why is the geometric mean less than the logarithmic mean? Can someone explain why the geometric mean is less than the logarithmic mean?
$$\sqrt{ab} \leq  \frac{b-a}{\log b-\log a}
$$
 A: Let $\sqrt{ab}=:g$. Then $a=ge^{-p}$, $\>b=ge^p$ for some $p\geq0$, and we obtain
$${b-a\over\log b-\log a}=g\>{2\sinh p\over 2p}\geq g\ ,$$
since $\sinh'(0)=1$ and $p\mapsto\sinh p$ is convex for $p\geq0$.
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.
A: This not an answer, but this remark cannot take place in a simple comment.
It should be said that, under the form $\dfrac{a-b}{ln(a)-ln(b)}$, the logarithmic mean may look rather artificial and difficult to extend to more than 2 variables unless one knows the formula 
$$L(a,b)=\int_0^1 \ \varphi(x)dx \ \ \text{with} \ \ \varphi(x)=a^xb^{1-x}=be^{x \ ln(a/b)}$$
which extends to 3D under the form:
$$L(a,b,c)=2\int_{x=0}^1\int_{y=0}^{1-x} a^x \ b^y \ c^{1-x-y} dx dy$$
and more generally in nD in the same way on the simplex defined by $x_1+x_2+\cdots +x_n=1$ (factor $2=2!$ will become $(n-1)!$). 
So, the next question is how one can prove for example that 
$$G(a,b,c)\leq L(a,b,c) \ ?$$
