How to prove that $|\ln(2+\sin(x)) - \ln(2+\sin(y))| <= |x-y| \space \forall \space x,y \in \mathbb R$

Question

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

Answer 1

Let $f(x)=\log(2+\sin x)$. Then $f'(x)=\frac{\cos x}{2+\sin x}$. So, if $x,y\in\mathbb R$ and $x\neq y$, then$$\frac{f(y)-f(x)}{y-x}=\frac{\cos c}{2+\sin c}$$for some $c$ between $x$ and $y$, and therefore$$\left|\frac{f(y)-f(x)}{y-x}\right|=\left|\frac{\cos c}{2+\sin c}\right|\leqslant\frac{|\cos c|}{2-|\sin c|}\leqslant1.$$

Answer 2

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 $$|f (x)-f (y)|=|x-y||f'(c)|\le |x-y|$$

Answer 3

Let $f(x)=log(2-sin(x))$, which is indeed continuous and differentiable on $\mathbb{R}$.

Note that $f'(x)=\frac{cos(x)}{2 + sin(x)}$ ,

and it's easy to show that $-1\leqslant f'(x)\leqslant 1$. (since $-1\leqslant cos(x),sin(x)\leqslant 1$)

By mean value theorem, for $x,y\in \mathbb{R}$,

$\frac{f(x)-f(y)}{x-y}=f'(z)$ for some $z\in(x,y)$,

thus $-1\leqslant \frac{f(x)-f(y)}{x-y}\leqslant 1$,

and $|\frac{f(x)-f(y)}{x-y}|\leqslant 1$,

finally, $|{f(x)-f(y)}|\leqslant |{x-y}|$.

Answer 4

I thought it might be instructive to present a way forward that relies on only elementary, pre-calculus tools. To that end we proceed.

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First, note that $\log(2+\sin(x))-\log(2+\sin(y))=\log\left(\frac{2+\sin(x)}{2+\sin(y)}\right)$.

Now, in THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the iequalities

$$\bbox[5px,border:2px solid #C0A000]{\frac{t-1}{t}\le \log(t)\le t-1}$$

for all $t>0$. Hence, with $t=\frac{2+\sin(x)}{2+\sin(y)}$, we have

$$\frac{\sin(x)-\sin(y)}{2+\sin(x)}\le \log(2+\sin(x))-\log(2+\sin(y))\le \frac{\sin(x)-\sin(y)}{2+\sin(y)}\tag1$$

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It is evident from $(1)$ that

$$\left|\log(2+\sin(x))-\log(2+\sin(y))\right|\le |\sin(x)-\sin(y)|\tag2$$

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Next, using the Prosthaphaeresis Identity, $\sin(x)-\sin(y)=2\cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)$ along with $|\cos(\theta)|\le 1$ and $|\sin(\theta)|\le \theta$ reveals

$$|\sin(x)-\sin(y)|\le |x-y|\tag 3$$

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Finally, using $(3)$ in $(2)$, we obtain the coveted inequality

$$\bbox[5px,border:2px solid #C0A000]{\left|\log(2+\sin(x))-\log(2+\sin(y))\right|\le |x-y|}$$