Which is bigger, $ \log_{1000} 1001$ or $\log_{999} 1000 $? 
Which is bigger, $ \log_{1000} 1001$ or $\log_{999} 1000 $?

I've tried using the identity of $\log_n x = \dfrac 1 {\log_x n} $, but didn't find a solution though. Any suggestions? or clues I can use?
 A: Use the fact that $\log_nx=\frac{\log x}{\log n}$ and then pick your favourite way to show that
$$\log(999)\log(1001)<(\log(1000))^2.$$
A: Note that $$\frac{d}{dx} \frac{\log(x+1)}{\log x} = \frac{\frac{\log x}{x+1} - \frac{\log (x+1)}{x}}{\log^2 x} < 0,$$ so $\log_{999} 1000$ is larger.
A: The question is, for $n=1000$, which is larger
$$\frac{\ln(n+1)}{\ln n}\qquad\text{or}\qquad\frac{\ln(n)}{\ln(n-1)}?$$
Or equivalently, is
$$\ln\ln(n+1)-2\ln\ln n+\ln\ln(n-1)$$
positive or negative?
Let $f(x)=\ln\ln x$. Then
$$f'(x)=\frac{1}{x\ln x}.$$
This is decreasing for $x>1$, and so $f$ is concave. Then
$f(n+1)-2f(n)+f(n-1)<0$
and so
$$\log_{1000}1001<\log_{999}1000.$$
A: Let  $$ f(x)= \log _{x}(x+1) = \frac {\ln (x+1)}{\ln (x)} \text { for } x>1$$  Upon differentiation we get 
$$ f'(x) = \frac {x\ln x -(x+1) \ln (x+1) }{x(x+1) (\ln x)^2} <0$$
Thus f(x) is decreasing.
As a result $$ log_{1000}1001<log_{900}1000 $$
A: By AM-GM
$$\log_{999}1000-\log_{1000}{1001}=\frac{\ln1000}{\ln999}-\frac{\ln1001}{\ln1000}=\frac{\ln^21000-\ln1001\ln999}{\ln999\ln1000}\geq$$
$$\geq\frac{\ln^21000-\left(\frac{\ln1001+\ln999}{2}\right)^2}{\ln999\ln1000}=\frac{\ln^21000-\ln^2\sqrt{1001\cdot999}}{\ln999\ln1000}=$$
$$=\frac{\ln^21000-\ln^2\sqrt{1000^2-1}}{\ln999\ln1000}>\frac{\ln^21000-\ln^2\sqrt{1000^2}}{\ln999\ln1000}=0.$$
A: Let $x\ge2$. If we write $a^2=x-1$ and $b^2=x+1$ with the understanding that $a,b\ge1$, then we see that
$$1\lt\sqrt3=\sqrt{4-1}\le\sqrt{x^2-1}=ab=\sqrt{x^2-1}\lt x$$
Thus we have $0\lt\log(ab)\lt\log x$ as well as $0\le\log a\lt\log b$. In particular, the positivity of the logs implies $(\log(ab))^2\lt(\log x)^2$, which we'll use shortly.
Using the Arithmetic-Geometric Mean inequality at a critical juncture (as in Michael Rozenberg's answer), we have
$$\begin{align}
\log(x-1)\log(x+1)
&=\log(a^2)\log(b^2)\\
&=4\log(a)\log(b)\\
&\le(\log a+\log b)^2\quad\text{(by AGM)}\\
&=(\log(ab))^2\\
&\lt(\log x)^2
\end{align}$$
Applying this with $x=1000$, we find
$$\log_{1000}1001={\log1001\over\log1000}\lt{\log1000\over\log999}=\log_{999}1000$$
