Proof of the following inequality $ \frac{x - y}{\log x - \log y} > \sqrt{xy} $, $x>y$.

Since $1/x$ is concave for $x>0$ then \begin{align*} \log b - \log a = \int_b^a\frac{\mathrm{d}x}{x} < \underbrace{ \frac{1}{2} \frac{b - a}{\sqrt{ba}} }_\text{Blue Area} +\underbrace{ \frac{1}{2}\frac{b - a}{\sqrt{ba}} }_{\text{Red area}} = \frac{b - a}{\sqrt{ba}}\end{align*} Implying $$ \hspace{4cm}\sqrt{ab} < \cfrac{b - a}{\log b - \log a} \hspace{4cm} \blacksquare$$

Hint: let $\dfrac{x}{y}=t>1$ and $$\Longleftrightarrow f(t)={t-1}-{\ln{t}}\cdot \sqrt{t}$$

and follow is easy to prove it

Writing $x=t^2y$ with $t\gt1$, the inequality to be proved can be written as

$$t-{1\over t}-2\log t\gt0$$

Letting $f(t)$ be the expression on the left, we see that $f(1)=0$ and

$$f'(t)=1+{1\over t^2}-{2\over t}={(t-1)^2\over t}\gt0 \text{ for }x\gt1$$

This means $f$ is strictly increasing, making it necessarily positive.

Note: This answer is along the lines of the approaches taken by math110 and egreg. The main difference was to write the inequality in a way that suggested a function that's easy to differentiate and show is always positive.

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: $$\left(e^{\frac{\ln a+\ln b}2}\right) (\ln b-\ln a) \le \int_{\ln a}^{\ln b} e^x \le \frac{e^{\ln a}+e^{\ln b}}2 (\ln b-\ln a)\\ \sqrt{ab} \le \frac{b-a}{\ln b-\ln a} \le \frac{a+b}2 $$

With the substitution $x/y=t^2$, the inequality becomes

$$ \frac{t^2-1}{\log t^2}>t $$ that, for $t>1$, is equivalent to $$ t^2-1>2t\log t. $$

Consider $f(t)=t^2-1-2t\log t$, defined for $t\ge1$; we have $f(1)=0$ and the derivative is $$ f'(t)=2t-2-2\log t. $$ I claim that $f'(t)>0$ for $t>1$, so the function $f$ is increasing. It's easy to see that $f'(1)=0$ and $\lim_{t\to\infty}f'(t)=\infty$.

Since $$ f''(t)=2-\frac{2}{t}>0 $$ for $t>1$, the function $f'$ is increasing, so it's everywhere positive (except at $1$).

Of course the proof with convexity is better.