Prove that $\sin x+2x\ge\frac{3x(x+1)}{\pi}$ for all $x\in [0,\frac{\pi}{2}]$ Question: Prove that $\sin x+2x\ge\frac{3x(x+1)}{\pi}$ for all $x\in [0,\frac{\pi}{2}]$
How did $\pi$ come in the expression? 
This is how I tried to solve.
$\sin x\le x$ so $\sin 2x\le 2x$
$\sin x+2x\le 3x$.
Applying AM GM inequality: $\sin x+2x\ge 2\sqrt{2x\sin x}$
$3x\ge 2\sqrt{2x\sin x}$
This how far I can go.
 A: Just expanding mastrok's answer, over $I=\left(0,\frac{\pi}{2}\right)$ we have:
$$ \frac{d^2}{dx^2}\left(\sin x+2x\right) = -\sin x<0 \tag{1}$$
Hence $f(x)=\sin(x)+2x$ is a concave function, and since:
$$ \frac{d^2}{dx^2}\left(\frac{3}{\pi}x(x+1)\right)=\frac{6}{\pi}>0\tag{2}$$
$g(x)=\frac{3}{\pi}x(x+1)$ is a convex function. We have $f(0)=g(0)=0$ and since $\pi>2$:
$$ f\left(\frac{\pi}{2}\right)=\pi + 1>\frac{3}{2}+\frac{3\pi}{4}=g\left(\frac{\pi}{2}\right)\tag{3}$$
proving that:
$$ \forall x\in I,\qquad f(x)>\left(2+\frac{2}{\pi}\right)x>\left(\frac{3}{2}+\frac{3}{\pi}\right)x>g(x).\tag{4}$$
A: $\displaystyle \sin x +2x \ge \left(2+\frac{2}{\pi}\right)x $ for $\displaystyle x\in \left[0,\frac{\pi}{2}\right]$ , Since $\sin x +2x$ is a concave function
To prove $\displaystyle \left(2+\frac{2}{\pi}\right)x\ge \frac{3x(x+1)}{\pi}$ in the range, just verify that $\displaystyle \frac{3x(x+1)}{\pi}$ is a convex function 
and the inequality holds at end points.
A: The $\bf{L.H.S}$ is Concave and its graph passes through $(0,0)$ and $\displaystyle \left(\frac{\pi}{2},1+\pi\right)$. 
Hence it is stay above the line Connecting These $2$ points.
And $\bf{R.H.S}$ is Convex and its graph passes through $(0,0)$ and $\displaystyle \left(\frac{\pi}{2},\frac{3}{2}+\frac{3\pi}{4}\right)$
Hence it is stay below the line connecting these $2$ points.
bcz $\displaystyle 1+\pi>\frac{3}{2}+\frac{3\pi}{4}$. So the first line is above the  second line , Giving the results.
