I am trying to prove that for any choice of reals $a$ and $b$ such that $a \neq b $ the function $f(x) := \sin(ax) - \sin(bx)$ has $|f(x)| \geq 1$ for some $x>0$
(This is not an exercise question but something I am trying to prove for myself. Plots with Wolfram Alpha seem to indicate it might be true, and so counterexamples are very much welcome)
Here is my attempt. For this I wrote $f$ of 3 variables $a,b,x$. We find that value of $x_0$ for which $f$ attains a maximum and minimum and then show that $|f(x_0)|$ greater than 1.
$g(a,b,x)=f(x)$ and then zeroing the gradient. (pending the 2nd derivative test)
$f_a = x \cos(ax)$
$f_b = -x \cos(bx)$
$f_x = a\cos(ax) - b\cos(bx)$
Since the point of minima/maxima are conjectured $x>0$ , this means $\cos(ax)$ and $\cos(bx)$ are both zero and the third equation is redundant.
If $a$ and $b$ are rational then obviously we can find such an $x$. But when one of them is irrational, I don't know how I would prove this.
Another attempt was using the $\sin(u)-\sin(v)$ product formula, but that led nowhere too.