How we can we prove that for any $b > a, x > 0$, $\frac{2}{\pi} (1-\frac{a}{b})<\sup|\frac{\sin(ax)}{ax} - \frac{\sin(bx)}{bx}|<4(1-\frac{a}{b})$ I want to prove this for any $x > 0$ and $b > a > 0$:
$$
\frac{2}{\pi} (1-\frac{a}{b})<\sup|\frac{\sin(ax)}{ax} - \frac{\sin(bx)}{bx}|<4(1-\frac{a}{b})
$$
I tried the derivation to find the maximum value, then I tried to insulate the $\frac{\sin(ax)}{ax}$ and I got this:
$$\sup\left|\frac{\sin(ax)}{ax} \right|\cdot \left|1 - \frac{a}{b}\cdot \frac{\sin(bx)}{\sin(ax)}\right| $$
I think this may be the key to prove that
Notice that the relation is true, I used Geogebra to plot the function with many configuration of a and b, and I get this relation true
any help, please.
 A: We can prove that
$$
1 - \frac ab \le \sup_{x > 0}\left|\frac{\sin(ax)}{ax} - \frac{\sin(bx)}{bx}\right|
\le 2\left(1 - \frac ab \right)
$$
for $0 < a < b$, which is a better estimate.
Proof of the lower bound: For $0 < a < b$ and $x = \pi/b$ is
$$
\left|\frac{\sin(ax)}{ax} - \frac{\sin(bx)}{bx}\right|
= \frac{\sin(\pi a/b)}{\pi a/b} \ge 1 - \frac ab \, .
$$
For the last inequality, see for example sinus estimation by quadratic polynomial or How prove this inequality $\pi<\frac{\sin{(\pi x)}}{x(1-x)}\le 4$ .
Proof of the upper bound: Let $0 < u < v$. We have
$$
\left|\frac{\sin(u)}{u} - \frac{\sin(v)}{v}\right| 
\le \left|\frac{\sin(u)}{u} - \frac{\sin(u)}{v} \right|+\left|\frac{\sin(u)}{v}- \frac{\sin(v)}{v}\right| \\
= |\sin(u)| \frac{v-u}{uv} + \frac{|\sin(u)-\sin(v)|}{v} \, .
$$
Using $|\sin(u)| \le |u|$ and $|\sin(u)-\sin(v) |\le |u-v|$ it follows that
$$
\left|\frac{\sin(u)}{u} - \frac{\sin(v)}{v}\right| \le 2\frac{v-u}{v}
= 2 \left(1 - \frac uv \right) \, .
$$
For $0< a < b$ and $x > 0$ we can now set
$u=ax$ and $v=bx$ and conclude that
$$
\left|\frac{\sin(ax)}{ax} - \frac{\sin(bx)}{bx}\right| \le 2 \left(1 - \frac ab \right) \, .
$$
