Solving $\cos(\log_{4x}(x+1))-\cos(\log_{4x}(4-x))\lt\log_{4x}(4-x)-\log_{4x}(x+1)$ There is an interesting inequality I've stumbled upon on the Internet. It has logarithms and trigonometry, but in contrast to something like this that uses trigonometry and logarithms separately, this one takes such kind of inequality to a whole new level!
See for yourself. The task is to solve the following inequality over reals:
$$\cos(\log_{4x}(x+1))-\cos(\log_{4x}(4-x))\lt\log_{4x}(4-x)-\log_{4x}(x+1)$$
I've found the domain of the inequality by applying restrictions to real-valued logarithms: its base must not be neither zero nor one, and its argument must be positive.
$\begin{cases}4x\gt0,\\4x\neq1,\\4-x\gt0,\\x+1\gt0.\end{cases}\Leftrightarrow\begin{cases}x\gt0,\\x\neq0.25,\\x\lt4,\\x\gt-1.\end{cases}\Leftrightarrow x\in(0;0.25)\cup(0.25;4)$
I tried to use the fact that the difference of cosines has the domain $[-2;2]$ no matter the arguments of the cosines. My assumption comes from the following facts:

*

*The domain of $\cos(x)$ is $[-1;1]$.

*The smaller the subtrahend, the bigger the difference.

*The bigger the subtrahend, the smaller the difference.

Thus, the domain of $\cos(a)-\cos(b)$ is $[\min(\cos(a))-\max(\cos(b));\max(\cos(a))-\min(\cos(b))]=[-1-1;1-(-1)]=[-2;2]$. QED.
This did not work, however, because the right hand side sadly isn't always greater than two.
From a mere observation, if we let $u=\log_{4x}(x+1)$, $v=\log_{4x}(4-x)$, and $f(x)=cos(x)$, then the inequality becomes $f(u)-f(v)\lt v-u$, leading to the inequality $f(u)+u\lt f(v)+v$. This may be useful in some way, I guess.
I've also tried to step away from arbitrary logs and only use natural ones:
$$\cos\left(\dfrac{\log (x+1)}{\log (4x)}\right)-\cos\left(\dfrac{\log (4-x)}{\log (4x)}\right)\lt\dfrac{\log (4-x)-\log(x+1)}{\log (4x)}$$
...and I'm stuck here. No way to manipulate those cosines of which I'm aware.
So, how to approach this kind of inequalities properly?

By inspection of log arguments the crucial roots/critical points of equality  are noted at:
$$ x= (0,\frac14,\frac32,4)$$
We compare the two component functions $ f(x),g(x)$ ( blue,red) defined/ listed in Mathematica, intervals are determined by above end points.
By evaluating the function signs in the neighborhood of the roots ( it suffices to track function behaviour) it is found :
In the interval $(\frac14,\frac32)\; f(x)< g(x) $.
Outside this full interval  $ f(x) , g(x) $ define all intervals of inequality. The plot also confirms the intervals.

