Prove that for `x in [0, (pi)/(2)], sin x + 2x ge (3x(x + 1))/(pi)`.

Explanation

Let `f(x)=sinx+2x-(3x(x+1))/(pi)`
`therefore f(x)=cosx+2-(3)/(pi)(2x+1)`
`=cosx-(6x)/(pi)+(3)/(pi)-2_`
Let f(x)=0
`therefore cosx=(6x)/(pi)+(3)/(pi)-2`
Graphs of y = cos x and y =`(6)/(pi)+(3)/(pi)-2` are as shown in the following figure.

There is only one point of intersection for which x=`x_(0)`
Clearly ofor `x lt x_(0), cos x gt (6x)/(pi)+(3)/(pi)-2.so f(x)gt0`
For `xgtx_(0)f(x)lt0`
So, `x=x_(0)` is point of maxima.
Now `f(0)=0 and f((pi)/(2))=(pi)/(2)-(1)/(2)`
So , graph of the function is as shown in the following figure

From the graph `f(x)ge0 for x in [0,(pi)/(2)]`

`sin x+2x-(3x)/(pi)(x+1)ge0 for x in [(0,(pi)/(2))]`