solutions. I was drawn to weigh on the problem by a post by Kunihiko Chikaya at the Short Mathematical Idea facebook group.

I found several solutions on the web, all nice and simple, using calculus, and most claiming to be the shortest beauty. My preference is for Kunihiko Chikaya's proof, although, as I discovered, it was not new (this is not to say the Kunihiko Chikaya did not do that independently.)

The answer is $\pi ^e \lt e^\pi.$ Several proofs are following.

This proof would unlikely be accepted in a SAT or any other exam, however, given the wide spread of calculators, computers, computer algebra systems, and the fact that we live in the 21st century, make the following argument reasonably convincing:

Since $3.14 \lt \pi \lt 3.15$ and $2.718 \lt e \lt 2.719,$ and using the monotonicity of the exponential and power functions,

$\pi ^e \lt 3.15^{2.719} \lt 2.718^{3.14} \lt e^\pi,$

the middle inequality being easily verifiable by any widely available technological means: $3.15^{2.719} \approx 22.6416 \lt 23.0963 \approx 2.718^{3.14}.$

This is the one posted by Kunihiko Chikaya, but it also can be found at the Stack Exchange.

Since $\displaystyle e^x=1+x+\frac{x^2}{2!}+\ldots,$ for $x\gt 0,$ we certainly have $e^x \gt 1+x,$ for positive $x.$ Taking $\displaystyle x=\frac{\pi}{e}-1$ leads to

$\displaystyle e^{\frac{\pi}{e}-1} \gt 1+(\frac{\pi}{e}-1)=\frac{\pi}{e},$

which is

$\displaystyle e^{\frac{\pi}{e}} \gt \pi,$ or $e^\pi \gt \pi ^e.$

Let $f(x)=x^{1/x}.$ Then

$\displaystyle f'(x)=\bigg[e^{\frac{\ln x}{x}}\bigg]'=x^{\frac{1}{x}}\frac{1-\ln x}{x^2},$

which vanishes at $x=e,$ being positive for $x\lt e$ and negative for $x\gt e.$ It follows that $x=e$ is the global maximum of $f(x)=x^{1/x}.$ In particular, $e^{1/e} \ge \pi ^{1/\pi },$ with equality impossible.

Let $\displaystyle f(x)=\frac{\ln x}{x}.$ This is the natural logarithm of the previous function. The derivative is even simpler

$\displaystyle f'(x)=\frac{1-\ln x}{x^2},$

with a unique zero for $x=e.$ The rest follows.

Define $f(x)=e^x - x^e.$ As a matter of fact, $f(x)\gt 0$ when $x\ge 0.$ He's the graph of that function

The derivative $f'(x)=e^x - ex^{e-1}$ vanishes at two points $x=1$ and $x=e,$ the latter being the global minimum. In particular, $f(\pi )\gt 0.$

Define $f(x)=x - e\ln x.$ This function has a particularly simple derivative to work with: $\displaystyle f'(x)=1-\frac{e}{x},$ with the only zero at $x=e.$ The function is monotone decreasing for $x\lt e$ and monotone increasing afterwards. It follows, since $\pi \gt e,$ that $0=f(e)\lt f(\pi ),$ i.e., $e\ln \pi\lt\pi.$ $\ln \pi ^e \lt \pi,$ same as $e^\pi \lt \pi ^e.$