Prove: $f(x) = \sin(x) + 2x$ is monotonically increasing

Question

Prove, without using derivatives, that

$$f(x) = \sin(x) + 2x$$

is monotonically increasing.

My intuition says it has something to do with the fact that $|\sin(x)| \leq |x|$ but I couldn't figure it out.

Answer 1

Let $x<y$. Then $|\sin y -\sin x | =2|\sin (\frac {y-x} 2) \cos (\frac {y+x} 2)|\leq 2(\frac {y-x} 2)=y-x$ from which the result follows.

Answer 2

Note:

$\sin x -\sin y =$

$2\cos (x+y)/2 \sin (x-y)/2$;

Let $x >y$:

$f(x)-f(y)=$

$2(x-y)+$

$2\cos (x+y)/2\sin (x-y)/2 \gt $

$2(x-y)-2\cdot 1 |\sin (x-y)/2| \gt$

$2(x-y)- |x-y| = x-y >0$.

Answer 3

The statement is true for $\sin x$ increasing, then consider the case with $\sin x$ strictly decreasing.

We have that forall $\epsilon>0$ arbitrarily small

$$\sin (x+\epsilon) + 2 (x+\epsilon) > \sin x + 2 x \iff \sin x -\sin (x+\epsilon)< 2\epsilon$$

and by sum to product identities

$$\sin x-\sin (x+\epsilon)=-2\sin\left(\frac \epsilon 2\right)\cos\left(x+\frac \epsilon 2\right)<2\epsilon$$

which is trivially true for $\cos\left(x+\frac \epsilon 2\right)\ge 0$ then consider wlog

$$\frac \pi 2 <x+\frac \epsilon 2<\pi$$

and by $x=\frac \pi 2+y$ with $ 0 <y+\frac \epsilon 2<\frac \pi 2$

then

$$\cos\left(x+\frac \epsilon 2\right)=\cos\left(\frac \pi 2+y+\frac \epsilon 2\right)=-\sin\left(y+\frac \epsilon 2\right)$$

that is

$$2\sin\left(\frac \epsilon 2\right)\sin\left(y+\frac \epsilon 2\right)\le 2\cdot \frac \epsilon 2\cdot 1=\epsilon<2\epsilon$$

Answer 4

Note : $|\sin x- \sin y|\leq |x-y|$ for all $x,y \in \mathbb{R}$.

For the function $f(x)=\sin x+2x$ $\quad$ and let $x<y$ then

$\sin x-\sin y \leq |\sin x- \sin y|\leq |x-y|$

$\implies \sin x-\sin y +2(x-y) \leq |\sin x- \sin y|+2(x-y) \leq |x-y|+2(x-y)$

$\implies f(x)-f(y) \leq |x-y|+2(x-y)$

$\implies f(x)-f(y)\leq -x+y+2(x-y)$

$\implies f(x)-f(y) \leq x-y<0$

$\implies f(x)<f(y)$.