Prove an inequality with a $\sin$ function: $\sin(x) > \frac2\pi x$ for $0<x<\frac\pi2$

$$\forall{x\in(0,\frac{\pi}{2})}\ \sin(x) > \frac{2}{\pi}x $$ I suppose that solving $ \sin x = \frac{2}{\pi}x $ is the top difficulty of this exercise, but I don't know how to think out such cases in which there is an argument on the right side of a trigonometric equation.

As one of the comments suggested, the easiest way is to draw a graph of sine and the line through $(0,0)$ and $(\frac{\pi}{2},1)$, and notice that one is above the other.

There's another way though; expanding on the hints above, consider the functions $f$ and $g$ defined by $$f(x) = \frac{\sin{x}}{x} \quad \text{and} \quad g(x) = x\cos{x} -\sin{x} $$ Then we have $$f'(x) = \frac{x\cos{x}-\sin{x}}{x^2} \quad \text{and} \quad g'(x) = -x\sin{x}$$ For $x \in [0,\frac{\pi}{2})$, we have $g'(x) \le 0$, so $g$ is decreasing. But we also have $g(0) = 0 $, so it follows that $g(x) \le 0$ on this interval. As a result, $f'(x) \le 0$ too, so $f$ is decreasing. As $x$ goes from (close to) $0$ to $\pi/2$, $f$ decreases from $1$ to $2/\pi$, and your result follows.

Hint: Consider the monotone property of $f(x)=\frac{\sin(x)}{x}$ on interval $[0, \pi/2]$.

Consider the properties of $\max$ and $\min$ of some function $f$,

$f = \dfrac{\sin x}{x}$ in this case.