How can one compare if a parametric curve lying in the Cartesian plane lies above or below a curve given by $y = f(x)$ on some interval?
Specifically, how can one show that the cycloid given parametrically by $$x(t) = \frac{1}{2}(t - \sin t) \quad \text{and} \quad y(t) = \frac{1}{2}(1 - \cos t),$$ for $t \in [0,2\pi]$, lies on or above the sine curve given by $y = \sin x$ on the interval $x \in [0,\pi]$?
Graphically it is obvious. Ideally I would like to set up some sort of inequality, but I really do not know how to proceed.
Let us restrict our study to $t \in I:=[0, \pi/2)$ which is enough due to the perodicity and (in)parities of both functions.
We have to show that $\forall t \in I$,
$$\underbrace{\frac12(1-\cos t)}_{y_C} \ge \sin(\underbrace{\frac12(t-\sin t)}_{x_C})$$
Let us prove that, on $I$,
$$f(t):=\frac12(1-\cos t)-\sin(\frac12(t-\sin t))$$ is a positive function.
$f$ is positive on $I$ under the following sufficient conditions:
its value in $0$ is $\ge 0$, which is true ($f(0)=0$), and
its derivative is positive on $I$, a fact that we are going to establish ; indeed:
$$f'(x)=\frac12\sin t-\frac12 (1- \cos t) \cos(\frac12(t-\sin t))$$
can be written, using classical trigonometric relationships:
$$f'(x)=\sin(t/2)\cos(t/2)-\sin(t/2)^2 \cos(\frac12(t-\sin t))$$
$$f'(x)=\sin(t/2)\cos(t/2)\left[1-\tan(t/2) \underbrace{\cos(\frac12(t-\sin t))}_{\le 1}\right]$$
which is positive for any $t \in I$ because, in this interval, $\tan (t/2) <1$