e^pi > pi^e

Let’s prove that e^{\pi} > \pi^e without a calculator. If you haven’t seen this before, give it a try before you read any further.

Consider the function f(x) = x/\ln(x) on the interval (1,\infty). This function shoots off to positive infinity as x tends towards either endpoint of the domain. Let’s find its minimum value via an application of good clean differential calculus.

Differentiating (remember the quotient rule?) with respect to x gives us that

\displaystyle f'(x) = \frac{\ln x - 1}{(\ln x)^2}.

Then, we have that f'(x) = 0 when \ln x = 1, so clearly the stationary point of this function occurs at x = e. It is also easy to check that this is a minimum using your favourite high-school test.

Therefore, as the minimum value of f(x) = x/ \ln(x) occurs when x=e, then if we choose any number a>1 which is not equal to e we have

\displaystyle \frac{a}{\ln a} > \frac{e}{\ln e}.

For instance, we can choose a =\pi. Using the fact that \ln(e)=1, this gives us that

\displaystyle \frac{\pi}{\ln \pi} > e.

This inequality can then be manipulated to get that e^{\pi} > \pi^e. The sharp reader will notice that any real number greater than 1 and not equal to e can replace \pi in the inequality and it shall still be true!

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