I want to check the monotonicity of the function for $x>0$ $$\frac{\cosh 2 x^3 }{3\cosh 5 x^3 }$$ Computing the first derivative, it can be proved that it is negative and then the function is decreasing.
My question is can we claim that since $\,\cosh x^3\,$ is an increasing function for $x>0$, and since the numerator is less than the denominator, then, the function is decreasing?
We have that by $y=x^3>0$
$$\frac{\cosh 2 x^3 }{3\cosh 5 x^3 }=\frac{\cosh 2y }{3\cosh 5 y }=\frac{e^{3y}(e^{4y}+1)}{3(e^{10y}+1)}$$
and $e^y=z>1$
$$\frac{e^{3y}(e^{4y}+1)}{e^{10y}+1}=\frac{z^3(z^4+1)}{z^{10}+1}=f(z)$$
and
$$f'(z)= \frac{z^2(-3z^{14}-7z^{10}+7z^4+3)}{(z^{10}+1)^2}$$
and
$$-3z^{14}-7z^{10}+7z^4+3=-3(z^{14}-1)-7z^4(z^6-1)<0$$