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I recently encountered the following inequality : if $a<b$ are two positive real numbers, then

$$ \frac{\ln(b)-\ln(a)}{b-a} \leq \frac{a+b}{2ab} $$

This inequality is not very hard to show : if we put $f(x)=\frac{x^2-a^2}{2ax}-\ln(x)+\ln(a)$, then $f'(x)=\frac{(x-a)^2}{2ax^2}$. I wonder if there are more "geometric" proofs ?

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up vote 8 down vote accepted

Yes there is! The average value of the function $\frac1x$ on the interval $[a,b]$ is $\frac{\ln b-\ln a}{b-a}$; the average of the two values at the endpoints, namely $\frac12(\frac1a+\frac1b) = \frac{a+b}{2ab}$, is larger because $\frac1x$ is convex. (Equivalently, the area under the curve itself on $[a,b]$ is smaller than the area under the trapezoid defined by its two endpoints.)

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Perhaps not as geometric as you hoped but for $t> 0$ with $\,e^t:=\dfrac ba\;$ (and thus $>1$) we have : \begin{align} t&<\sinh(t)\\ 2\;\log e^t&<e^t-e^{-t}\\ 2\;\log\frac ba&<\frac ba-\frac ab\\ \log b-\log a&<\frac {(b-a)(a+b)}{2ab}\\ \end{align} (of course Greg Martin's answer is more appropriate here (+1) but I like this interpretation too)

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