Find the smallest $a > 1$ such that $$\frac{a + \sin{x}}{a + \sin{y}} \leq \exp(y-x)$$ for all $x\leq y$.
I'm finding this tricky. I got $a = \displaystyle{\frac{e^\pi +1}{e^\pi -1}}$ but it's probably incorrect. My method was to maximise the LHS by letting $\sin{x}=1$ and $\sin{y}=-1$ and then minmise the RHS by letting $y-x = \pi$, and then I solved to find $a$.
Equivalently, you want $ae^x+\sin xe^x\leq ae^y+\sin y e^y$. Therefore, the function $f(x)=ae^x+\sin xe^x$ is increasing. But, $f'(x)=ae^x+\cos x e^x+\sin x e^x$, so $$ae^x+\cos x e^x+\sin x e^x\geq 0\Rightarrow a\geq-\cos x-\sin x=-\sqrt{2}\cos\left(x-\frac{\pi}{4}\right)\Rightarrow$$$$a\geq-\sqrt{2}\cos\left(\frac{5\pi}{4}-\frac{\pi}{4}\right)=\sqrt{2}$$ Conversely, if $a=\sqrt{2}$ then $f'\geq 0$, so the function is increasing, therefore your inequality holds. So the minimum value of $a$ is $\sqrt{2}$.
