Show that the variance of a binomial distribution with parameters n and p cannot exceed $\displaystyle \frac{n}{4}$
If you dont know Analytical geometry, there are other ways to see it as well:
You can try:
1) Completing the squares.
$\displaystyle np(1-p) = n(p - p^2) = n\left(\frac14 - \left( p - \frac12\right)^2\right)$. But $\displaystyle \left( p - \frac12\right)^2 \geq 0$, so what can you conclude?
2) AM-GM inequality,
You can apply it(since p and q are non-negative), to get $\displaystyle \frac{p+q}2 \geq \sqrt{pq} \Rightarrow \sqrt{pq} \leq \frac12 \Rightarrow npq \leq \frac{n}4$