Mean value on log function For $0<a\leq 1$ and $0<b\leq 1$, let 
$$f(a,b) = \frac{\log a - \log b}{a-b} \qquad a\ne b \\
= \frac{1}{a} \qquad a=b$$
I encountered this expression while working out a contour integral. Now it turns out that
$$f(a,b) \geq \frac{2}{a+b}$$
If you apply mean value theorem, you can argue for some $c\in (b,a)$, (assuming w.l.o.g $b < a$),
$$f(a,b) = \frac{1}{c}$$ and due to the sharp slope of $\log$ in the given region, $c < \frac{a+b}{2}$. My question is that, is there a better way at arriving at the result? Like a simple convexity argument etc...
 A: If $0<a<b$ we have
$$ f(a,b) = \frac{1}{b-a}\int_{a}^{b}\frac{dx}{x} $$
and since $\frac{1}{x}$ is a convex function on $\mathbb{R}^+$, by the Hermite-Hadamard inequality
$$ f(a,b) \leq \frac{1}{b-a}\left[\frac{b-a}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\right] = \frac{a+b}{2ab}$$
and
$$ f(a,b) \geq \left.\frac{1}{x}\right|_{x=\frac{a+b}{2}}=\frac{2}{a+b}.$$
You may use the HHI to prove the stronger bound:
$$\boxed{\frac{\log b-\log a}{b-a}\geq \frac{8(a+b)}{(a+3b)(3a+b)}}$$
by writing the LHS as 
$$ \frac{1}{b-a}\int_{0}^{b-a}\frac{2\,dt}{(a+b)-\frac{t^2}{a+b}}. $$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Note that $\ds{\ln\pars{x} \leq x - 1 \implies \ln\pars{1/x} \geq 1 - x}$.

\begin{align}
{\ln\pars{a} - \ln\pars{b} \over a - b} & = 
{\ln\pars{1/\bracks{b/a}} \over a - b} \geq {1 - b/a \over a - b} = {1 \over a}
\\[5mm]
\mbox{Similarly,}\quad
{\ln\pars{a} - \ln\pars{b} \over a - b} & = 
{\ln\pars{b} - \ln\pars{a} \over b - a}  \geq {1 \over b}
\\[5mm]
\implies
{\ln\pars{a} - \ln\pars{b} \over a - b} & \geq {1 \over \min\braces{a,b}} =
{1 \over \pars{a + b - \verts{a - b}}/2} \geq \bbx{2 \over a + b} 
\end{align}
