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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.
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Take the function f:R+↦R, defined by : f(x)=xlogx.
It derivative is : f′(x)=x21−logx, therefore f(e) is the maximum.
Then : f(e)>f(π), which means that : πloge>elogπ⇔eπ>πe.
How did I choose this function : just try to get π is one side and e is another to get the idea.
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Brilliant! Great solution.
This solution exists in the math book of highschool's terminal year in Algeria. I loved it ;)
I took the function f(x)=xx1 and found that it has a maximum at x=e.So f(e)>f(π). ee1>ππ1. eπ>πe.
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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.
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Yeah..I also solved this one a long time ago ....before i joined Brilliant!!
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A well-known and important inequality: en>1+n given n>0. Now if we let n=eπ−1 we will immediately get after rearranging our inequality eπ>πe.