Hyperbolic Trig Inequality 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.
