How to show that $\frac{\sinh(b)}{\sinh(a)} \leq \frac{b}{a}$ for $a\geq b>0$.

Question

I want to prove that $\frac{\sinh(b)}{\sinh(a)} \leq \frac{b}{a}$ where $a\geq b>0$ and $a,b \in \mathbb{R}$. I have tried to set $a = bx$ for $x\geq 1$ but did not manage to show that the inequality holds. Also, I did not manage to prove it via the identity $\frac{\sinh(b)}{\sinh(a)} = \frac{\sinh(b)}{\int_{0}^{a} \cosh(x) dx} = \frac{\sinh(b)}{\int_{0}^{b} \cosh(x) dx + \int_{b}^{a} \cosh(x) dx}$.