Proving that one of $a(1-b), b(1-c), c(1-a) \le \frac{1}{4}$ how can a prove that at least one of those is less than or equal to 1/4.
$$\forall a,b,c\in \mathbb R^+, \ a(1-b)\leq 1/4 \lor b(1-c) \leq 1/4 \lor c(1-a) \leq 1/4.$$
help please!

We can assume $1-a, 1-b, 1-c \geq 0$, since otherwise we are done.
By the AM-GM inequality (see http://en.wikipedia.org/wiki/AM-GM_inequality), we have $abc(1-a)(1-b)(1-c) \leq (\frac{a+b+c+(1-a)+(1-b)+(1-c)}{6})^6= (\frac{1}{2})^6 = \frac{1}{64}$. 
Then, if $a(1-b)> 1/4, b(1-c) > 1/4$ and $ c(1-a) > 1/4$, multiplying together we get $abc(1-a)(1-b)(1-c)> (\frac{1}{4})^3 = \frac{1}{64}$, which is a contradiction, and thus the result follows.
EDIT:
If you do not want to use AM-GM: Let $x \in \mathbb{R}^+ $. We have $0\leq (\sqrt x - \sqrt{(1-x)})^2 = x +(1-x) -2\sqrt{x(1-x)}$, and thus $ 2\sqrt{x(1-x)} \leq 1$, which implies $x(1-x) \leq \frac{1}{4}$. Apply this for $a,b$ and $c$, multiply together, and you get the inequality of the first paragraph. 
