I suppose I might have just let this be, but I thought there might be a reasonably nice non-inductive proof for this proposition. I'd been thinking about this occasionally since this was posted and was finally able now to complete the step I was missing.
The function sin(nx) has a period 2(pi)/n and certainly equals
(n sin x) at x = 0 and x = pi. On the interval (0, pi), (n sin x) will complete a single non-negative half-cycle with its maximum y = n at x = (pi)/2 , while |sin(nx)| will have a set of n non-negative "bumps" with unit amplitude. The first local maximum for |sin(nx)| occurs at
x = (pi)/2n , while (n sin x) will have reached the value
[n sin({pi}/2n)] > 1 . (In fact, in the limit as n approaches infinity, the ratio of the values of these two functions at that point approaches (pi)/2 .) So we can conclude that (n sin x) > |sin(nx)| for the interval ( {pi}/2n , {(2n-1)·pi}/2n ).
It remains to show that this same inequality holds on (0, (pi)/2n) ; a similar argument will hold on ( {(2n-1)·pi}/2n , pi ) by symmetry. Consider the function f(x) = (n sin x) - |sin(nx)| = (n sin x) - sin(nx) on this interval. We already know f(0) = 0 and f( {pi}/2n ) > 0 , so we want to assure that there are no other zeroes for f(x) in the interval. We find
f'(x) = (n cos x) - (n cos(nx) ) = n·( cos x - cos (nx) ) .
Since we have integral n > 1 , we can just look at cos x - cos (nx) . But, by the sum-to-product trigonometric relations,
cos x - cos (nx) =
-2 · sin[ {(n+1)/2}·x ] · sin[ {(1-n)/2}·x ] =
2 · sin[ {(n+1)/2}·x ] · sin[ {(n-1)/2}·x ] > 0 .
So f(x) is only increasing on the interval ( 0 , (pi)/2n ), thus there are no zeroes there. Therefore, the proposition for the inequality holds everywhere on (0, pi).
