Prove that $x+\sin x$ is strictly increasing

I have a function $f(x)=x+\sin x$ and I want to prove that it is strictly increasing. A natural thing to do would be examine $f(x+\epsilon)$ for $\epsilon > 0$, and it is equal to $(x+\epsilon)+\sin(x+\epsilon)=x+\epsilon+\sin x\cos \epsilon + \sin \epsilon \cos x$.

Now all I need to prove is that $x+\epsilon+\sin x\cos \epsilon + \sin \epsilon \cos x - \sin x - x$ is always greater than $0$ but it's a dead end for me as I don't know how to proceed. Any hints?

Let $f(x)=x+\sin x$. Then $f'(x)=1+\cos x\geq 0$ and:
$$f(x+h)-f(x) = h\, f'(\xi),\quad \xi\in(x,x+h)$$ by Lagrange's theorem, hence $f(x+h)-f(x)\geq 0$.
In order to prove that the inequality is strict, we can notice that: $$f(x+h)-f(x-h) = 2h + 2\cos x \sin h$$ can be zero only if $\cos x=-1$ and $\sin h=h$, i.e. for $h=0$.
• Could you please show why $\cos x=-1$ (I see why $h=0$.) Commented Mar 8, 2015 at 20:35
• @user84413: if $h>0$ and $|\cos x|<1$ then the RHS is positive for sure, since $|\sin h|\leq h$. So the only possibilities are given by $\cos x=\pm 1$, and it is quite easy to check that $\cos x=+1$ does not lead to any solution different from $h=0$, too. Commented Mar 8, 2015 at 20:37
differentiate: you get $1 + \cos (x)$ this is positive except on a discrete set of points. Integrate it and you get a strictly increasing function.