WLOG, a>b. We will prove the right inequality first $ \frac {a - b}{\ln a - \ln b} < \frac {a + b}{2} \Rightarrow \ln x > 2 \left(\frac {x - 1}{x + 1}\right)$ where $ x = \ln \left(\frac {a}{b} \right) > 1$ Consider $ f(x) = \ln x - 2 \left(\frac {x - 1}{x + 1}\right) = \ln x - 2\left(1 - \frac {2}{x + 1} \right)$ $ f'(x) = \frac {1}{x} - \frac {4}{(x + 1)^2} > 0$ by AM-GM. Further f(1) = 0. And thus we have for all x>1, $ \ln x > 2 \left(\frac {x - 1}{x + 1}\right)$ as desired Likewise, for the left inequality, we have to prove that $ \ln x < \frac {x - 1}{\sqrt x}$. Let $ y = \sqrt x$ Then the inequality becomes $ \ln y < \frac{1}{2} \left(y - \frac {1}{y} \right)$ Again let $ g(y) = \frac{1}{2} \left(y - \frac {1}{y} \right) - \ln y$. We have $ g'(y) = \frac{1}{2} + \frac {1}{2y^2} - \frac {1}{y} > 0$ Also $ g(1) = 0$ Hence for all y>1, we have $ \ln y < \frac{1}{2} \left(y - \frac {1}{y} \right)$ and the left inequality is also proved

$1^\circ$ Suppose that $a>b$ then dividing by $b$ both sides we will obtain: $\sqrt \frac{a}{b} < \frac{\frac{a}{b}-1}{\ln(\frac{a}{b})} < \frac{\frac {a}{b} + 1}{2}$. $2^\circ$ Let $t=\frac{a}{b}$, rewriting our statement we get: $\sqrt t\cdot \ln t < t-1 < \frac{t+1}{2}\cdot \ln t$. Now we consider the function: $f(t)=\frac{t-1}{\sqrt t} - \ln t$ , $f(1)=0$ and $f'(t)>0$ $\Longrightarrow$ $f(t)$-increasing it follows: $t-1>\sqrt t\cdot \ln t$. $3^\circ$ Similarly we show the right side of this inequality.