TITLE: Is there a probabilistic proof of the inequality $4p(1-p) \leq 1$ for a probability $p$?
QUESTION [12 upvotes]: Let $p\in(0,1)$. The inequality $4p(1-p)\leq 1$ is very easy and elementary,
but I wonder if there is a probabilistic proof of it. By that, I mean constructing
a “natural” probability space and an event in it with probability $4p(1-p)$.
This is easy to do if $4p(1-p)$ is replaced by $3p(1-p)$ : consider
three i.i.d. variables $(X_1,X_2,X_3)$ with Bernoulli distribution
${\cal B}(p)$, and consider the event “The $X_i$ are not all equal”.
Update : One thing that makes this problem hard is that there is no “discrete,finite” solution, involving only a set of $n$ i.i.d. variables $X_1,X_2,X_3,\ldots,X_n$ with Bernoulli distribution ${\cal B}(p)$ (this is because the inequality becomes an equality exactly when $p=\frac{1}{2}$).

REPLY [3 votes]: Alternative solution:
If $p \leq \frac{1}{2}$, then let be a binomial random variable with 2 trials and probability of success 2p. Then $P(X > 0) = 1 - (1-2p)^2 = 1 - (1 - 4p + 4p^2) = 4p - 4p^2 = 4p(1-p)$.
If $p \geq \frac{1}{2}$, then let $q = 1-p$ and let X be a binomial random variable with 2 trials and probability of success 2q. Then $P(X > 0) = 1 - (1-2q)^2 = 4q(1-q) = 4(1-p)p$.