I have the following inequality:
$x < \dfrac{-z(x-1)(1+\cosh(xy))}{(1-z)(1+\cosh(y-xy))}$
where we know $x \in [0,1], y > 0, z \in (0,1)$ and $x, y, z \in \mathbb{R}$.
Since $x \not<0 \implies (x-1) \neq 0 \implies (x-1) < 0$, this can also be written as:
$\dfrac{x}{x-1} > \dfrac{-z(1+\cosh(xy))}{(1-z)(1+\cosh(y-xy))} \iff \dfrac{x}{1-x} < \dfrac{z(1+\cosh(xy))}{(1-z)(1+\cosh(y-xy))}$
although I'm not sure this helps.
I need to rewrite this in terms of just $x$. So something like $x < f(y,z)$ is what I am looking for.
I am completely stuck on how to deal with $\cosh(xy)$ and $\cosh(y-xy)$. I don't even need a perfect bound for $x$, if I can get any improvement on the given bound of 1, that would be significant progress.
Is there any straightforward way to deal with products inside hyperbolic trig functions?