I think I would approach it this way: minimize the function
on the interval $[0,\pi/2]$ using the standard calculus techniques. There's one critical point near $0.88$, and then you have the endpoints. The left endpoint, I think, will end up being the smallest point on the graph. The second derivative might come in handy.
Using f(x) as defined in post #3, show that \( f''(x) < 0 \) for all x, so f is concave. By computation, show that \( f(0) = 0 \) and \( f(\pi / 2) > 0 \). This is enough to conclude that \( f(x) > 0 \) on \( [0 , \pi/2] \).