Proving $f(x) := \sin(ax) - \sin(bx)$ has absolute value greater or equal to 1 for some $x>0$ 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.
Any pointers?

Hint : 
For $x_k=\pi/2a + 2*k\pi/a $, $\sin(ax_k) =1$ You "just" have to prove that for a certain $x_k$ $\sin(bx_k)$ is inferior than 0.
By this method, you could also prove a greater result : There exists $x$ where $f(x)>A$ for every $A<2$
