Power trip (part 3)

On further thought the inequality

\displaystyle a<\frac{b-a}{\ln(b/a)}<b 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)

which leads to

\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.

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