fall 2010 problem 2

fall 2010 problem 2
Show that sin(x) >= x-(x^2)/pi.  x belongs to [0,pi].
This problem seems very easy, but spend a lot of my time to finish it.  But I post this solution due to the deadline.
let f(x)=sin(x)-x+(x^2)/pi, and going to prove f(x)>=0, which is a necessary and sufficient condition of statement.


[ English is not my first language, so it may hard to read my prove. Then you be better guess what I was going to say ]

pf:
let f(x)=sin(x)-x+(x^2)/pi. then let x=t+pi/2, t belongs to [-pi/2, pi/2].
g(t)=f(t+pi/2)=sin(t+pi/2)+((t+pi/2-pi/2)^2)/pi-pi/4. 
so g(t) defines on (-pi/2, pi/2), which is a even function.
Simplify g(t)=cos(t)+((t+pi/2-pi/2)^2)/pi-pi/4.
Here going to prove g(t) >= 0 on [0,pi/2].
Consider g'(t)= -sin(t)+2*t/pi. 

Jordan's inequality said sin(x)>= 2x/pi, for 0< x < pi/2.

g'(t) is strictly decreasing on the interval of (0,pi/2). Because g(t) -> 0 as t->pi/2. That means g(t)>=0 on the interval of (0,pi/2). 
Since g(t) is a even function on (0,pi/2). So g(t)>=0 on (-pi/2,pi/2). Then f(x)>=0.  
 #


I also think about the f'(t)=cos(x)-1+2x/pi= -2(sin(x/2))^2+2x/pi. but it seems hard to separate the interval [0, pi] into [0,pi/2] and [pi/2, pi]. Then prove f'(x) is greater or equal to zero on [0,pi/2],  and less or equal to zero on [pi/2,pi/2]. I think it is easy to follow. But that will look a little complicate to show. If there is somebody know how to, please tell me.