Finding solution set of $\frac{1}{\log_4 \left(\frac{x+1}{x+2}\right)}<\frac{1}{\log_4(x+3)}$ without using derivatives

Question

$$\frac{1}{\log_4\left(\frac{x+1}{x+2}\right)}\lt \frac{1}{\log_4(x+3)}$$

This inequality can be solved by using the monotonicity of $f(x)$ on $x\in(-1 ,\infty)$ where $f(x)=\frac{1}{\log_4\left(\frac{x+1}{x+2}\right)}-\frac{1}{\log_4(x+3)}$ and using the value of the function at any random point. Using this, the solution set comes out to be $(-1,\infty)$. Can the inequality be solved more simply and without using methods of brute force? Any hints are appreciated. Thanks

Answer 1

The domain gives $$(-3,-2)\cup(-1,+\infty).$$

Now, $$\frac{1}{\log_4\left(\frac{x+1}{x+2}\right)}=\frac{1}{\log_4(x+3)}$$ gives $$(x+2)^2=0,$$ which is impossible.

Thus, it's enough to check our inequality for some value $-3<x<-2$

and for some value $x>-1$.

$x=-2.5$ does not play, but $x=0$ is valid, which gives the answer: $$(-1,+\infty).$$