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!
 A: 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. 
A: If you don't want to use AM-GM you can do it this way.
Let's asume without loss of generalization that $a\leq b\leq c$.
Trivially, If $1 \leq c$ then $b(1-c) \leq 0 \leq 1/4$.
If $a\leq 1/2 \leq b$ or $b\leq 1/2 \leq c$ then $a(1-b)\leq 1/4$ or $b(1-c)\leq 1/4$. (respectively)
If $1/2 \leq a$ then let $a' = a-1/2$ and $b' = 1/2 - (1-b) = b - 1/2$ Note that $a' \leq b'$
Then $a(1-b) = (1/2 + a')(1/2 - b') = 1/4 - (b'-a')/2 -a'b' \leq 1/4$.
Similarly, if $c \leq 1/2$. Let $a' = 1/2 - a$ and $b' = (1-b) - 1/2 = 1/2 - b$. Note that this time, $a' \geq b'$.
Then $a(1-b) = (1/2 - a')(1/2 + b') = 1/4 - (a'-b')/2 -a'b' \leq 1/4$.
A: First assume $a, b, c \lt 1$.
without loss of generality, assume $a \le b$
Now consider the quadratic
$$f(x) = x^2 - x + a(1-b)$$
$f(0) = a(1-b) \gt 0$
$f(b) = b^2 - b + a(1-b) = (a-b)(1-b) \le 0$
Thus the quadratic has a real root (ok, some calculus used, but elementary proofs exist), and thus the discriminant =$ 1 -4a(1-b) \ge 0 \implies a(1-b) \le \frac{1}{4}$
