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

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?

• The inequality can be written as $\int_y^x \frac{1}{t}dt < \frac{x-y}{\sqrt{xy}}$. I wonder if one can prove it by comparing areas, and the shape of $\frac{1}{x}$. – N. S. Nov 22 '13 at 15:46
• This seems related to Logarithmic mean. – Martin Sleziak Nov 22 '13 at 15:49
• @N.S. This demonstration on WA site seems to be based on geometric idea. It was the first search result when I googled for logarithmic geometric mean inequality. – Martin Sleziak Nov 22 '13 at 16:22
• Very nice, that does it! – N3buchadnezzar Nov 22 '13 at 16:36

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

• Yes, the inequality becomes $(t-1)/\log t>\sqrt{t}$ for $t>1$. A possible simplification is to set $x/y=t^2$, instead, just to get rid of the square root. – egreg Nov 22 '13 at 15:11
• And how does one show the last equality? I proved that they both are zero as $t\to 1$. But showing that the derivative was increasing greater than zero was hard. – N3buchadnezzar Nov 22 '13 at 15:18

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.

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.

• I think you need $t^2$'s on the left hand side of the first inequality. – Barry Cipra Nov 22 '13 at 22:37
• @BarryCipra Yes, thanks. – egreg Nov 22 '13 at 22:38

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

Inequality has many demonstrations. Some are made ​​using logarithmic representation with integral. Give here an example: $$\frac{1}{L( a, b )} = \frac{1}{b - a}\int^{\frac{b}{a}}_{1}\frac{dx}{x}$$ Integrating inequality: $$\frac{4}{(x+1)^2}\leq\frac{1}{x}\leq\frac{x+1}{2x\sqrt{x}}$$ is obtained $$\frac{b-a}{b+a}\leq\frac{1}{2}\ln\frac{b}{a} \leq\frac{b-a}{2\sqrt{ab}}$$ hence the inequality in question, but more logarithmic mean framing between the arithmetic mean and geometric mean:$$\sqrt{ab} < \frac{b-a}{\ln b-\ln a}<\frac{a+b}{2}.$$ $(0<a<b)$