Can You Prove ln(e)/e > ln(pi)/pi Without Calculations?

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Discussion Overview

The discussion revolves around proving the inequality ln(e)/e > ln(pi)/pi without performing any calculations. Participants explore various approaches and hints related to the properties of the function f(x) = ln(x)/x.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant expresses difficulty in proving the inequality without calculations and seeks assistance.
  • Another participant suggests considering the function f(x) = ln(x)/x and determining its maximum value as a potential approach.
  • A different participant indicates interest in proving a related inequality, e^pi > pi^e, instead of the original one.
  • One participant asserts that since ln(e)/e is the maximum value of f, it leads to the conclusion that ln(e^pi) > ln(pi^e), which implies the original inequality.
  • A participant expresses realization and gratitude for the hints provided, indicating a moment of clarity.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method to prove the original inequality without calculations, as different approaches and related inequalities are discussed.

Contextual Notes

Some assumptions about the behavior of the function f(x) and its maximum value are implied but not explicitly stated or proven within the discussion.

Guero
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I haven't been able to prove:

ln(e)/e > ln(pi)/pi

without calculating any of the values. Help would be much appreciated.
 
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Hint:
Consider the function
f(x)=\frac{ln(x)}{x}},
with domain the positive real half-axis.

Determine the function's maximum value.
 
Last edited:
mm, I can see that, but I was looking for a proof that shows that e^pi > pi^e
 
Well, since you can prove that ln(e)/e is the maximum value for f, we also have:
\pi(ln(e))>eln(\pi)\to{ln}(e^{\pi})>ln(\pi^{e})
wherefrom your inequality follows.
 
Argh! I get it, Thanks!

I feel pretty stupid now.
 

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