Show that $ x< \sin(\frac{\pi x}2)\forall x\in (0,1)$ I don't know how to prove this inequality, some help would be appreciated, thank you!

$x&lt; \sin(\frac{\pi x}2)\quad\quad \forall x \in (0,1)$
Let we have to prove $$t&lt;\sin \left(\frac{\pi t}{2}\right)\;\forall t \in (0,1)$$

Now put $\displaystyle \frac{\pi t}{2} = x\Rightarrow t = \frac{2x}{\pi}$

So we have to prove $$\frac{2x}{\pi}&lt;\sin x\;\forall x \in \left(0,\frac{\pi}{2}\right)$$



Now Slope of $$\displaystyle \bf{OB} = m_{0B} = \frac{\sin x-0}{x-0} = \frac{\sin x}{x}.$$

and Slope of $$\displaystyle \bf{OA} = m_{0A} = \frac{1-0}{\frac{\pi}{2}-0} = \frac{2}{\pi}.$$

So Here in above graph $$\displaystyle \bf{m_{OB}&gt;m_{OA}}$$

So we get $$\displaystyle \frac{\sin x}{x}&gt;\frac{2}{\pi}\Rightarrow \frac{2x}{\pi}&lt;\sin x\;\forall x\in \left(0,\frac{\pi}{2}\right)$$
For $x\in(0,1)$ the function $\sin( \frac\pi2x) $ is an increasing function and we have
$$1&lt; \frac\pi2x$$
so $1&lt; \sin (\frac\pi2x) $