## Power trip (part 2)

That proof that $e^a>a^e$ for any positive $a\ne e$ relies on $\ln(e)=1$, so doesn’t generalize to anything nearly as simple relating $a^b$ to $b^a$. But let’s see what we can do.

Here’s an inequality which, if you’re like me, looks pretty mysterious:

$\displaystyle b>\frac{b-a}{\ln(b/a)}>a$ for $a$ and $b>a$ positive real numbers.

I mean, the difference between $a$ and $b$ doesn’t “know” anything about $a$ or $b$, right? And neither does their ratio. But the difference over the log of the ratio is always between $a$ and $b$? Weird!

(Okay, maybe you’re not so easily impressed, since if you know $b-a$ and $b/a$ you can get $a$ and $b$. Fine. I think it’s mysterious anyway.)

Well, let’s consider a function $f(x)$ with derivative $df(x)/dx = f'(x)$ monotonically decreasing (we could do increasing too, similarly). Let $a$ and $b>a$ be in the range where $f(x)$ and $f'(x)$ exist and $f'(x)>0$. Then

$\displaystyle \int_a^b f'(x)dx = f(b)-f(a)$

But that integral is the area under the curve from $a$ to $b$ and so

$\displaystyle f'(a)(b-a)>f(b)-f(a)>f'(b)(b-a)$.

Then $(b-a)>[f(b)-f(a)]/f'(a)$ and $[f(b)-f(a)]/f'(b)> (b-a)$ or

$\displaystyle \frac{f(b)-f(a)}{f'(b)}> (b-a)>\frac{f(b)-f(a)}{f'(a)}$.

But that makes sense given the definition of the derivative and the monotonic property of $f'(x)$; with a little thought you could’ve written this down directly. As obvious as it is, it takes a decidedly less obvious form if you specify $f(x) = \ln(x)$, $f'(x) = 1/x$, which gives

$\displaystyle b\ln(b/a)>(b-a)>a\ln(b/a)$

or the formerly mysterious

$\displaystyle b>\frac{b-a}{\ln(b/a)}>a$ QED.

I called the original post “Power trip” and this one “Power trip (part 2)”, and there haven’t been any powers here. Sorry. Here:

$\displaystyle a\ln(b/a) < b-a < b\ln(b/a)$

so

$\displaystyle (b/a)^a < e^{b-a} < (b/a)^b$.

That form is… neither obvious nor interesting, though. Unless $a=e$ in which case the left part is

$\displaystyle (b^e)/(e^e) < (e^b)/(e^e)$

or

$\displaystyle b^e

which is where we came in.

## Power trip

Here’s a mathematical paper of which I can smugly say I understood every word.

Ha ha. No, seriously, I like it. A visual proof that $e^\pi>\pi^e$.

But as Math with Bad Drawings pointed out someone else pointed out, the proof doesn’t depend on $\pi$ but can be generalized to any number $> e$, and as Math with Bad Drawings pointed out, a similar proof works for any positive number $< e$. That is, for all $a > 0$ and $a\ne e$, $e^a>a^e$.

Proof here if you don’t want to click through to MwBD:

If $a > e$,

$\displaystyle \frac{1}{e}(a-e) = \frac{a}{e}-1 > \int_e^a \frac{1}{x}dx = \ln(a)-\ln(e) = \ln(a)-1$

so

$\displaystyle \frac{a}{e} > \ln(a)$

or

$\displaystyle a > \ln(a^e)$

or

$\displaystyle e^a > a^e$

and similarly, if $b < e$,

$\displaystyle \frac{1}{e}(e-b) = 1-\frac{b}{e} < \int_b^e \frac{1}{x}dx = \ln(e)-\ln(b) = 1-\ln(b)$

so

$\displaystyle \frac{b}{e} > \ln(b)$

or

$\displaystyle b > \ln(b^e)$

or

$\displaystyle e^b > b^e$

QED