$$e^ \pi > \pi ^ e$$.

Prove that $$e^ \pi > \pi ^ e$$.

This is a list of Calculus proof based problems that I like. Please avoid posting complete solutions, so that others can work on it.

Note by Calvin Lin
4 years, 6 months ago

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Take the function $$f:\mathbb{R_+}\mapsto \mathbb{R}$$, defined by : $$f(x)=\frac{\log x}{x}$$.

It derivative is : $$f'(x)= \frac{1-\log x}{x^2}$$, therefore $$f(e)$$ is the maximum.

Then : $$f(e)>f(\pi)$$, which means that : $$\pi \log e >e\log \pi \Leftrightarrow e^{\pi} >\pi^e$$.

How did I choose this function : just try to get $$\pi$$ is one side and $$e$$ is another to get the idea.

- 4 years, 6 months ago

This solution exists in the math book of highschool's terminal year in Algeria. I loved it ;)

- 4 years, 4 months ago

Brilliant! Great solution.

- 4 years, 6 months ago

I took the function $$f(x) = x^{\frac{1}{x}}$$ and found that it has a maximum at $$x = e$$.So $f(e) > f(\pi)$. $e^{\frac{1}{e}} > \pi^{\frac{1}{\pi}}$. $e^{\pi} > \pi^{e}$.

- 4 years, 5 months ago

This is the way I did it a long time ago. This one is an oldie. But this approach takes longer than Haroon's proof.

- 4 years, 5 months ago

Yeah..I also solved this one a long time ago ....before i joined Brilliant!!

- 4 years, 5 months ago

Hey, me too.

- 4 years, 5 months ago

A well-known and important inequality: $$e^n>1+n$$ given $$n>0$$. Now if we let $$n=\dfrac \pi e-1$$ we will immediately get after rearranging our inequality $$e^\pi>\pi^e$$.

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