Find the smallest $a>1$ such that $\frac{a+\sin x}{a+\sin y} \leq e^{(y-x)}$ for all $x \leq y$ Can anyone please help me with the following question:
Find the smallest $a>1$ such that $$\frac{a+\sin x}{a+\sin y} \leq e^{(y-x)}$$ for all $x \leq y$
My attempt:
I think we should rearrange to get $a$. So I multiply both sides by $a+\sin y$ which is always positive since $a>1$, we get
$$a(1- e^{(y-x)}) \leq (\sin y)(e^{(y-x)}) - \sin x$$
we can divide by $1 - e^{(y-x)}$ because $x \leq y$, now we get
$$a \geq  \frac{(\sin y)(e^y) - (\sin x)(e^x)} {e^x - e^y}$$
so we have a necessary and sufficient condition.
I think we need to maximise the right hand side to find the answer, but I don't know how! This is from a undergraduate admission test (Trinity College Cambridge) so we can use only high school mathematics.
 A: Let $b = e^x$ and $c = e^y$. Then the problem is to find the supremum of all numbers of the form $f(c) - f(b)/(c-b)$ for $c > b > 0$, where $f(t) = - t \sin \log t$.
By the mean value theorem, such numbers are always of the form $f'(d)$. Conversely, any value $f'(d)$ of the derivative is itself the limit of numbers of the given form.
So the problem amounts to finding the supremum of $f'(t) = -\sin \log t - \cos \log t$ for $t > 0$. That means the maximum value of $-\cos x - \sin x = - \sqrt{2}\sin(x + \pi/4)$ for all $x$. 
Therefore take $a = \sqrt{2}$.
EDIT: A comment questioned the use of the mean value theorem. An alternative is to prove the following lemma: If $f'(t) \leq M$ for all $t$, then $(f(c) - f(b))/(c-b) \leq M$, which can fulfill the same purpose in this problem. This can be proved using only the fact that a function with nonnegative derivative is increasing (i.e., nondecreasing). 
A: $y=x+t \implies t\ge 0 \iff a\ge \dfrac{\sin{x}-e^t\sin{(x+t)}}{e^t-1}=\dfrac{\sin{x}-e^t\sin{x}\cos{t}-e^t\sin{t}\cos{x}}{e^t-1}=\dfrac{(1-e^t\cos{t})\sin{x}-e^t\sin{t}\cos{x}}{e^t-1}=A(t)\sin{(x-g(t))}$
$g(t)=\arcsin{\dfrac{e^t\sin{t}}{\sqrt{(1-e^t\cos{t}))^2+(e^t\sin{t})^2}}} , \sin{(x-g(t))} \le 1 $
$A(t)=\dfrac{\sqrt{(1-e^t\cos{t}))^2+(e^t\sin{t})^2}}{e^t-1}=\sqrt{\dfrac{1-2\cos{t}e^t+e^{2t}}{(e^t-1)^2}}$
now $a\ge A_{max}(t)$ means to find max of $f(t)=\dfrac{1-2\cos{t}e^t+e^{2t}}{(e^t-1)^2}$
$f(t)=1+\dfrac{2e^t(1-\cos{t})}{(e^t-1)^2}=1+\dfrac{4e^t\sin^2{\dfrac{t}{2}}}{(e^t-1)^2}\le 1+\dfrac{4\sin^2{\dfrac{t}{2}}}{t^2} \le 2 \iff \dfrac{\sin{\frac{t}{2}}}{\frac{t}{2}} \le 1 \cap \dfrac{(e^t-1)^2}{e^t} \ge t^2$
$\dfrac{(e^t-1)^2}{e^t} \ge t^2 \iff h(t)=e^{2t}+1-2e^t-t^2e^t \ge 0 \iff h'(t) \ge 0\cap h(0)=0 \iff h'(t)=2e^t(e^t-t-1) \ge0 \iff e^t \ge t+1$
the last one is trivial or one can do $k(t)=e^t-t-1\ge 0 \iff k'(t)= e^t-1 \ge 0 \cap k(0)=0 $
so $a \ge \sqrt{2}$ and "=" will hold when $t \to 0^+\cap x-g(t)=2n\pi+\dfrac{\pi}{2} $
