TITLE: Proof of the following inequality $ \frac{x - y}{\log x - \log y} > \sqrt{xy} $, $x>y$.
QUESTION [11 upvotes]: I have seen the following inequality 
$$
\frac{x - y}{\log x - \log y} > \sqrt{xy} \ , \quad \forall x>y
$$
be stated as a near "obvious" fact in another question, on the site. 
The inequality is very cute, but so far I have not been able to prove it. 
It reminds me of the Lipschitz inequality, but has some minor differences.
Also Jensens inequality comes to mind. Is there something obvious I am missing, or is this ineqality not as easy to prove as it looks?

REPLY [2 votes]: Another proof based on geometry is mentioned in 
Frank Burk: The Geometric, Logarithmic, and Arithmetic Mean Inequality, The American Mathematical Monthly , Vol. 94, No. 6 (Jun. - Jul., 1987), pp. 527-528, jstor, link. (It was among the first hits in the google search for logarithmic geometric mean inequality.)
We know that function $e^x$ is convex. Let us have a look on this function on the interval $\ln a$, $\ln b$. The line joining the points $(\ln a,a)$ and $(\ln b,b)$ lies above the graph of this function, so we have a trapezoid which contains the whole area lying below the graph. On the other hand, if we make a tangent in the midpoint $(\ln a+\ln b)/2$, we get a trapezoid which lies below the graph of this function. (See the picture in the pdf linked above.)
So we have:
$$\left(e^{\frac{\ln a+\ln b}2}\right) (\ln b-\ln a) \le \int_{\ln a}^{\ln b} e^x \le \frac{e^{\ln a}+e^{\ln b}}2 (\ln b-\ln a)\\
\sqrt{ab} \le \frac{b-a}{\ln b-\ln a} \le \frac{a+b}2
$$