powers – MathematRec

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