Inequality: showing that $p(1-p)\leq \frac{1}{4}$ if $0I was wondering if there was another way to derive the inequality shown in the title of this question apart from calculus.
Doing it with calculus is pretty straightforward: 
Let $f:(0,1)\to\mathbb R$ be a function defined by $f(p)=p(1-p)$.
Taking the first derivative $f'(p) = 1-2p=0\Rightarrow p=1/2$ is a critical point. Then the maximum is $1/4$ and it is reached when $p=1/2$. Hence $p(1-p)\leq 1/4,$ $\forall p\in(0,1)$.

Another high school solution is given by solving the quadratic equation $4p^2-4p-1=4(p-1)p-1=0$. The zeroes of this polynomial represent the points where the original equation switches from true to false or vice versa, and then simply examining the interval that $(0,1)$ falls into shows that it's true.
