How to show $ \sin x \geq \frac{2x}{\pi}, x \in [0, \frac{\pi}{2}]$? [duplicate]

I have tried the following: $$ f(x) = \sin x-\frac{2x}{\pi} \\ f'(x)= \cos x-\frac{2}{\pi} \\ f''(x) = -\sin x \leq 0 $$ But this doesn't seem to be heading in the right direction as it would appear $f'(x)$ is decreasing and has a zero within the domain, so I'm not sure how this might be used to prove the inequality.

Solution 1:

$\sin(x)$ is concave on that range and the line pass through $(0,0)$ and $(\pi/2,1)$ is $y=\frac{2}{\pi}x$

Solution 2:

Let $$f(x) = \frac{\sin{x}}{x} \quad\implies f'(x) = \frac{g(x)}{x^2}~~ \text{with} \quad g(x) = x\cos{x} -\sin{x} $$ and $$g'(x) = -x\sin{x}\le 0$$ For $x \in [0,\frac{\pi}{2})$, we have $g'(x) \le 0$, then $g$ is decreasing whereas $g(0) = 0 $.

Thus $g(x) \le 0$ on this interval. As a result, $f'(x) \le 0$ too, hence $f$ is decreasing on $[0,\frac{\pi}{2}]$. That is $$\frac{\sin{x}}{x} =f(x) \ge f(\pi/2) =\frac{2}{\pi}~~for ~~~0\le x\le \pi/2$$

Solution 3:

Note that $f$ first increases from $0$ to some value and then decreases again to $0$, so it must be greater than $0$ all this time. The precise value is where $f'(x)=0$ and the graph of $f(x)$ will roughly look like a parabola.BTW, i.e. $f(\arccos(2/\pi))=\sin(\arccos(2/\pi))-(2/\pi)\arccos(2/pi)=\sqrt{\pi^2-4}/\pi-(2/\pi)\arccos(2/pi)$.