Showing that $x+ cos x - 1 > 0$ for all $x > 0$ I got this problem:
Show that for all $0<x$, $0<x+cos x - 1$
I tried to show it several times but none worked.
I showed that $lim_{x\to\infty} (x+cos x - 1) = \infty$ by using the Squeeze theorm for infinite limits. And so I got that for each $0<M$ there exist $0<N$ such that for all $N<x$, $M<x+cos x -1$. But I was not able to show that for all $0<x$, $0<x+cos x - 1$.
Then I tried to show it using derivative but it seems not to help.
 A: Let $f(x)=x+\cos x-1$. We have $f(0)=0$. Show that $f'(x)\ge 0$ for $x\gt 0$. It will follow that $f$ is non-decreasing on the interval $[0,\infty)$.
Note that $f'(x)\gt 0$ in, say, the interval $[0,\pi/6)$. So $f$ is strictly increasing in this interval, and is therefore positive for all $x\gt 0$. (Actually, $f$ is strictly increasing on the whole interval $[0,\infty)$.)
Remark: The strategy we have used is a standard one. To show that $g(x)\gt h(x)$ on an interval, it is often useful to let $f(x)=g(x)-h(x)$, and study the increasing/decreasing behaviour of $f(x)$ with the usual tools. 
A: If $f\left(x\right)=x+\cos x-1$ then $f'\left(x\right)=1-\sin x\geq 0$
showing that the function is nowhere decreasing. Secondly $f'(x)>0$  on
$\left(0,\pi\right)$, showing that it is increasing on that interval. This together implies that $f\left(x\right)>f\left(0\right)=0$ for $x>0$.
A: Actually let f(x)=x+cos(x)-1,f '(x)=1-sin(x)$\geqslant$0,so f(x) is an increasing function,noticed that f(0)>0,so,f(x)>0 is true when x>0.
