Show that there exists a constant $0<c<1$ such that
$$
\frac{\sin x}{x} < c,\quad\textrm{for all }x\ge1.
$$
–Context–
In Probability Theory Lévy's Theorem is crucial to uncover probability measures from certain functions which are obtained as limits of characteristic functions (Fourier transforms). The hard load of the work comes from Prokhorov's Theorem plus one key inequality involving the sinc function. In the 1-dimensional case it is enough to observe that ${\rm sinc}(x) \le \min(1,1/|x|)$. However, in the multivariate case a more accurate upper bound is needed, namely ${\rm sinc}(x) < c$ for some positive constant $c<1$.
It was in the middle of trying to adapt a proof of single-dimensional Lévy's Theorem to the multidiminsional case that I came across the need for this "detail". Since this function seemed related to the Laplace and Fourier transforms and I had found it in other contexts I decided to google it so to learn a little bit more. Here was where I learned how relevant this function is in e.g., Signal Processing (Kalman filters?). So I thought than instead of trying to proof the inequality myself it would be good to have it in SE so that others could start a "relationship" with this interesting function and its many properties. Obviously, I was plain wrong and I regret being so naïve.
Best Answer
$\frac {\sin x} x \to \sin 1 <1$ as $x \to 1$. Hence there exists $r>0$ such that $1 \leq x \leq 1+r$ implies $\frac {\sin x} x <a$ where $a=\frac {1+\sin 1} 2$. Also $\frac {\sin x} x \leq \frac 1 {1+r}$ for $x \geq 1=r$. Can you finish?