For $a\gt b\gt 0$, prove that $\dfrac{a+b}{2}\gt \dfrac{a-b}{\ln a-\ln b}$.

For $a\gt b\gt 0$, prove that $\dfrac{a+b}{2}\gt \dfrac{a-b}{\ln a-\ln b}$.

There sure is many way to prove this problem, but I am going to show you one graphical method that if we recognize some function is always greater than the other in certain interval, then it makes the problem all that easier for us to crack.

Note that the graph of $y=\ln (x)$ and $y=2\left(1-\dfrac{2}{x+1}\right)$ are concave down on the interval $(0,+\infty)$, therefore, they only intersect at one and only one point, i.e. $(1,\,0)$.

If we can prove at any point that is to the right of $x=1$ that $y=\ln (x)$ lies above $y=2\left(1-\dfrac{2}{x+1}\right)$, then we have proved $\ln(x) >2\left(1-\dfrac{2}{x+1}\right)$. At $x=e>1$, $y=\ln(e)=1$, $y=2\left(1-\dfrac{2}{e+1}\right)≈0.924234$. Therefore, for $x\gt 1$, we have $\ln(x) >2\left(1-\dfrac{2}{x+1}\right)$ and $x$ can be replaced by $\dfrac{a}{b}$ since $a\gt b\gt 0$.

It's equivalent to the form $\dfrac{\ln \left(\dfrac{a}{b}\right)}{2}>1-\dfrac{2}{(\dfrac{a}{b}-1)}$, after rearranging it takes also the form

$\dfrac{a+b}{2}\gt \dfrac{a-b}{\ln a-\ln b}$ and we're hence done.