Question: The question is

$$\frac{1}{\log\_4{\left(\frac{x+1}{x+2}\right)}}<\frac{1}{\log\_4{(x+3)}}$$

I did the first step for defining the arguments of both sides and got $x\in(-3,-2)\cup (-1,\infty)$

next I did reciprocal both sides and then what to do?

Answer: 
a solution is $$g(x) = \sum\_{n=-\infty, n \ne 0}^\infty e^{-n^4 (x-n)^2}$$

* it is real analytic since it is complex analytic
* it is $L^1$ since $$\|g\|\_{L^1} = \sum\_{n=-\infty}^\infty \int\_{-\infty}^\infty e^{-n^4 (x-n)^2} dx = \sum\_{n=-\infty, n \ne 0}^\infty \frac{1}{n^2}\int\_{-\infty}^\infty e^{-x^2} dx = 2\frac{\pi^2}{6}\sqrt{2\pi}$$
* and clearly $g(x) \ge 0$ and $\displaystyle\underset{|x| \to \infty}{\lim \sup} g(x) = 1$

[![enter image description here](https://i.stack.imgur.com/FLGxD.png)](https://i.stack.imgur.com/FLGxD.png)
The function

$$f(x) = \left ( \frac{2+\cos x}{3} \right) ^{x^4}$$

has this property. This follows from the fact that $\int\_0^{2\pi} [(2+\cos x)/3]^n\, dx \sim 1/\sqrt n$ as $n\to \infty$ (by Laplace's method).