prove or find counterexample for the following theorem: $\forall x_1,x_2 \in (0,\frac{\pi }{2}):$$x_1<x_2 \Rightarrow \frac{\sin x_2}{x_2}< \frac{\sin x_1}{x_1}$
Since I can't find any counterexample, it's probably a proof I need to give. So, I thought about declare a new function $f$ such that $f$:$[0,\pi/2]\to\mathbb R$, $f\left(x\right)\:=\:\frac{\sin x}{x}$ and show what the theorem want, but I'm stuck. Can someone help me here?
On
$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{0 < x_{1} < x_{2} < {\pi \over 2}}$: $$ {\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}}} $$
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 $(0,\pi/2)$.