The following hyperbolic trig inequality came up. $$0 \leq y \leq x \leq 2 \implies \sinh(x)-\sinh(y) \leq \sinh(x-y)\cdot e^{xy/2}.$$
I spent many hours trying to prove it. The first few terms of the Taylor series work out, but I couldn't get a general proof. I passes the usual sanity checks ($y=0$, $y=x$, $x=2$) and I also verified it numerically by plotting and checking about $10^9$ random $x$ and $y$ values.
Any ideas?
Completing Brevan Ellefsen's solution.
First, since $\sinh x - \sinh y = 2 \sinh \frac{x-y}{2} \cosh \frac{x+y}{2}$, and $\sinh (x-y) = 2 \sinh \frac{x-y}{2} \cosh \frac{x-y}{2}$, we have $$ r\equiv\frac{\sinh x - \sinh y}{\sinh (x-y)} = \frac{\cosh \frac{x+y}{2} } {\cosh \frac{x-y}{2}} = \frac{1 + \tanh\frac x 2 \tanh \frac y 2 } { 1 - \tanh\frac x 2 \tanh\frac y 2}. $$ For $0 \le \dfrac{x}{2}, \dfrac{y}{2} \le 1$, we have $$ 0 \le \tanh\frac{x}{2} \cdot \tanh\frac{y}{2} \le \frac{x}{2} \cdot \tanh\frac{y}{2} \le \tanh\left(\frac{x}{2} \cdot \frac{y}{2}\right) \le 1. $$ Since $\dfrac{1+t}{1-t}$ is an increasing function of $t$ for $t \in (0, 1)$, we get $$ r \le \frac{1 + \tanh\frac{xy}{4}}{1-\tanh\frac{xy}{4}} =\exp\frac{xy}{2}. $$ This completes the proof.