Balti Tee 2018 · Ülesanne 17

We can assume that $p$ and $q$ are coprime, and since both sides of first inequality are positive, we can change it to $11q^2 > p^2$. The same way we can change second inequality:

$$11p^2q^2 > p^4 + p^2 + \frac{1}{4}.$$

To see this one holds, we will prove stronger one:

$$11p^2q^2 \geq p^4 + 2p^2.$$

Indeed, dividing this inequality by $p^2$ we get $11q^2 \geq p^2+2$, and since we already know that $11q^2 > p^2$ we only have to see, that $11q^2$ can't be equal to $p^2+1$. Since we know that the only reminders of squares (mod 11) are 0, 1, 3, 4, 5 and 9, $p^2+1$ can't be divisible by 11, and therefore $11q^2 \neq p^2+1$.