Raise both e^pi and pi^e to the power 1/(pi*e). This is monotonic in the positive reals, so it doesn't affect which of them is greater. But now the question becomes, is e^(1/e) greater or less than pi^(1/pi)? And the nice thing about the question in that form is that we've turned it into a function of one variable which we can investigate using the usual calculus tools: differentiating x^(1/x) gives (if I haven't made a mistake) x^(1/x-2)*(1-log(x)), which is positive for x < e, negative for x > e, and zero at x = e itself. Thus, it has a unique global maximum at x=e, and hence e^(1/e) > pi^(1/pi), and therefore we can raise that to the power e*pi and find the answer to the original question is that e^pi > pi^e. And, as Jack says, this proof demonstrates that the key thing is that one of the numbers is e; pi is arbitrary and could have been any other number you like.
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Date: 2007-07-23 03:14 pm (UTC)Raise both e^pi and pi^e to the power 1/(pi*e). This is monotonic in the positive reals, so it doesn't affect which of them is greater. But now the question becomes, is e^(1/e) greater or less than pi^(1/pi)? And the nice thing about the question in that form is that we've turned it into a function of one variable which we can investigate using the usual calculus tools: differentiating x^(1/x) gives (if I haven't made a mistake) x^(1/x-2)*(1-log(x)), which is positive for x < e, negative for x > e, and zero at x = e itself. Thus, it has a unique global maximum at x=e, and hence e^(1/e) > pi^(1/pi), and therefore we can raise that to the power e*pi and find the answer to the original question is that e^pi > pi^e. And, as Jack says, this proof demonstrates that the key thing is that one of the numbers is e; pi is arbitrary and could have been any other number you like.