question from a test prove or find counterexample for the following theorem: $\forall x_1,x_2 \in (0,\frac{\pi }{2}):$$x_1<x_2 \Rightarrow  \frac{\sin x_2}{x_2}< \frac{\sin x_1}{x_1}$ 
Since I can't find any counterexample, it's probably a proof I need to give. So, I thought about declare a new function $f$ such that $f$:$[0,\pi/2]\to\mathbb R$, $f\left(x\right)\:=\:\frac{\sin x}{x}$ and show what the theorem want, but I'm stuck. Can someone help me here?
 A: Hint: Let $f(x)=\frac{\sin x}{x}$. We want to show $f$ is decreasing on our interval. That's a job that can be handled by the derivative. You will need to know that $\tan x\gt x$ in the interval $(0,\pi/2)$. There is a geometric argument for this, or else it is another job for the derivative. 
Added: From a comment, it seems that OP has trouble proving that $f'(x)\lt 0$ in the interval $(0,\pi/2)$. It is enough to show that $x\cos x-\sin x\lt 0$ in this interval, or equivalently (divide by $\cos x$) that $x\lt \tan x$. Let $g(x)=\tan x-x$. We have $g(0)=0$, and $g'(x)=\sec^2 x-1\gt 0$ for $x$ in the interval $(0,\pi/2)$. 
A: $\newcommand{\+}{^{\dagger}}
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$\ds{0 < x_{1} < x_{2} < {\pi \over 2}}$:
$$
{\sin\pars{x_{1}} \over x_{1}}
=\int_{0}^{1}\cos\pars{kx_{1}}\,\dd k
\quad >\quad 
\int_{0}^{1}\cos\pars{kx_{2}}\,\dd k
={\sin\pars{x_{2}} \over x_{2}}
$$

$$\color{#66f}{\large%
{\sin\pars{x_{2}} \over x_{2}} < {\sin\pars{x_{1}} \over x_{1}}}
$$

