How to prove that $|\ln(2+\sin(x)) - \ln(2+\sin(y))| <= |x-y| \space \forall \space x,y \in \mathbb R$ Question states
Prove that for all $x$ and $y$ $\in R$, the following inequality is true:
$\lvert \ln(2+\sin(x)) - \ln(2+\sin(y))\rvert \le \lvert x-y\rvert$
i've gotten to the point that
$\frac{y-x}{2+\sin(c)}  = \ln\frac{2+\sin(y)}{2+\sin(x)}$  (y-x divided by 2+sin(c) is my f dash c from the mean value theorem
I asked my teacher that i should use mean value theorem here so please don't use anything other than this, but i have no idea how to push this problem further.
Also this is my first post so i'm sorry for the f dash (c) thing, mathjax is hard

Let $f (t)=\ln (2+\sin (t)) $.
$f $ is differentiable at $\Bbb R, $ thus
by MVT,  for $x <y,$
$$f (x)-f (y)=(x-y)f'(c) $$
with  $x <c <y $ and
$$f'(c)=\frac {\cos (c)}{2+\sin (c)} $$
observe that
$$\frac {1}{1+(1+\sin (c))}\le 1$$
which gives
$$|f'(c)|\le |\cos (c)|\le 1$$
and
$$|f (x)-f (y)|=|x-y||f'(c)|\le |x-y|$$
