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$\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.

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    See also: http://math.stackexchange.com/questions/596634/mean-value-theorem-frac2-pi-frac-sin-xx12015-11-22

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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.

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    Perhaps it would have been better for me to have defined $g$ after finding $f'$. The motivation for choosing $g$ to be what I did is that then $f'(x) = \frac{g(x)}{x^2}$. Basically we want to conclude that $f$ is decreasing, but we don't see that immediately from its derivative, so we introduce $g$ to figure things out.2012-10-15
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Hint: Consider the monotone property of $f(x)=\frac{\sin(x)}{x}$ on interval $[0, \pi/2]$.

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

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