<s>[INST] 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.
 [/INST] Suppose that for all $a,b,c$ $$a(1-b) > \frac{1}{4}\\b(1-c) > \frac{1}{4}\\c(1-a) > \frac{1}{4}$$ We then have $$a+b+c\gt \frac 34+(ab+ac+bc)=\frac 34+\frac{(a+b+c)^2-(a^2+b^2+c^2)}{2}$$ It follows that if $a=b=c$ we have $$12a\gt 15a^2+3\Rightarrow(15a-6)^2+45\lt 0$$ This is absurde.
</s>