Rephrase:
Given p > 0, h > 0
Square roots discarded for simplicity.
\(\displaystyle p^{2}+[sin(x)-sin(x+p)]^{2} > (p+h)^{2}+[sin(x)-sin(x+p+h)]^{2} = f(x,p,h)\)
Rephrase:
\(\displaystyle \frac{\partial{f(x,p,h)}}{\partial{h}} = 2\cdot h + 2\cdot p + 2\cdot\cos(x+p+h)\cdot (\sin(x+p+h)-sin(x)) < 0\) if this decreases.
The '2' doesn't help.
\(\displaystyle h + p + \cos(x+p+h)\cdot (\sin(x+p+h)-sin(x)) < 0\) ??
Worst Case Scenario: \(\displaystyle \sin(x+p+h)-sin(x) = -2\)
Worst Case Scenario: \(\displaystyle \cos(x+p+h) = 1\)
Thus, if \(\displaystyle h + p > 2\), we have \(\displaystyle h + p + \cos(x+p+h)\cdot (\sin(x+p+h)-sin(x)) > 0\), quite contrary to our premise!
With a little more thought, one probably could tighten up that bound and reduce h + p a little more. In any case, this leads us to wonder, what is it about sin(x) that would be any different when comparing an horizontal span greater than 2 with an horizontal span less than 2?
Ah, there's a little bit better. \(\displaystyle \cos(x+p+h)*(\sin(x+p+h)-sin(x)) = \frac{1}{2}\sin(2\cdot(x+p+h))-\cos(x+p+h)\cdot\sin(x) \le \frac{1}{2} - (-1) = \frac{3}{2}\)
Now we have, if \(\displaystyle p + h > 3/2\), it's increasing, not decreasing.
Where does that leave us?
