Lower bound for $2\sin^2(y\pi)$ I was trying to understand the proof of a theorem, and the author uses the fact that if $y \in \mathbb{Q} \cap(0, \frac{1}{2}]$, then 
$$2\sin^2(y\pi) \geq \frac{8}{n^2},$$
where $y=\frac{p}{q}$, with $\gcd(p,q)=1$ and $n \in \mathbb{N}, n \geq q$.
I can "see" it's true, because the closer $y$ is to $\frac{1}{2}$ the closer to $1$, $\sin^2(y\pi)$ gets, and the closer to $0$ $y$ is, the closer to $\pi\frac{p}{q}$ $\sin^2(y\pi)$ is, however, if $\pi\frac{p}{q}$ is close to $0$, then $q$ must be big, and then $\frac{4}{n^2}$ will be very small, smaller then $\sin^2(y\pi)$ because of the quadratic term...
But I can't prove this formally ... Can someone give a little help? :)
Thanks!
 A: First, because $\sin \pi y$ is concave on $(0,1/2]$, we have
$$\sin \pi y\geq 2y,\quad\text{for }y\in(0,1/2].$$
This picture illustrates this step:

Then, because both functions are non-negative on $(0,1/2]$, squaring both sides yields to
$$\sin^2\pi y\geq 4y^2=\frac{4p^2}{q^2}\geq\frac{4}{q^2}\geq\frac{4}{n^2}.$$
Multiply both sides by $2$ and you are done.
A: Since the left-hand side is an increasing function of $y$ on the set $(0, \frac{1}{2}]$, it suffices to prove the claim for $p = 1$, and since the right-hand side decreases with increasing $n$, it suffices to prove the claim for $n = q$. In fact, we might as well prove the corresponding claim for real numbers:
$\sin^2 \left(\dfrac{\pi}{x}\right) \geq \dfrac{4}{x^2}$ for $x$ such that $\frac{1}{x} \in (0, \frac{\pi}{2}]$
or, equivalently, substituting $x = \frac{1}{t}$, that
$\sin^2 (\pi t) \geq 4 t^2$ for $t \in (0, \frac{1}{2}]$,
or that
$\sin(\pi t) \geq 2t$ on that interval.
Now, $\sin(\pi (0)) = 0 = 2(0)$ and $\sin\left(\pi \left(\frac{1}{2} \right) \right) = 1 = 2\left(\frac{1}{2}\right)$ so the two sides of the inequality agreed at the endpoints of $[0, \frac{1}{2}]$. Their difference $\sin(\pi t) - 2t$ is concave down on the interval, so the inequality holds there.
