Prove that $\sin(x)\geq\ln(x+1)$ for any $x\in[0,\frac{\pi}{2}]$ Prove that $\sin(x)\geq\ln(x+1)$ for any $x\in[0,\frac{\pi}{2}]$
I tried to consider $f(x)=\sin(x)-\ln(x+1)$ for $x\in[0,\frac{\pi}{2}]$, $f(0)=0,f(\frac{\pi}{2}) > 0$. Then $f'(x)=\cos(x)-\frac{1}{x+1}$, to anser my question i need to show, that exists some $c\in(0,\frac{\pi}{2})$: $f'(c)=0$, for any $x\in(0,c)$: $f'(x) > 0$ and  for any $x\in(c,\frac{\pi}{2})$: $f'(x) < 0$. I'm having trouble with this. 
Maybe there is simpler way to prove that $\sin(x)\geq\ln(x+1)$ for any $x\in[0,\frac{\pi}{2}]$?
 A: Actually you don't NEED to show that, but it is SUFFICIENT to show that (if you can).
As near $0$, $f'(x) = x +o(x)$, there exist $\epsilon$ such that $f(x) > 0$ for $x \in ]0;\epsilon[$., and $f'(\frac{\pi}{2}) < 0$, you know, using the Intermediate Value Thereom, that there is $\alpha > 0$ such that $f'(\alpha) = 0$. The shapes of the functions let you think that there is a unique solution in $]0;\frac{\pi}{2}[$.
To show that, we are gonna play by the fact that when we derive multiple time, the sign of the second term changes every time, while the first on changes every two times. 
We compute $f''(x) = -sin(x) + \frac{2}{(x+1)^2}$, $f'''(x) = -cos(x) - \frac{6}{(x+1)^3}$. So, $f''' < 0$, hence $f''$ is decreasing. $f''(0) > 0, f''(\pi/2)  < 0$ so $f''$ is increasing then decreasing. $f'(0) > 0$, so $f'(x) = 0$ has only one solution. Which completes the proof. 
A: Consider $f'(x) = \cos x - \frac1{x+1}$. Substitute $x = \frac\pi3$,
$$f'\left(\frac\pi3\right) = \cos \frac\pi3 - \frac{3}{\pi+3} > \frac12  - \frac{3}{3+3} = 0$$
Substitute $x=\frac\pi2$,
$$f'\left(\frac\pi2\right) = \cos\frac\pi2 - \frac{2}{\pi+2} < 0$$
By the intermediate value theorem, there is at least one $c \in \left(0,\frac\pi2\right)$ such that $f'(c) = 0$.
Consider $\cos x$ and $\frac1{x+1}$ separately. In the range $x\in\left[0,\frac\pi2\right]$, there are at least two intersections:
$$\cos 0 =1 = \frac1{0+1},\\
\cos c = \cos c - f'(c) = \frac1{c+1}$$
But $\cos x$ is concave downwards (in that range) and $\frac1{x+1}$ is concave upwards. This eliminates the possibility of other $x$ where $\cos x = \frac1{x+1}$, or $f'(x) = 0$.
Since $f'(x)$ is continuous, $f'(x) > 0$ for $x\in(0,c)$, and $f'(x) < 0$ for $x\in\left(c,\frac\pi2\right]$.
