## Power trip (part 3)

On further thought the inequality $\displaystyle a<\frac{b-a}{\ln(b/a)} for $a$ and $b>a$ positive real numbers

can be tightened up on the left side. I used $\displaystyle \left(\frac{1}{b}\right)(b-a)<\ln(b/a)<\left(\frac{1}{a}\right)(b-a)$

but in fact this can be improved to $\displaystyle \left(\frac{1}{b}\right)(b-a)<\ln(b/a)<\left(\frac{1}{b}\right)(b-a)+\frac{1}{2}\left(\frac{1}{a}-\frac{1}{b}\right)(b-a)$ $\displaystyle a\left(\frac{2b}{b+a}\right) < \frac{b-a}{\ln(b/a)} < b$.

For $b\gg a$, the left side is $\approx 2a$, while for $b = a(1+2\epsilon)$ where $\epsilon<<1$, the left side is $\approx a(1+\epsilon) = (a+b)/2$.

## 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.