why is the geometric mean less than the logarithmic mean? Can someone explain why the geometric mean is less than the logarithmic mean?
$$\sqrt{ab} \leq  \frac{b-a}{\log b-\log a}
$$

There might be a geometric interpretation that you are looking for, but I still prefer an algebraic approach. So let's suppose $0 < a < b$, and put $b = ta, t > 1$. Thus: $LHS = \dfrac{1}{\sqrt{ab}} = \dfrac{1}{\sqrt{ta^2}} = \dfrac{1}{a\sqrt{t}}$,and $RHS = \dfrac{\log(at) - \log a}{at- a}= \dfrac{\log t}{a(t-1)}$. Thus you prove: $\dfrac{\log t}{t-1} < \sqrt{t}\iff f(t) =\log t - t\sqrt{t} + \sqrt{t} < 0$. Taking first derivative: $f'(t) = \dfrac{1}{t} - \dfrac{3\sqrt{t}}{2} + \dfrac{1}{2\sqrt{t}} < 0, t > 1\Rightarrow f(t) < f(1) = 0$, and the inequality follows.
