Show that this inequality doesn't hold Given  $(a,b,c) \in \mathbb R^3_+$ show that atleast one of the real numbers $a(1-b)$, $ b(1-c)$ and $c(1-a)$ is less than or equal to 1\4.
I tried to show it by contradiction i.e Suppose that $$a(1-b) > \frac{1}{4} $$
$$b(1-c) > \frac{1}{4} $$$$c(1-a) > \frac{1}{4} $$
So as to end with a contradiction but it ended up in a mess. Please help.

Note that if $a,b,c$ are all equal to $\frac{1}{2}$, then the given products are all  $\frac{1}{4}$.  So it is not true that at least one of $a(1-b)$, $b(1-c)$, and $c(1-a)$ is $\lt \frac{1}{4}$. But we can prove that at least one is $\le \frac{1}{4}$. 
If one or more of $1-a$, $1-b$, $1-c$ is $\le 0$, the result is obvious. So suppose they are all positive. 
Their product  is $a(1-a)(b)(1-b)(c)(1-c)$ (note that we rearranged). Recall, or prove using AM/GM, or in some other way, that if $0\lt x\lt 1$, then $x(1-x)\le \frac{1}{4}$. Thus $a(1-a)(b)(1-b)(c)(1-c)\le \frac{1}{4^3}$, and therefore at least one of the given  products is $\le \frac{1}{4}$.
