Prove $\frac{\sin (2016x)}{2016x}<\frac{\sin x}{x}$ How can we prove that $$\frac{\sin (2016x)}{2016x}<\frac{\sin x}{x}$$ with $x$ close to/near $0$
I don't know where to start from, but I think that we need to examine distinct cases of $x>0$ and $x<0$.
 A: We'll treat the more general problem: compare $\;\dfrac{\sin cx}{cx}$ and $\dfrac{\sin x}x$ for any number $c>0$ in a neighbourhood of $0$.
Since $\dfrac{\sin x}x$ is an even function, it is enough to consider the case $x>0$. We'll  denote $f$ the continuous extension of  $\dfrac{\sin x}x$ to $\mathbf R$. 
One easily checks that $f$ has a local maximum, $1$, at $0$, and is decreasing on $[0, \pi]$, hence its derivative is negative on $(0,\pi]$.
Set $g(x) =\begin{cases}\dfrac{\sin cx}{cx}&\text{if}\enspace x\ne 0,\\1&\text{if}\enspace x=0.\end{cases}$
Actually, we have $g(x)=f(cx) $, so that $g'(x)=cf'(x)$.
We have $g(0)=f(0)$ and, since $f'(x)<0$ on $(0,\pi]$, we have
 $$\begin{cases}g'(x)<f'(x)&\text{if}\enspace c>1,\\
g'(x)>f'(x))&\text{if}\enspace c<1\end{cases}$$
Now, a classical corollary of the Mean Value theorem states that, on $(0,\pi]$,


*

*$g(x)<f(x)$ in the first case,

*$g(x)>f(x)$ in the second case.

A: Using Taylor at $x=0$, $$\;\dfrac{\sin (cx)}{cx}=1-\frac{c^2 x^2}{6}+O\left(x^4\right)$$ Is this sufficient for your needs ?
