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

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
6 years ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
• Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

## Comments

Sort by:

Top Newest

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.

- 6 years ago

Log in to reply

Brilliant! Great solution.

Log in to reply

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

- 5 years, 11 months ago

Log in to reply

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

- 6 years ago

Log in to reply

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.

- 5 years, 12 months ago

Log in to reply

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

- 5 years, 12 months ago

Log in to reply

Hey, me too.

- 5 years, 12 months ago

Log in to reply

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

- 5 years, 10 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...